MTH401 — Midterm Summary (Lectures 1–22)

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📘 Lecture 1 — DIFFERENTIAL EQUATIONS

📖 Overview: This lecture introduces the fundamental concepts of differential equations, which are essential for modeling and solving problems in science and engineering. It establishes the key definitions for classifying equations by type, order, and form of solution, providing the foundational vocabulary for the entire course.

🗂️ Topics Covered

The lecture begins with an introduction to the theory of differential equations, describing what they are and how they arise. It then covers the elements of the theory, providing specific examples and defining the order of an equation. The distinction between ordinary and partial differential equations is explained, along with basic properties and the principle of superposition. The lecture concludes by defining explicit and implicit solutions.

📝 Lecture Summary

1 Introduction

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. Such relations are common in engineering, physics, and other disciplines.

💡 Why this matters: Understanding differential equations is crucial for predicting how systems change over time or space.

2 Fundamentals

2.1 Elements of the Theory

The theory involves studying the equation's structure and finding functions that satisfy it. A key goal is to determine the solution of the equation, which is a function that, when substituted into the equation, makes it true for all values in its domain.

2.2 Specific Examples of ODE’s

Several classical examples of Ordinary Differential Equations (ODEs) are introduced:

1. Newton’s Second Law (mechanics): ( m \frac{d^2x}{dt^2} = F(t, x, \frac{dx}{dt}) )
2. Radioactive Decay (nuclear physics): ( \frac{dN}{dt} = -\lambda N ), where N is the number of atoms and (\lambda) is the decay constant.
3. Newton’s Law of Cooling (thermodynamics): ( \frac{dT}{dt} = k(T - T_m) ), where T is the temperature of the body and (T_m) is the ambient temperature.
4. Simple Harmonic Motion (physics): ( \frac{d^2y}{dt^2} = -ky )
5. Fluid Flow (fluid dynamics): ( \frac{dV}{dt} = kV^{2/3} ), where V is the volume of fluid.
6. Electric Circuits (RLC) (electrical engineering): ( L \frac{d^2i}{dt^2} + R \frac{di}{dt} + \frac{1}{C} i = E'(t) )

2.3 The order of an equation

The order of a differential equation is the order of the highest derivative that appears in the equation.

• Example 1: ( y' = y + 5 ) is a first-order equation.
• Example 2: ( y'' + 2y' + 5y = 0 ) is a second-order equation.

📌 Example: For the equation ( \frac{d^3y}{dx^3} + 2x\frac{dy}{dx} = \cos x ), the highest derivative is ( \frac{d^3y}{dx^3} ), so the order is 3.

2.4 Ordinary Differential Equation

An Ordinary Differential Equation (ODE) is a differential equation that contains one or more derivatives of a dependent variable with respect to a single independent variable.

🔑 Definition — ODE: An equation containing only ordinary derivatives of one or more unknown functions with respect to a single independent variable.

2.5 Partial Differential Equation

A Partial Differential Equation (PDE) contains partial derivatives of a dependent variable with respect to more than one independent variable.

🔑 Definition — PDE: An equation containing partial derivatives of one or more unknown functions with respect to two or more independent variables.

📌 Example: The heat equation ( \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} ) is a PDE because the dependent variable u depends on both time (t) and space (x).

2.6 Results from ODE data

An ODE can be thought of as a condition that a function must satisfy. Solving it means finding all functions that satisfy that condition. While a solution may exist, finding a formula for it is not always possible.

2.7 BVP Examples

A Boundary Value Problem (BVP) is a differential equation together with a set of additional constraints, called boundary conditions, on the solution at more than one point.

🔑 Definition — BVP: A problem where a differential equation must be solved for a function that must satisfy conditions (boundary conditions) at two or more distinct points.

📌 Example: Find a function y(x) such that ( y'' + y = 0 ) and ( y(0) = 0, y(\pi) = 0 ).

2.8 Properties of ODE’s

ODEs can be classified as linear or nonlinear. A linear ODE is one in which the dependent variable and all its derivatives appear to the first power only and are not multiplied together.

2.9 Superposition

The superposition principle applies to linear, homogeneous differential equations. It states that if ( y_1 ) and ( y_2 ) are two solutions, then any linear combination ( c_1 y_1 + c_2 y_2 ) is also a solution.

2.10 Explicit Solution

An explicit solution is a solution in which the dependent variable is expressed solely in terms of the independent variable and constants. The function is written in the form ( y = f(x) ).

🔑 Definition — Explicit Solution: A function where the dependent variable is isolated on one side of the equation.

📌 Example: ( y = e^{2x} ) is an explicit solution to the equation ( y' = 2y ).

2.11 Implicit Solution

An implicit solution is a relation ( G(x, y) = 0 ) that defines one or more explicit solutions. It is a solution to the differential equation even though the dependent variable is not isolated.

🔑 Definition — Implicit Solution: A relationship of the form ( G(x, y) = 0 ) that can be reduced to one or more explicit solutions.

📌 Example: The relation ( x^2 + y^2 = 1 ) is an implicit solution to the equation ( y' = -\frac{x}{y} ).

⭐ Key Takeaways

The most critical concept from this lecture is the precise definition of a differential equation and its fundamental classification by order and type (ODE vs. PDE). Students must be able to identify the order of any given equation and distinguish between ordinary and partial differential equations. Understanding the difference between linear and nonlinear equations, as well as the principle of superposition for linear homogeneous equations, is essential for future solution methods. Finally, one must be able to differentiate between an explicit solution, where the dependent variable is isolated, and an implicit solution, which is a relation between variables.

🧠 Quick Revision Questions

1. What is the order of a differential equation, and how do you find it?
2. What is the key difference between an Ordinary Differential Equation (ODE) and a Partial Differential Equation (PDE)?
3. Give an example of an explicit solution and write the general form of an implicit solution.
4. What is a Boundary Value Problem (BVP)?
5. Does the principle of superposition apply to all differential equations? Explain.

📘 Lecture 2 — Applications of First Order Differential Equations

📖 Overview: This lecture explores how first-order differential equations model real-world phenomena, focusing on orthogonal trajectories in geometry, population growth, radioactive decay, Newton's law of cooling, carbon dating, and nonlinear equations like the logistic model. It bridges theoretical methods with practical applications in physics, biology, and chemistry.

🗂️ Topics Covered

Orthogonal trajectories and their geometric interpretation, finding orthogonal trajectories methodically, population dynamics using exponential and logistic models, radioactive decay and half-life calculations, Newton's law of cooling, carbon dating for archaeological dating, the logistic equation and its special cases, chemical reaction kinetics, and miscellaneous applications.

📝 Lecture Summary

10 Orthogonal Trajectories (OT)

An orthogonal trajectory is a curve that intersects every member of a given family of curves at right angles (90°). This concept is used in fields like electromagnetism, fluid dynamics, and heat transfer where orthogonal paths represent flow lines or equipotential lines.

🔑 Definition — Orthogonal Trajectory (OT): A curve that intersects every curve of a given family at right angles.

10.1 Orthogonal Trajectories

The method uses the fact that if two curves are orthogonal at an intersection point, the product of their slopes is -1. Given a family of curves ( F(x, y, c) = 0 ), we find its differential equation ( \frac{dy}{dx} = f(x, y) ). The orthogonal trajectories satisfy ( \frac{dy}{dx} = -\frac{1}{f(x, y)} ).

📐 Formula: ( m_1 \times m_2 = -1 ) → ( \frac{dy}{dx} ) (trajectory) ( = -\frac{1}{\frac{dy}{dx} \text{ (original)}} )

📌 Example: Find orthogonal trajectories of ( y = cx^2 ).

1. Differentiate: ( \frac{dy}{dx} = 2cx )
2. Eliminate ( c ) from original: ( c = \frac{y}{x^2} ), so ( \frac{dy}{dx} = 2\left(\frac{y}{x^2}\right)x = \frac{2y}{x} )
3. For OT: ( \frac{dy}{dx} = -\frac{x}{2y} )
4. Solve: ( 2y , dy = -x , dx ) → ( y^2 = -\frac{x^2}{2} + C ) or ( x^2 + 2y^2 = K ). Answer: family of ellipses.

10.2 Method of finding Orthogonal Trajectory

The systematic procedure: (1) Write family of curves ( F(x, y, c) = 0 ). (2) Differentiate to find ( \frac{dy}{dx} ). (3) Eliminate parameter ( c ). (4) Replace ( \frac{dy}{dx} ) with ( -dx/dy ) (i.e., negative reciprocal). (5) Solve the new differential equation.

💡 Why this matters: This method connects differential equations to geometry and physical field theory.

10.3 Population Dynamics

Population growth often follows a first-order differential equation where the growth rate is proportional to the current population: ( \frac{dP}{dt} = kP ), where ( k ) is the growth constant. This gives exponential growth ( P(t) = P_0 e^{kt} ).

10.4 Radioactive Decay

Radioactive substances decay at a rate proportional to the amount present: ( \frac{dA}{dt} = -kA ). Solution: ( A(t) = A_0 e^{-kt} ). The half-life is the time for half the substance to decay.

📐 Formula: Half-life ( t_{1/2} = \frac{\ln 2}{k} )
📌 Example: If a sample decays from 100g to 25g in 10 years, find half-life.
( 25 = 100 e^{-k(10)} ) → ( \frac{1}{4} = e^{-10k} ) → ( \ln(0.25) = -10k ) → ( k = \frac{\ln 4}{10} ).
Half-life ( = \frac{\ln 2}{k} = \frac{\ln 2}{\ln 4 / 10} = 10 \times \frac{\ln 2}{\ln 4} = 10 \times \frac{\ln 2}{2\ln 2} = 5 ) years.

11.1 Newton's Law of Cooling

The rate of change of an object's temperature ( T ) is proportional to the temperature difference between the object and its environment (ambient temperature ( T_m )): ( \frac{dT}{dt} = -k(T - T_m) ).

📐 Formula: ( T(t) = T_m + (T_0 - T_m) e^{-kt} )
📌 Example: A body at 100°C is placed in room at 20°C. After 10 min, temperature is 60°C. Find after 20 min.
( 60 = 20 + (100 - 20) e^{-10k} ) → ( 40 = 80 e^{-10k} ) → ( e^{-10k} = 0.5 ) → ( -10k = \ln 0.5 ) → ( k = 0.06931 ).
After 20 min: ( T = 20 + 80 e^{-20(0.06931)} = 20 + 80(0.25) = 40^\circ C ).

11.2 Carbon Dating

The radioactive isotope carbon-14 decays with a known half-life of about 5730 years. By measuring the remaining ( ^{14}C ) in an organic sample, we can determine its age using ( A(t) = A_0 e^{-kt} ) with ( k = \frac{\ln 2}{5730} ).

12 Applications of Non-linear Equations

Nonlinear differential equations model more complex phenomena, especially when growth or reaction rates depend on additional factors.

12.1 Logistic equation

The logistic equation models population growth with limited resources: ( \frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right) ), where ( K ) is the carrying capacity (maximum sustainable population).

12.1.1 Solution of the Logistic equation

Using separation of variables:
( \frac{dP}{P(1 - P/K)} = k , dt )
Partial fractions: ( \frac{1}{P} + \frac{1/K}{1 - P/K} = \frac{1}{P} + \frac{1}{K - P} )
Integrate: ( \ln |P| - \ln |K - P| = kt + C ) → ( \ln\left|\frac{P}{K-P}\right| = kt + C )
Solution: ( P(t) = \frac{K}{1 + Be^{-kt}} ), where ( B = \frac{K - P_0}{P_0} ).

📐 Formula: ( P(t) = \frac{K}{1 + Be^{-kt}} ), ( B = \frac{K - P_0}{P_0} )

12.1.2 Special Cases of Logistic Equation

• If ( P_0 = K ), population remains constant at carrying capacity.
• If ( P_0 < K ), population grows asymptotically toward ( K ).
• If ( P_0 > K ), population decreases toward ( K ).

12.1.3 A Modification of LE

Modified logistic equations may include a threshold population or time-dependent carrying capacity, such as ( \frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right) - h ), where ( h ) is harvesting rate.

12.2 Chemical reactions

In first-order chemical reactions, the rate of reaction is proportional to the concentration of the reactant: ( \frac{d[A]}{dt} = -k[A] ). For second-order reactions: ( \frac{d[A]}{dt} = -k[A]^2 ).

📌 Example: For a first-order reaction with initial concentration ([A]_0 = 2.0,M) and rate constant (k = 0.1,s^{-1}), find concentration after 10 seconds.
( [A] = 2.0 e^{-0.1 \times 10} = 2.0 e^{-1} = 0.736,M ).

12.3 Miscellaneous Applications

Other applications include mixing problems (salt in tanks), electric circuits (RC circuits), and interest compounding in finance, all modeled by first-order linear or separable differential equations.

⭐ Key Takeaways

The most critical concepts include: (1) orthogonal trajectories are found by replacing ( dy/dx ) with its negative reciprocal and solving; (2) exponential growth/decay models come from ( dy/dx = ky ); (3) logistic equation introduces carrying capacity for realistic growth; (4) Newton's law of cooling and radioactive decay follow similar exponential forms with different physical interpretations; (5) chemical reaction rates depend on reaction order. Always identify the physical context to select the correct model and initial conditions.

🧠 Quick Revision Questions

1. What is the geometric condition for two curves to be orthogonal at an intersection point?
2. Write the differential equation for radioactive decay and its general solution.
3. What are the three cases for the logistic equation based on initial population relative to carrying capacity?
4. In Newton's law of cooling, what does the constant ( k ) physically represent?
5. How does the half-life of carbon-14 relate to its decay constant?

📘 Lecture 3 — Fundamentals (continued) & Introduction to Differential Equations

📖 Overview: This lecture covers the fundamental theory of differential equations, including classification, solutions, and initial value problems. It then introduces key concepts of vibration analysis including simple harmonic motion, damped motion, and forced motion, followed by advanced topics in power series solutions and systems of differential equations. This material is essential for understanding how differential equations model real-world phenomena in physics and engineering.

🗂️ Topics Covered

The lecture spans from basic definitions and classification of differential equations through specific examples including growth equations and pendulum equations, then progresses through vibration analysis (undamped, damped, forced), Cauchy-Euler equations, power series solutions, Bessel's and Legendre's equations, systems of linear differential equations, matrix theory, eigenvalue problems, and homogeneous/non-homogeneous systems.

📝 Lecture Summary

2.1 Elements of the Theory

A differential equation involves an unknown function and one or more of its derivatives. These equations are applicable across chemistry, physics, engineering, medicine, biology, and anthropology. An ordinary differential equation (ODE) is one where the unknown function depends on only one independent variable.

🔑 Definition — Differential Equation: An equation that involves an unknown function and one or more of its derivatives. 🔑 Definition — Ordinary Differential Equation (ODE): A differential equation where the unknown function depends on only one independent variable.

Examples of differential equations include:

• dy/dx - 5y = 1
• (y - x)dx + 4xdy = 0
• d²y/dx² + 5(dy/dx)³ - 4y = eˣ
• Partial differential equations like ∂u/∂y + ∂v/∂x = 0

2.2 Specific Examples of ODE's

The growth equation is du/dt = F(t)·G(u), which models population growth or decay. The pendulum equation is d²θ/dt² + (g/l)sinθ = F(t), describing the motion of a pendulum. The van der Pol equation is d²y/dt² + ε(y²-1)dy/dt + y = 0, which models oscillatory systems with nonlinear damping.💡 Why this matters: These specific equations appear frequently in physics and engineering applications.

22.1.4 Differential Equation

For simple harmonic motion, the differential equation governing an undamped vibrating system is derived from Newton's second law and Hooke's law. The restoring force is proportional to displacement: F = -kx. The resulting differential equation is d²x/dt² + (k/m)x = 0, where k is the spring constant and m is the mass.

📐 Formula: d²x/dt² + ω²x = 0 → where ω = √(k/m) is the angular frequency

22.1.5 Initial Conditions

To obtain a unique solution, initial conditions are required. Typically, these specify the initial displacement x(0) = x₀ and initial velocity dx/dt(0) = v₀. These conditions determine the arbitrary constants in the general solution.

🔑 Definition — Initial Conditions: The specified values of the dependent variable and its derivatives at a single point (usually t=0) that determine a unique solution to a differential equation.

22.1.6 Solution and Equation of Motion

The general solution to d²x/dt² + ω²x = 0 is x(t) = c₁cos(ωt) + c₂sin(ωt), where c₁ and c₂ are constants determined by initial conditions. Applying initial conditions x(0) = x₀ and x'(0) = v₀ gives c₁ = x₀ and c₂ = v₀/ω.

📌 Example: For a mass-spring system with k = 16 N/m, m = 4 kg, initial displacement x₀ = 0.5 m, and initial velocity v₀ = 2 m/s:

• Step 1: Find ω = √(k/m) = √(16/4) = 2 rad/s
• Step 2: Solution is x(t) = c₁cos(2t) + c₂sin(2t)
• Step 3: x(0) = 0.5 = c₁
• Step 4: x'(t) = -2c₁sin(2t) + 2c₂cos(2t), so x'(0) = 2 = 2c₂, giving c₂ = 1
• Step 5: Equation of motion: x(t) = 0.5cos(2t) + sin(2t)

22.1.7 Alternative form of Solution

The solution can be written in amplitude-phase form: x(t) = A sin(ωt + φ) or x(t) = A cos(ωt - φ), where A is the amplitude and φ is the phase angle.

🔑 Definition — Amplitude-Phase Form: An alternative representation of the solution where A = √(c₁² + c₂²) and φ = arctan(c₁/c₂) (or similar depending on the form used).

22.1.8 Amplitude

The amplitude A represents the maximum displacement from equilibrium. It is given by A = √(c₁² + c₂²). The amplitude depends on both initial conditions.

📐 Formula: A = √(x₀² + (v₀/ω)²) → maximum displacement from equilibrium

22.1.9 A Vibration or a Cycle

One complete vibration or cycle occurs when the system returns to its original state (same position and velocity direction). For simple harmonic motion, this corresponds to the argument ωt increasing by 2π.

🔑 Definition — Cycle: One complete repetition of the motion, from a given point back to the same point moving in the same direction.

22.1.10 Period of Vibration

The period T is the time required to complete one full cycle. For the system d²x/dt² + ω²x = 0, the period is T = 2π/ω = 2π√(m/k).

📐 Formula: T = 2π/ω → time for one complete cycle

22.1.11 Frequency

The frequency f is the number of cycles per unit time, given by f = 1/T = ω/(2π). The natural (circular) frequency ω is measured in radians per second.

📐 Formula: f = 1/T = ω/(2π) → number of cycles per second (Hertz)

23 Damped Motion

Damped motion includes a damping force that opposes motion, typically proportional to velocity: F_d = -c(dx/dt), where c is the damping constant.

23.1 Damping Force

The damping force is F_d = -c(dx/dt), representing friction or resistance. The negative sign indicates the force opposes the direction of motion.

23.2 The Differential Equation

Combining spring force -kx and damping force -c(dx/dt) via Newton's Second Law: m(d²x/dt²) = -kx - c(dx/dt), giving m(d²x/dt²) + c(dx/dt) + kx = 0, or d²x/dt² + (c/m)(dx/dt) + (k/m)x = 0.

📐 Formula: d²x/dt² + 2β(dx/dt) + ω²x = 0 → where β = c/(2m) is the damping coefficient and ω = √(k/m)

23.2.1 Solution of the Differential Equation

The characteristic equation is r² + 2βr + ω² = 0, giving roots r = -β ± √(β² - ω²). Three cases arise:

• Overdamped (β² > ω²): Real distinct negative roots, x(t) = c₁e^(r₁t) + c₂e^(r₂t)
• Critically damped (β² = ω²): Real equal roots, x(t) = (c₁ + c₂t)e^(-βt)
• Underdamped (β² < ω²): Complex roots, r = -β ± i√(ω² - β²)

🔑 Definition — Damping Cases:

• Overdamped: System returns to equilibrium slowly without oscillation
• Critically damped: System returns to equilibrium fastest without oscillation
• Underdamped: System oscillates with decreasing amplitude

23.2.2 Alternative form of the Solution

For underdamped motion: x(t) = e^(-βt)[c₁cos(ω₁t) + c₂sin(ω₁t)] or x(t) = Ae^(-βt)sin(ω₁t + φ), where ω₁ = √(ω² - β²) is the damped natural frequency.

📐 Formula: ω₁ = √(ω² - β²) → frequency of damped oscillation

23.2.3 Quasi Period

The quasi period T₁ = 2π/ω₁ is the time between successive maxima of the damped oscillation.

24 Forced Motion

Forced motion occurs when an external force F(t) is applied to the system. The equation becomes m(d²x/dt²) + c(dx/dt) + kx = F(t).

24.1 Forced motion with damping

The non-homogeneous differential equation d²x/dt² + 2β(dx/dt) + ω²x = F(t)/m describes forced damped motion. The solution consists of the complementary function (transient) plus a particular integral (steady-state).

24.2 Transient and Steady-State Terms

The transient term is the solution of the homogeneous equation and decays to zero as t→∞ (due to damping). The steady-state term is a particular solution that persists as t→∞.

🔑 Definition — Transient: The part of the solution that decays to zero over time 🔑 Definition — Steady-State: The part of the solution that persists indefinitely, having the same form as the forcing function

24.3 Motion without Damping

For undamped forced motion (c=0): d²x/dt² + ω²x = F₀cos(γt)/m. If γ = ω (forcing frequency equals natural frequency), resonance occurs, leading to unbounded amplitude growth. Resonance is described by x(t) = (F₀/(2mω))tsin(ωt).

24.4 Electric Circuits

Differential equations also model electric circuits, where charge q(t) and current i(t) = dq/dt are analogous to displacement and velocity in mechanical systems. Voltage across components follows known laws.

24.5 The LRC Series Circuits

An LRC series circuit contains a resistor (R), inductor (L), and capacitor (C) in series. The differential equation relates charge q or current i to the applied voltage E(t).

24.5.1 Resistor

Voltage across a resistor: V_R = iR (Ohm's law)

24.5.2 Inductor

Voltage across an inductor: V_L = L(di/dt)

24.5.3 Capacitor

Voltage across a capacitor: V_C = q/C, where q is charge

24.6 Kirchhoff's Voltage Law

Kirchhoff's Voltage Law states that the sum of voltage drops around a closed loop equals zero: V_L + V_R + V_C = E(t) or L(di/dt) + Ri + q/C = E(t).

24.6.1 The Differential Equation

Substituting i = dq/dt gives L(d²q/dt²) + R(dq/dt) + q/C = E(t), which has the same form as the mechanical system.

24.6.2 Solution of the differential equation

The characteristic equation is Lr² + Rr + 1/C = 0 with roots r = [-R ± √(R² - 4L/C)]/(2L).

Case 1 Real and distinct roots: R² > 4L/C → overdamped, q(t) = c₁e^(r₁t) + c₂e^(r₂t)

Case 2 Real and equal: R² = 4L/C → critically damped, q(t) = (c₁ + c₂t)e^(rt)

Case 3 Complex roots: R² < 4L/C → underdamped, q(t) = e^(-Rt/(2L))[c₁cos(ω₁t) + c₂sin(ω₁t)]

25 Forced Motion (Examples)

Forced motion with sinusoidal input E(t) = E₀sin(γt) leads to steady-state solution of the form q_p(t) = A sin(γt - φ). Resonance occurs when γ ≈ 1/√(LC) (the natural frequency of the circuit).

26 Differential Equations with Variable Coefficients

Variable coefficient differential equations have coefficients that depend on the independent variable, unlike constant coefficient equations.

26.1 Cauchy-Euler Equation

The Cauchy-Euler (or Euler) equation has the form ax²(d²y/dx²) + bx(dy/dx) + cy = f(x). It is characterized by coefficients that are powers of x matching the order of the derivative.

🔑 Definition — Cauchy-Euler Equation: A differential equation of the form ax²y″ + bxy′ + cy = f(x) where the power of x matches the order of the derivative.

26.1.1 Method of Solution

Assume solution of the form y = x^r for the homogeneous equation. Substituting gives the indicial (auxiliary) equation: ar(r-1) + br + c = 0. The roots determine the form of the solution.

26.1.2 Case-I (Distinct Real Roots)

If roots r₁ and r₂ are real and distinct: y_h = c₁x^(r₁) + c₂x^(r₂)

26.1.3 Case II (Repeated Real Roots)

If r₁ = r₂: y_h = (c₁ + c₂ln|x|)x^r

26.1.4 Case III (Conjugate Complex Roots)

If r = α ± iβ: y_h = x^α[c₁cos(βln|x|) + c₂sin(βln|x|)]

📌 Example: Solve x²y″ - 2xy′ + 2y = 0

• Step 1: Assume y = x^r, then y′ = rx^(r-1), y″ = r(r-1)x^(r-2)
• Step 2: Substitute: x²·r(r-1)x^(r-2) - 2x·rx^(r-1) + 2x^r = 0
• Step 3: Simplify: [r(r-1) - 2r + 2]x^r = 0, so r² - 3r + 2 = 0
• Step 4: Factor: (r-1)(r-2) = 0, so r₁ = 1, r₂ = 2
• Step 5: General solution: y = c₁x + c₂x²

26.2 Exercises

Practice problems involve applying the Cauchy-Euler method to various equations with different root types.

27 Cauchy-Euler Equation (Alternative Method of Solution)

An alternative approach uses the substitution x = e^t (or t = ln x), transforming the Cauchy-Euler equation into a constant coefficient equation in terms of t. This converts x(dy/dx) = Dy and x²(d²y/dx²) = D(D-1)y, where D = d/dt.

📐 Formula: Under x = e^t: x(dy/dx) = dy/dt and x²(d²y/dx²) = d²y/dt² - dy/dt

28 Power Series (An Introduction)

A power series is an infinite series of the form ∑aₙ(x-x₀)ⁿ from n=0 to ∞, where aₙ are coefficients and x₀ is the center.

28.1 Power Series

The general form is ∑aₙ(x-x₀)ⁿ = a₀ + a₁(x-x₀) + a₂(x-x₀)² + ... Power series can represent functions and solve differential equations.

28.2 Convergence and Divergence

A power series may converge (approach a finite value) for some x-values and diverge for others. The set of x for which the series converges is the interval of convergence.

28.2.1 The Ratio Test

The ratio test determines convergence: lim|aₙ₊₁/aₙ| as n→∞. If the limit is L, the series converges when |x-x₀|L < 1 and diverges when |x-x₀|L > 1.

28.2.2 Interval of Convergence

The interval of convergence is the set of x-values where the series converges, centered at x₀.

28.2.3 Radius of Convergence

The radius of convergence R is found from R = 1/L = lim|aₙ/aₙ₊₁| (if the limit exists). The series converges absolutely for |x-x₀| < R.

📐 Formula: R = lim|aₙ/aₙ₊₁| as n→∞ → distance from center within which the series converges

28.2.4 Convergence at an Endpoint

At x = x₀ ± R, convergence must be checked separately using standard tests (e.g., alternating series test, p-series test).

28.3 Absolute Convergence

A series converges absolutely if ∑|aₙ||x-x₀|ⁿ converges. Absolute convergence implies ordinary convergence.

28.4 Power Series Representation of Functions

Many functions can be represented by power series, such as eˣ = ∑xⁿ/n!, sin x = ∑(-1)ⁿx^(2n+1)/(2n+1)!, and cos x = ∑(-1)ⁿx^(2n)/(2n)!.

28.4.1 Theorem

If a function has a power series representation, that representation is unique for a given center. The series is the Taylor series of the function.

28.4.2 Series that are Identically Zero

If ∑aₙ(x-x₀)ⁿ = 0 for all x in some interval, then all coefficients aₙ = 0. This is crucial for the power series method.

28.5 Analytic at a Point

A function f(x) is analytic at x₀ if it has a power series representation f(x) = ∑aₙ(x-x₀)ⁿ with a positive radius of convergence. This means f is infinitely differentiable near x₀ and its Taylor series converges to f.

28.6 Arithmetic of Power Series

Power series can be added, subtracted, multiplied, and divided (with care for the radius of convergence). Term-by-term differentiation and integration are also valid within the interval of convergence.

29 Power Series Solution of a Differential Equation

To solve y″ + P(x)y′ + Q(x)y = 0 using power series:

1. Assume y = ∑aₙ(x-x₀)ⁿ
2. Compute y′ = ∑naₙ(x-x₀)^(n-1) and y″ = ∑n(n-1)aₙ(x-x₀)^(n-2)
3. Substitute into the DE
4. Shift indices to combine sums
5. Set coefficients of each power to zero (equating coefficients)
6. Find recurrence relation for aₙ

📌 Example: Solve y′ = y using power series about x₀=0

• Step 1: Assume y = ∑aₙxⁿ, y′ = ∑naₙx^(n-1)
• Step 2: Substitute: ∑naₙx^(n-1) = ∑aₙxⁿ
• Step 3: Shift index: let k = n-1 in left sum, so ∑(k+1)aₖ₊₁xᵏ = ∑aₖxᵏ
• Step 4: Equate coefficients: (k+1)aₖ₊₁ = aₖ for all k≥0
• Step 5: Recurrence: aₖ₊₁ = aₖ/(k+1)
• Step 6: a₁ = a₀, a₂ = a₁/2 = a₀/2, a₃ = a₂/3 = a₀/6, giving aₙ = a₀/n!
• Step 7: Solution: y = a₀∑xⁿ/n! = a₀eˣ

30 Solution about Ordinary Points

A point x₀ is an ordinary point of y″ + P(x)y′ + Q(x)y = 0 if both P(x) and Q(x) are analytic at x₀.

30.1 Analytic Function

A function is analytic at x₀ if it has a convergent power series representation centered at x₀.

30.2 Ordinary and singular points

• Ordinary point: Both P(x) and Q(x) are analytic at x₀
• Singular point: At least one of P(x) or Q(x) is not analytic at x₀

30.2.1 Polynomial Coefficients

For polynomial coefficients P(x) and Q(x): if the denominator of P or Q becomes zero at x₀, that point is singular.

30.3 Theorem (Existence of Power Series Solution)

If x₀ is an ordinary point, there exists a power series solution y = ∑aₙ(x-x₀)ⁿ that converges in some interval |x-x₀| < R. The coefficients are determined by the recurrence relation from the DE.

30.4 Non-polynomial Coefficients

For non-polynomial coefficients (like sin x, eˣ), the same method applies: expand P(x) and Q(x) as power series, then find the recurrence relation by equating coefficients.

31 Solutions about Singular Points

Near singular points, the power series method may fail, requiring the Method of Frobenius.

31.1 Regular and Irregular Singular Points

A singular point x₀ is regular if (x-x₀)P(x) and (x-x₀)²Q(x) are analytic at x₀. Otherwise, it is irregular.

🔑 Definition — Regular Singular Point: A singular point x₀ such that (x-x₀)P(x) and (x-x₀)²Q(x) are analytic at x₀.

31.1.1 Polynomial Coefficients

For polynomial coefficients: factor the denominator; if P(x) has a pole of order at most 1 and Q(x) has a pole of order at most 2, the point is a regular singular point.

31.2 Method of Frobenius

The Method of Frobenius assumes a solution of the form y = x^r∑aₙxⁿ = ∑aₙx^(n+r), where r is determined by the indicial equation.

31.2.1 Frobenius' Theorem

If x₀=0 is a regular singular point, there exists at least one Frobenius series solution y = x^r∑aₙxⁿ valid for 0<|x|<R. The exponent r satisfies the indicial equation obtained from the lowest power of x.

31.3 Cases of Indicial Roots

The indicial equation is a quadratic in r, giving two roots r₁ and r₂. Three cases arise.

31.3.1 Case I (Roots not Differing by an Integer)

If r₁ and r₂ are distinct and their difference is NOT an integer, two linearly independent solutions exist: y₁ = x^(r₁)∑aₙxⁿ and y₂ = x^(r₂)∑bₙxⁿ

32 Solutions about Singular Points

Further solution methods for Frobenius cases.

32.1 Method of Frobenius (Cases II and III)

Additional cases for the Method of Frobenius.

32.1.1 Case II (Roots Differing by a Positive Integer)

If r₁ - r₂ = N (a positive integer), the second solution may contain a logarithmic term: y₂ = y₁ln|x| + x^(r₂)∑bₙxⁿ

33 Bessel's Differential Equation

Bessel's differential equation is x²y″ + xy′ + (x² - ν²)y = 0, where ν is a constant. Its solutions are Bessel functions.

33.1 Series Solution of Bessel's Differential Equation

Using the Method of Frobenius about x₀=0 (a regular singular point), the indicial equation is r² - ν² = 0, giving r = ±ν.

33.2 Bessel's Function of the First Kind

Bessel's function of the first kind of order ν is J_ν(x) = ∑(-1)ⁿ(x/2)^(2n+ν)/[n!Γ(n+ν+1)]. For integer ν=n, J_n(x) = ∑(-1)ⁿ(x/2)^(2n+k)/[k!(n+k)!].

34 Legendre's Differential Equation

Legendre's differential equation is (1-x²)y″ - 2xy′ + n(n+1)y = 0, where n is a constant (usually a non-negative integer).

34.1 Legendre's Polynomials

For integer n, the power series solution terminates, giving Legendre polynomials P_n(x). These are polynomials of degree n, with P₀(x)=1, P₁(x)=x, P₂(x)=(3x²-1)/2, etc.

34.2 Rodrigues Formula for Legendre's

📘 Lecture 4 — Differential Equations (MTH401)

📖 Overview: This lecture introduces the foundational concepts of differential equations, from classification by order and type to methods for solving separable equations. It covers definitions of ordinary and partial differential equations, properties of linearity, and the explicit and implicit forms of solutions, providing essential tools for analyzing and solving these equations.

🗂️ Topics Covered

The lecture covers the classification of differential equations by order, type (ordinary vs. partial), and linearity; the concepts of general and particular solutions, initial value problems (IVPs), and boundary value problems (BVPs); and the superposition principle. It defines explicit and implicit solutions, then concludes with the method for solving separable differential equations.

📝 Lecture Summary

2.3 The order of an equation

The order of a differential equation is defined as the order of the highest derivative appearing in the equation. For instance, d²y/dx² + 5(dy/dx)³ - 4y = eˣ is a second-order equation.

2.4 Ordinary Differential Equation

An Ordinary Differential Equation (ODE) contains only ordinary derivatives of one or more dependent variables with respect to a single independent variable. For example, d²y/dx² + 5(dy/dx)³ - 4y = eˣ is an ODE.

2.5 Partial Differential Equation

A Partial Differential Equation (PDE) involves partial derivatives of one or more dependent variables with respect to two or more independent variables. For example, a² ∂⁴u/∂x⁴ + ∂²u/∂x² = 0 is a PDE. 💡 Why this matters: This distinction is fundamental for selecting the correct solution method.

2.6 Results from ODE data

A solution of a differential equation f(t, y, y', ..., yⁿ) = 0 is defined over some interval I and has the following properties:

• y(t) and its first n derivatives exist for all t in I.
• y(t) and its first n-1 derivatives are continuous in I.
• y(t) satisfies the differential equation for all t in I.

Key terms include:

• General Solution: Represents all solutions to the differential equation for all arbitrary constants.
• Particular Solution: Contains no arbitrary constants.
• Initial Condition: A condition that specifies the value of the solution at a specific point.
• Boundary Condition: A condition specified at the boundaries of the domain.
• Initial Value Problem (IVP): A differential equation together with initial conditions.
• Boundary Value Problem (BVP): A differential equation together with boundary conditions.

2.7 BVP Examples

Examples of Boundary Value Problems include:

• y'' + 9y = sin(t) with boundary conditions y(0) = 1, y'(2π) = -1. The solution is y(t) = (1/8) sin(t) + cos(3t) + sin(3t).
• y'' + π²y = 0 with boundary conditions y(0) = 2, y(1) = -2. The solution is y(t) = 2cos(πt) + (c)sin(πt).

2.8 Properties of ODE’s

A differential equation is linear if it can be written in the form a_n(t)yⁿ + a_{n-1}(t)y^{n-1} + ... + a_1(t)y' + a_0(t)y = h(t). It is nonlinear if it cannot be written in this form. An example of a nonlinear equation is x³(y''')³ - x²y(y'')² + 3xy' + 5y = eˣ.

2.9 Superposition

The superposition principle allows us to decompose a problem into smaller, simpler parts and then combine their solutions to find a solution to the original problem. This principle applies to linear homogeneous differential equations.

2.10 Explicit Solution

An explicit solution of a differential equation can be written in the form y = f(x). For example, y = xeˣ is an explicit solution of d²y/dx² - 2 dy/dx + y = 0.

2.11 Implicit Solution

An implicit solution is a relation G(x, y) = 0 that defines one or more explicit solutions on an interval I. For example, x² + y² - 4 = 0 is an implicit solution of y' = -x/y because it defines two explicit solutions: y = +√(4 - x²) and y = -√(4 - x²). 💡 Why this matters: Implicit solutions are common because solving for y explicitly is often difficult or impossible.

3 Separable Equations

A differential equation of the form dy/dx = f(x, y) is called separable if it can be written as dy/dx = h(x)g(y).

3.1 Solution steps of Separable Equations

To solve a separable equation, perform the following steps:

1. Solve g(y) = 0 to find any constant solutions.
2. For non-constant solutions, write the equation as dy / g(y) = h(x)dx. Integrate both sides: ∫ dy / g(y) = ∫ h(x)dx. This yields a solution of the form G(y) = H(x) + C.
3. List all constant and non-constant solutions to avoid repetition.
4. If given an IVP, use the initial condition to find the particular solution.

Note: Only one constant of integration is needed, as C₁ - C₂ = C. The constant may be relabeled.

Example 1: Find the particular solution of dy/dx = (y² - 1)/x, y(1) = 2.

1. Constant solutions are found by solving y² - 1 = 0 → y = ±1.
2. Separate variables: dy / (y² - 1) = dx / x. Integrate using partial fractions: (1/2) [1/(y-1) - 1/(y+1)] dy = dx / x.
3. Integrating gives (1/2) ln |(y-1)/(y+1)| = ln |x| + C.
4. Using the initial condition y(1) = 2: (1/2) ln(1/3) = C. The explicit solution is y = (3 + x²) / (3 - x²).

Example 2: Solve dy/dt = 1 + 1/y².

1. No constant solutions exist because 1 + 1/y² = 0 has no real roots.
2. Separate variables: dy / (1 + 1/y²) = dt. Since dy / (1 + 1/y²) = y - tan⁻¹(y), the implicit solution is y - tan⁻¹(y) = t + C.

Example 3: Solve the IVP dy/dt = 1 + t² + y² + t²y², y(0) = 1.

1. The equation is separable: dy/dt = (1 + t²)(1 + y²). No constant solutions because 1 + y² = 0 has no real roots.
2. Separate variables: dy / (1 + y²) = (1 + t²) dt. Integrate: tan⁻¹(y) = t + t³/3 + C.
3. The explicit solution is y = tan(t + t³/3 + C).
4. Using the initial condition y(0) = 1: C = tan⁻¹(1) = π/4. The particular solution is y = tan(t + t³/3 + π/4).

Example 4: Solve (1 + x) dy - y dx = 0.

1. The only constant solution is y = 0.
2. Separate variables: dy / y = dx / (1 + x). Integrate: ln |y| = ln |1 + x| + c₁. This simplifies to y = C(1 + x), where C = ±eᶜ¹.
3. All solutions are y = C(1 + x) and y = 0.

Example 5: Solve xy⁴ dx + (y² + 2) e⁻³ˣ dy = 0.

1. The constant solution is y = 0.
2. Separate variables: xe³ˣ dx + (y⁻² + 2y⁻⁴) dy = 0. Integrate, using integration by parts on the first term, to get (1/3)xe³ˣ - (1/9)e³ˣ - y⁻¹ - (2/3)y⁻³ = c₁. This simplifies to e³ˣ(3x - 1)/9 = y⁻¹ + 2/(3y³) + c.
3. All solutions are e³ˣ(3x - 1)/9 = y⁻¹ + 2/(3y³) + c and y = 0.

Example 6: Solve the IVPs dy/dx = (y - 1)², y(0) = 1 and dy/dx = (y - 1)², y(0) = 1.01, and compare.

1. The constant solution is y = 1.
2. Separate variables: dy / (y - 1)² = dx. Integrate: -1/(y - 1) = x + c. The general solution is y = 1 - 1/(x + c).
3. For y(0) = 1:
   • Plugging in yields -1/0 = c, so c is infinite. Thus, the particular solution is the constant solution y = 1.
4. For y(0) = 1.01:
   • Plugging in yields -1/0.01 = c → c = -100. Thus, the particular solution is y = 1 - 1/(x - 100). 💡 Why this matters: The first IVP is unstable; a small change in the initial condition leads to a drastically different, blow-up solution.

⭐ Key Takeaways

The order of a differential equation is the highest derivative it contains, while its type (ODE vs. PDE) depends on the kind of derivative used. Solutions can be general (with constants) or particular (no constants), and explicit (solved for the dependent variable) or implicit (defined by a relation). Separable equations are solved by isolating variables and integrating each side, but always check for constant solutions, as they can be lost during the separation process.

🧠 Quick Revision Questions

1. What is the order of the equation (y''')² + 2y' - y = 0?
2. Is the equation x³y''' + x²y'' - xy' + y = 0 linear or nonlinear?
3. Explain the difference between an explicit solution and an implicit solution.
4. What are the two key steps (constant and non-constant) for solving a separable differential equation?
5. For the equation dy/dx = y² - 1, identify all constant solutions before attempting to solve for non-constant solutions.

📘 Lecture 5 — Comparison of Solutions: Sensitivity in Differential Equations

📖 Overview: This lecture explores how very small changes in initial conditions or differential equations can produce radically different solutions. Through detailed examples, it demonstrates the concept of sensitivity in differential equations and introduces homogeneous differential equations as the next major topic. Understanding this sensitivity is crucial for recognizing the limitations of mathematical modeling.

🗂️ Topics Covered

The lecture examines two comparison examples showing radical changes in solutions due to tiny alterations in initial conditions (Example 6) or in the differential equation itself (Example 7). It then transitions to a new section on Homogeneous Differential Equations, covering their definition, how to identify homogeneous functions, and the step-by-step method of solution using the substitution v = y/x. The lecture concludes with Exercise problems and an Example 2 demonstrating the solution process.

📝 Lecture Summary

Comparison: A radical change in the solutions

The lecture presents two examples showing how extremely small changes in initial conditions or the differential equation can lead to dramatically different solution behaviors.

Example 6: Comparison of Initial Value Problems

Consider the differential equation dy/dx = (y - 1)². Two initial value problems are solved:

(a) dy/dx = (y - 1)², y(0) = 1.01 Using separation of variables, the non-constant solution is found:

• ∫ dy/(y - 1)² = ∫ dx → -1/(y - 1) = x + c
• Applying y(0) = 1.01: -1/(0.01) = 0 + c → c = -100
• Solution: -1/(y - 1) = x - 100 → y = 1 + 1/(100 - x)
• This solution has a vertical asymptote at x = 100.

(b) dy/dx = (y - 1)², y(0) = 1

• Applying y(0) = 1: -1/(1 - 1) = -∞ = 0 + c → y - 1 = 0
• Solution: y = 1 (constant solution)

🔑 Definition — Sensitivity: A radical change in the solutions of the differential equation has occurred corresponding to a very small change in the initial condition (from 1.01 to 1).

Example 7: Comparison of Differential Equations

Solve the initial value problems with the same initial condition but a tiny change in the equation:

(a) dy/dx = (y - 1)² + 0.01, y(0) = 1 Separating variables:

• ∫ dy/[(y - 1)² + 0.01] = ∫ dx → ∫ d(y-1)/[(y-1)² + (0.01)²] = ∫ dx
• Using the formula ∫ du/(u² + a²) = (1/a)tan⁻¹(u/a):
  • (1/0.01)tan⁻¹[(y-1)/0.01] = x + c
• Applying y(0) = 1: tan⁻¹(0) = 0.01(0 + c) → 0 = c
• Solution: y = 1 + 0.01 tan(0.01x)

(b) dy/dx = (y - 1)² - 0.01, y(0) = 1 Separating variables:

• ∫ dy/[(y-1)² - 0.01] = ∫ dx → ∫ d(y-1)/[(y-1)² - (0.01)²] = ∫ dx
• Using the formula ∫ du/(u² - a²) = (1/2a)ln|(u-a)/(u+a)|:
  • (1/2×0.01)ln|(y-1-0.01)/(y-1+0.01)| = x + c
  • (1/0.02)ln|(y-1-0.01)/(y-1+0.01)| = x + c
• Applying y(0) = 1: (1/0.02)ln(1) = 0 + c → c = 0
• Simplifying using the property a/b = c/d → (a+b)/(a-b) = (c+d)/(c-d):
  • (y-1-0.01 + y-1+0.01)/(y-1-0.01 - (y-1+0.01)) = (e^(0.02x) + 1)/(e^(0.02x) - 1)
  • 2(y-1)/(-2×0.01) = (e^(0.02x) + 1)/(e^(0.02x) - 1)
  • -(y-1)/0.01 = (e^(0.02x) + 1)/(e^(0.02x) - 1)
• Solution: y = 1 - 0.01[(e^(0.02x) + 1)/(e^(0.02x) - 1)]

Comparison: The solutions are: (a) y = 1 + 0.01 tan(0.01x) (b) y = 1 - 0.01[(e^(0.02x) + 1)/(e^(0.02x) - 1)]

Again a radical change has occurred corresponding to a very small change in the differential equation (adding 0.01 vs subtracting 0.01).

💡 Why this matters: These examples demonstrate that differential equations can be extremely sensitive to small changes in either initial conditions or the equation itself. This sensitivity has profound implications for mathematical modeling — tiny measurement errors can lead to completely different predictions.

4 Homogeneous Differential Equations

🔑 Definition — Homogeneous Differential Equation: A differential equation of the form dy/dx = f(x,y) is said to be homogeneous if the function f(x,y) is homogeneous, meaning f(tx, ty) = tⁿf(x,y) for some real number n, for any number t.

Example: Determining homogeneity For f(x,y) = xy/(x² + y²):

• f(tx,ty) = (t²xy)/(t²(x² + y²)) = xy/(x² + y²) = f(x,y)
• Therefore f is homogeneous (n = 0)

For g(x,y) = ln(-3x²y/(x³ + 4xy²)):

• g(tx,ty) = ln(-3t³x²y/(t³(x³ + 4xy²))) = ln(-3x²y/(x³ + 4xy²)) = g(x,y)
• Therefore g is also homogeneous

Method of Solution

To solve a homogeneous differential equation dy/dx = f(x,y), use the substitution v = y/x.

If f(x,y) is homogeneous of degree zero (n=0), then:

• f(x,y) = f(1, v) = F(v)
• Since y = xv, differentiation gives: dy/dx = x(dv/dx) + v
• The differential equation becomes: x(dv/dx) + v = f(1, v)
• This is a separable equation — solve for v, then substitute back y = xv

Summary — Step-by-step method:

1. Identify the equation as homogeneous by checking f(tx, ty) = tⁿf(x,y)
2. Write out the substitution v = y/x
3. Through differentiation, find the new equation satisfied by v
4. Solve the new equation (which is always separable) to find v
5. Go back to the old function y through y = vx
6. If we have an IVP, use the initial condition to find the constant of integration

⚠️ Caution:

• Since we have to solve a separable equation, we must be careful about constant solutions
• If the substitution y = vx does not reduce the equation to separable form, then the equation is not homogeneous or something is wrong along the way

Example 2

Solve the differential equation: dy/dx = (-2x + 5y)/(2x + y)

Step 1: Check homogeneity — f(x,y) = (-2x + 5y)/(2x + y)

• f(tx,ty) = (-2tx + 5ty)/(2tx + ty) = t(-2x + 5y)/t(2x + y) = (-2x + 5y)/(2x + y) = f(x,y)
• Therefore f is homogeneous of degree zero

Step 2: Substitute v = y/x, so y = xv Step 3: dy/dx = x(dv/dx) + v The equation becomes: x(dv/dx) + v = (-2x + 5xv)/(2x + xv) = x(-2 + 5v)/x(2 + v) = (-2 + 5v)/(2 + v)

Step 4: x(dv/dx) = (-2 + 5v)/(2 + v) - v = (-2 + 5v)/(2 + v) - v(2+v)/(2+v) = (-2 + 5v - 2v - v²)/(2+v) = (-2 + 3v - v²)/(2+v)

• x(dv/dx) = (-v² + 3v - 2)/(2+v)
• Separate: (2+v)/[-(v² - 3v + 2)] dv = dx/x
• (2+v)/[-(v-1)(v-2)] dv = dx/x
• Using partial fractions and integrating gives the solution for v, then substitute back y = xv

⭐ Key Takeaways

The most critical concept from this lecture is the extreme sensitivity of differential equations — tiny changes in initial conditions (e.g., from y(0)=1 to y(0)=1.01) or in the equation itself (adding or subtracting 0.01) can produce fundamentally different solutions, such as switching from a constant solution to an unbounded one or changing the entire functional form. This sensitivity is demonstrated through detailed comparison examples where solutions range from constant functions to those with vertical asymptotes or exponential forms. The lecture also introduces homogeneous differential equations, defined by f(tx,ty)=tⁿf(x,y), which are solved using the substitution v=y/x to convert them into separable equations. The step-by-step method involves checking homogeneity, substituting, differentiating, solving the separable equation, and back-substituting — with caution about constant solutions and ensuring the substitution actually works.

🧠 Quick Revision Questions

1. In Example 6, how does changing the initial condition from y(0)=1 to y(0)=1.01 radically alter the solution of dy/dx=(y-1)²?
2. In Example 7, why do the solutions differ so dramatically when the differential equation changes from (y-1)²+0.01 to (y-1)²-0.01?
3. What is the definition of a homogeneous function, and how do you check if f(x,y)=xy/(x²+y²) is homogeneous?
4. What substitution is used to solve a homogeneous differential equation, and what form does the resulting equation take?
5. What caution must be observed when using the substitution v=y/x to solve homogeneous equations?

📘 Lecture 6 — Equations Reducible to Homogeneous Form & Exact Differential Equations

📖 Overview: This lecture covers two important advanced topics in solving first-order differential equations. First, it explains how to reduce non-homogeneous equations with specific coefficient relationships to homogeneous form through appropriate substitutions. Second, it introduces exact differential equations, providing a systematic method for identifying and solving them using the condition of exactness and the potential function approach. These techniques significantly expand the range of solvable differential equations.

🗂️ Topics Covered

The lecture begins by completing a worked example of solving a homogeneous differential equation reducible from a non-homogeneous form. It then introduces two cases for reducing non-homogeneous equations: Case 1 where the coefficients of (x) and (y) in the numerator and denominator are proportional (solved via substitution (z = a_1 x + b_1 y)), and Case 2 where they are not proportional (solved via translation (x = X + h, y = Y + k) to make the equation homogeneous). Detailed examples demonstrate both cases. The lecture then transitions to exact differential equations, defining the exactness condition (\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}), and outlining a step-by-step method for finding the potential function (F(x, y)) and the general solution (F(x, y) = C). Several solved examples and applications to initial value problems are provided.

📝 Lecture Summary

4.2 Equations reducible to homogenous form

The lecture continues the discussion of equations that can be reduced to homogeneous form. The general non-homogeneous equation considered is: [ \frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2} ]

4.2.1 Case 1: (\frac{a_1}{a_2} = \frac{b_1}{b_2})

When the coefficients satisfy (\frac{a_1}{a_2} = \frac{b_1}{b_2}), we use the substitution (z = a_1 x + b_1 y). This reduces the equation to a separable equation in the variables (x) and (z). Solving the separable equation and then replacing (z) with (a_1 x + b_1 y) gives the solution of the original differential equation.

🔑 Definition — Separable equation: A differential equation that can be written in the form (g(y) dy = f(x) dx), allowing direct integration of both sides.

📌 Example 3: Solve (\frac{dy}{dx} = -\frac{2x + 3y - 1}{2x + 3y + 2})

Here, (a_1 = 1, a_2 = 1) and (b_1 = 1, b_2 = 1), so (\frac{a_1}{a_2} = \frac{b_1}{b_2}). We substitute (z = 2x + 3y).

Step 1: Differentiate: (\frac{dy}{dx} = \frac{1}{3}\left(\frac{dz}{dx} - 2\right))

Step 2: Substitute into the equation: [ \frac{1}{3}\left(\frac{dz}{dx} - 2\right) = -\frac{z - 1}{z + 2} ]

Step 3: Rearrange to separable form: [ \frac{dz}{dx} = \frac{-z + 7}{z + 2} ]

Step 4: Separate variables: [ \frac{z + 2}{-z + 7} dz = dx ]

Step 5: Integrate both sides: [ \int \frac{z + 2}{7 - z} dz = \int dx ]

Step 6: Using long division or partial fractions: [ \int \left(-1 + \frac{9}{7 - z}\right) dz = \int dx ]

Step 7: Integrating: [ -z - 9\ln|z - 7| = x + A ]

Step 8: Replace (z = 2x + 3y): [ -(2x + 3y) - 9\ln|2x + 3y - 7| = x + A ]

Step 9: Simplify: [ -9\ln|2x + 3y - 7| = 3x + 3y + A ]

Step 10: Exponentiate: [ (2x + 3y - 7)^{-9} = ce^{3(x+y)}, \text{ where } c = e^A ]

📌 Final implicit solution: ((2x + 3y - 7)^{-9} = ce^{3(x+y)})

4.2.2 Case 2: (\frac{a_1}{a_2} \neq \frac{b_1}{b_2})

When the coefficients are not proportional, we use the substitution: [ x = X + h, \quad y = Y + k ] where (h) and (k) are constants to be determined.

Step 1: Substitute into the equation: [ \frac{dY}{dX} = \frac{a_1 X + b_1 Y + (a_1 h + b_1 k + c_1)}{a_2 X + b_2 Y + (a_2 h + b_2 k + c_2)} ]

Step 2: Choose (h) and (k) such that: [ \begin{cases} a_1 h + b_1 k + c_1 = 0 \ a_2 h + b_2 k + c_2 = 0 \end{cases} ]

Step 3: This reduces the equation to: [ \frac{dY}{dX} = \frac{a_1 X + b_1 Y}{a_2 X + b_2 Y} ] which is a homogeneous differential equation in (X) and (Y).

Step 4: Solve the homogeneous equation using (Y = VX) substitution.

Step 5: Return to original variables (x) and (y) using (x = X + h, y = Y + k).

📌 Example 4: Solve (\frac{dy}{dx} = \frac{x + 2y - 4}{2x + y - 5})

Here, (a_1=1, b_1=2, a_2=2, b_2=1), and (\frac{1}{2} \neq \frac{2}{1}), so we use Case 2.

Step 1: Substitute (x = X + h, y = Y + k): [ \frac{dY}{dX} = \frac{(X + h) + 2(Y + k) - 4}{2(X + h) + (Y + k) - 5} = \frac{X + 2Y + (h + 2k - 4)}{2X + Y + (2h + k - 5)} ]

Step 2: Solve for (h, k): [ \begin{cases} h + 2k - 4 = 0 \ 2h + k - 5 = 0 \end{cases} ]

Solving: From first equation, (h = 4 - 2k). Substituting into second: (2(4 - 2k) + k - 5 = 0 \Rightarrow 8 - 4k + k - 5 = 0 \Rightarrow 3 - 3k = 0 \Rightarrow k = 1). Then (h = 4 - 2(1) = 2).

Step 3: The equation becomes: [ \frac{dY}{dX} = \frac{X + 2Y}{2X + Y} ]

Step 4: This is homogeneous. Substitute (Y = VX): [ \frac{dY}{dX} = V + X\frac{dV}{dX} = \frac{X + 2VX}{2X + VX} = \frac{1 + 2V}{2 + V} ]

Step 5: Rearrange: [ X\frac{dV}{dX} = \frac{1 + 2V}{2 + V} - V = \frac{1 + 2V - 2V - V^2}{2 + V} = \frac{1 - V^2}{2 + V} ]

Step 6: Separate variables: [ \frac{2 + V}{1 - V^2} dV = \frac{dX}{X} ]

Step 7: Decompose into partial fractions: [ \frac{2 + V}{(1-V)(1+V)} = \frac{3/2}{1-V} + \frac{1/2}{1+V} ]

Step 8: Integrate: [ \int \left(\frac{3/2}{1-V} + \frac{1/2}{1+V}\right) dV = \int \frac{dX}{X} ]

[ -\frac{3}{2}\ln|1-V| + \frac{1}{2}\ln|1+V| = \ln|X| + \ln|A| ]

Step 9: Combine logarithms: [ \ln\left(\frac{(1+V)^{1/2}}{(1-V)^{3/2}}\right) = \ln|AX| ]

[ \frac{(1+V)^{1/2}}{(1-V)^{3/2}} = AX ]

Step 10: Square both sides: [ \frac{1+V}{(1-V)^3} = A^2 X^2 ]

Let (C = A^{-2}): [ (1-V)^3/(1+V) = CX^{-2} ]

Step 11: Replace (V = Y/X): [ \frac{(1 - Y/X)^3}{1 + Y/X} = C X^{-2} ]

Step 12: Multiply numerator and denominator by (X^3): [ \frac{(X - Y)^3}{X^3} \cdot \frac{X}{X + Y} = C X^{-2} ]

[ \frac{(X - Y)^3}{X^2(X + Y)} = C X^{-2} ]

[ \frac{(X - Y)^3}{X + Y} = C ]

Step 13: Return to original variables: (X = x - 2, Y = y - 1): [ \frac{(x - 2 - y + 1)^3}{x - 2 + y - 1} = C ]

📌 Final implicit solution: (\frac{(x - y - 1)^3}{x + y - 3} = C)

💡 Why this matters: The translation method ((x = X + h, y = Y + k)) geometrically shifts the coordinate system so that the intersection point of the lines in the numerator and denominator moves to the origin, making the equation homogeneous. This is a powerful technique that transforms a seemingly complicated equation into one we already know how to solve.

5 Exact Differential Equations

A differential equation written in the form (M(x, y)dx + N(x, y)dy = 0) is called exact if there exists a function (F(x, y)) such that: [ \frac{\partial F}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial F}{\partial y} = N(x, y) ]

🔑 Definition — Exact differential equation: A differential equation (Mdx + Ndy = 0) for which there exists a function (F(x, y)) whose total differential (dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy) equals (Mdx + Ndy).

Condition for exactness: The equation (Mdx + Ndy = 0) is exact if and only if: [ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ]

5.1 Method of Solution

Step 1: Check that the equation is exact by verifying (\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}).

Step 2: Write down the system: [ \frac{\partial F}{\partial x} = M(x, y), \quad \frac{\partial F}{\partial y} = N(x, y) ]

Step 3: Integrate the first equation with respect to (x) (treating (y) as constant): [ F(x, y) = \int M(x, y) dx + \theta(y) ] where (\theta(y)) is an arbitrary function of (y) only.

Step 4: Differentiate the result with respect to (y) and equate to (N(x, y)): [ \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}\left(\int M(x, y) dx\right) + \theta'(y) = N(x, y) ]

Step 5: Solve for (\theta'(y)): [ \theta'(y) = N(x, y) - \frac{\partial}{\partial y}\int M(x, y) dx ]

⚠️ Caution: The expression for (\theta'(y)) should be independent of (x). If (x) does not cancel out, something is wrong (either the equation is not exact or an integration error was made).

Step 6: Integrate (\theta'(y)) to find (\theta(y)).

Step 7: Write (F(x, y) = \int M(x, y) dx + \theta(y)).

Step 8: The general solution is given implicitly by: [ F(x, y) = C ]

Step 9: If given an initial value problem, plug in the initial condition to find the constant (C).

📌 Example 1: Solve ((3x^2 y + 2)dx + (x^3 + y)dy = 0)

Step 1: Identify (M = 3x^2 y + 2), (N = x^3 + y)

Check exactness: [ \frac{\partial M}{\partial y} = 3x^2, \quad \frac{\partial N}{\partial x} = 3x^2 ] Since (\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}), the equation is exact.

Step 2: We need (F) such that: [ \frac{\partial F}{\partial x} = 3x^2 y + 2, \quad \frac{\partial F}{\partial y} = x^3 + y ]

Step 3: Integrate (\frac{\partial F}{\partial x}) with respect to (x): [ F(x, y) = \int (3x^2 y + 2) dx = x^3 y + 2x + h(y) ] where (h(y)) is an arbitrary function of (y).

Step 4: Differentiate with respect to (y): [ \frac{\partial F}{\partial y} = x^3 + h'(y) ]

Step 5: Equate to (N = x^3 + y): [ x^3 + h'(y) = x^3 + y ] [ h'(y) = y ]

Step 6: Integrate: (h(y) = \frac{y^2}{2})

Step 7: The function is: [ F(x, y) = x^3 y + 2x + \frac{y^2}{2} ]

📌 Final solution: (x^3 y + 2x + \frac{y^2}{2} = C)

📌 Example 2 (Initial Value Problem): Solve ((2y \sin x \cos x + y^2 \sin x)dx + (\sin^2 x - 2y \cos x)dy = 0), (y(0) = 3)

Step 1: (M = 2y \sin x \cos x + y^2 \sin x), (N = \sin^2 x - 2y \cos x)

Check exactness: [ \frac{\partial M}{\partial y} = 2\sin x \cos x + 2y \sin x ] [ \frac{\partial N}{\partial x} = 2\sin x \cos x + 2y \sin x ] Since (\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}), the equation is exact.

Step 2: We need (F) such that: [ \frac{\partial F}{\partial x} = M, \quad \frac{\partial F}{\partial y} = N ]

Step 3: Integrate (\frac{\partial F}{\partial y} = N) with respect to (y): [ F(x, y) = \int (\sin^2 x - 2y \cos x) dy = y \sin^2 x - y^2 \cos x + h(x) ]

Step 4: Differentiate with respect to (x): [ \frac{\partial F}{\partial x} = y(2\sin x \cos x) - y^2(-\sin x) + h'(x) = 2y \sin x \cos x + y^2 \sin x + h'(x) ]

Step 5: Equate to (M = 2y \sin x \cos x + y^2 \sin x): [ 2y \sin x \cos x + y^2 \sin x + h'(x) = 2y \sin x \cos x + y^2 \sin x ] [ h'(x) = 0 \Rightarrow h(x) = C_1 ]

Step 6: The function is: [ F(x, y) = y \sin^2 x - y^2 \cos x ]

Step 7: General solution: (y \sin^2 x - y^2 \cos x = C)

Step 8: Apply initial condition (y(0) = 3): [ 3 \cdot \sin^2 0 - 3^2 \cdot \cos 0 = C \Rightarrow 0 - 9 = C \Rightarrow C = -9 ]

📌 Final solution: (y \sin^2 x - y^2 \cos x = -9) or equivalently (y^2 \cos x - y \sin^2 x = 9)

💡 Why this matters: When solving exact equations, you can choose to integrate either (\frac{\partial F}{\partial x} = M) (with respect to (x)) or (\frac{\partial F}{\partial y} = N) (with respect to (y)). Choose whichever integral is simpler. In Example 2, integrating (N) with respect to (y) was easier because it avoided the trigonometric integral.

📌 Example 3: Solve ((e^{2y} - y\cos xy)dx + (2xe^{2y} - x\cos xy + 2y)dy = 0)

Step 1: Identify (M = e^{2y} - y\cos xy), (N = 2xe^{2y} - x\cos xy + 2y)

Check exactness: [ \frac{\partial M}{\partial y} = 2e^{2y} - \cos xy + xy\sin xy ] [ \frac{\partial N}{\partial x} = 2e^{2y} - \cos xy + xy\sin xy ] Since they are equal, the equation is exact.

Step 2: We need (F) such that: [ \frac{\partial F}{\partial x} = e^{2y} - y\cos xy, \quad \frac{\partial F}{\partial y} = 2xe^{2y} - x\cos xy + 2y ]

Step 3: Let's integrate (\frac{\partial F}{\partial y}) with respect to (y): (Note: when integrating with respect to (y), treat (x) as constant)

[ F(x, y) = \int (2xe^{2y} - x\cos xy + 2y) dy ]

[ F(x, y) = 2x \cdot \frac{e^{2y}}{2} - x \cdot \frac{\sin xy}{x} + 2 \cdot \frac{y^2}{2} + h(x) ]

[ F(x, y) = xe^{2y} - \sin xy + y^2 + h(x) ]

Step 4: Differentiate with respect to (x): [ \frac{\partial F}{\partial x} = e^{2y} - y\cos xy + h'(x) ]

Step 5: Equate to (M = e^{2y} - y\cos xy): [ e^{2y} - y\cos xy + h'(x) = e^{2y} - y\cos xy ] [ h'(x) = 0 \Rightarrow h(x) = C_1 ]

📌 Final solution: (xe^{2y} - \sin xy + y^2 = C)

📌 Example 4: Solve (2xy dx + (x^2 - 1)dy = 0)

Step 1: (M = 2xy), (N = x^2 - 1)

Check exactness: [ \frac{\partial M}{\partial y} = 2x, \quad \frac{\partial N}{\partial x} = 2x ] Since they are equal, the equation is exact.

Step 2: We need (F) such that: [ \frac{\partial F}{\partial x} = 2xy, \quad \frac{\partial F}{\partial y} = x^2 - 1 ]

Step 3: Integrate (\frac{\partial F}{\partial x}) with respect to (x): [ F(x, y) = \int 2xy dx = x^2 y + g(y) ]

Step 4: Differentiate with respect to (y): [ \frac{\partial F}{\partial y} = x^2 + g'(y) ]

Step 5: Equate to (N = x^2 - 1): [ x^2 + g'(y) = x^2 - 1 ] [ g'(y) = -1 \Rightarrow g(y) = -y ]

📌 Final solution: (x^2 y - y = C) or (y(x^2 - 1) = C)

📌 Example 5 (Initial Value Problem): Solve ((\cos x \sin x - xy^2)dx + y(1 - x^2)dy = 0), (y(0) = 2)

Step 1: (M = \cos x \sin x - xy^2), (N = y(1 - x^2) = y - x^2 y)

Check exactness: [ \frac{\partial M}{\partial y} = -2xy, \quad \frac{\partial N}{\partial x} = -2xy ] Since they are equal, the equation is exact.

Step 2: We need (F) such that: [ \frac{\partial F}{\partial x} = \cos x \sin x - xy^2, \quad \frac{\partial F}{\partial y} = y - x^2 y ]

Step 3: Integrate (\frac{\partial F}{\partial y}) with respect to (y): [ F(x, y) = \int (y - x^2 y) dy = \frac{y^2}{2} - \frac{x^2 y^2}{2} + h(x) = \frac{y^2}{2}(1 - x^2) + h(x) ]

Step 4: Differentiate with respect to (x): [ \frac{\partial F}{\partial x} = \frac{y^2}{2}(-2x) + h'(x) = -xy^2 + h'(x) ]

Step 5: Equate to (M = \cos x \sin x - xy^2): [ -xy^2 + h'(x) = \cos x \sin x - xy^2 ] [ h'(x) = \cos x \sin x ]

Step 6: Integrate: [ h(x) = \int \cos x \sin x dx = \frac{1}{2}\int \sin 2x dx = -\frac{1}{4}\cos 2x + C_1 ] Alternatively: (\int \cos x \sin x dx = \frac{\sin^2 x}{2} + C_1)

Step 7: The function is: [ F(x, y) = \frac{y^2}{2}(1 - x^2) + \frac{\sin^2 x}{2} ]

Step 8: General solution: (\frac{y^2}{2}(1 - x^2) + \frac{\sin^2 x}{2} = C)

Multiply by 2: (y^2(1 - x^2) + \sin^2 x = C)

Step 9: Apply initial condition (y(0) = 2): [ 2^2(1 - 0) + \sin^2 0 = C \Rightarrow 4 = C ]

📌 Final solution: (y^2(1 - x^2) + \sin^2 x = 4)

⭐ Key Takeaways

1. Recognizing when to use each reduction: For non-homogeneous equations (\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}), check if (\frac{a_1}{a_2} = \frac{b_1}{b_2}) (Case 1: use (z = a_1 x + b_1 y)) or not (Case 2: use translation (x = X + h, y = Y + k)). This classification is critical for selecting the correct substitution.

2. Exactness condition is non-negotiable: Always verify (\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}) before attempting the solution. If the condition fails, the equation is not exact, and you must use other methods (integrating factors, which will be covered later).

3. The (\theta(y)) or (h(x)) function is the key: After integrating, the arbitrary function (\theta(y)) (if integrating with respect to (x)) or (h(x)) (if integrating with respect to (y)) is found by differentiating and equating to the other partial derivative. The resulting expression for (\theta'(y)) or (h'(x)) must be independent of the other variable.

4. Choose the simpler integration path: You can integrate either (\frac{\partial F}{\partial x} = M) with respect to (x) or (\frac{\partial F}{\partial y} = N) with respect to (y). Choose the one that gives the simpler integral to save time and reduce errors.

5. The solution is always implicit: Exact equations give solutions in

📘 Lecture 7 — Integrating Factor Technique

📖 Overview: This lecture addresses the situation when a differential equation is not exact, introducing the concept of an integrating factor (IF) to transform it into an exact equation. It explains the theory behind IFs and provides systematic methods for finding them in four special cases, making non-exact equations solvable.

🗂️ Topics Covered

The lecture begins with an exercise section reviewing exact equations, then introduces the concept of an integrating factor for non-exact equations. It presents a general partial differential equation for finding IFs and details four special cases: Case 1 (IF as a function of x only), Case 2 (IF as a function of y only), Case 3 (for homogeneous equations), and Case 4 (for equations of the form y f(xy) dx + x g(xy) dy = 0). A worked example demonstrates the application of Case 1, and a step-by-step summary is provided.

📝 Lecture Summary

5.2 Exercise

This section lists several differential equations for practice. The first problem involves verifying exactness and solving: (sin y – y sin x) dx + (cos x + x cos y) dy = 0. Another problem asks to solve an initial value problem: (eˣ + y) dx + (2 + x + yeʸ) dy = 0, y(0) = 1. There are also problems requiring finding the value of k so that a given differential equation is exact, such as (2xy³ – y sin xy + ky⁴) dx – (20x³ + x sin xy) dy = 0.

6 Integrating Factor Technique

If the equation M(x,y) dx + N(x,y) dy = 0 is not exact, meaning ∂M/∂y ≠ ∂N/∂x, we can sometimes find a function u(x,y), called an integrating factor (IF), such that the new equation u M dx + u N dy = 0 becomes exact. The IF must satisfy a partial differential equation: ∂/∂y (uM) = ∂/∂x (uN). This equation is generally very difficult to solve directly.

🔑 Definition — Integrating Factor (IF): A function u(x,y) that, when multiplied by a non-exact differential equation M dx + N dy = 0, transforms it into an exact equation.

💡 Why this matters: The IF technique extends the method for exact equations to a much larger class of differential equations, making them solvable.

Example

Show that 1/(x²+y²) is an integrating factor for (x² + y² – x) dx – y dy = 0 and solve.

• Step 1: Check exactness: M = x² + y² – x, N = –y. ∂M/∂y = 2y, ∂N/∂x = 0. Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
• Step 2: Multiply by the IF u = 1/(x²+y²): M_new = 1 – x/(x²+y²), N_new = –y/(x²+y²). Now ∂M_new/∂y = 2xy/(x²+y²)² = ∂N_new/∂x, so the new equation is exact.
• Step 3: Solve. It is simpler to rewrite the original equation as: dx – (x dx + y dy)/(x²+y²) = 0 → dx – (1/2) d[ln(x²+y²)] = 0.
• Step 4: Integrate: x – (1/2) ln(x²+y²) = C.

📌 Example: The equation (x² + y² – x) dx – y dy = 0 is not exact because ∂M/∂y = 2y and ∂N/∂x = 0. Multiplying by 1/(x²+y²) yields [1 – x/(x²+y²)] dx – [y/(x²+y²)] dy = 0, which is exact because ∂M_new/∂y = 2xy/(x²+y²)² = ∂N_new/∂x. The solution is x – (1/2) ln(x²+y²) = C.

6.1 Case 1: IF is a function of x only, u(x)

This case applies if the expression [ (∂M/∂y – ∂N/∂x) / N ] is a function of x only. Then the integrating factor is given by: 📐 Formula: u(x) = exp( ∫ [(∂M/∂y – ∂N/∂x) / N] dx ) → The IF is the exponential of the integral of that x-only function with respect to x.

6.2 Case 2: IF is a function of y only, u(y)

This case applies if the expression [ (∂N/∂x – ∂M/∂y) / M ] is a function of y only. Then the integrating factor is given by: 📐 Formula: u(y) = exp( ∫ [(∂N/∂x – ∂M/∂y) / M] dy ) → The IF is the exponential of the integral of that y-only function with respect to y.

6.3 Case 3: IF for Homogeneous Equations

If the given differential equation is homogeneous, and xM + yN ≠ 0, then an integrating factor exists: 📐 Formula: u = 1 / (xM + yN)

6.4 Case 4: IF for Equations of the form y f(xy) dx + x g(xy) dy = 0

If the equation can be written as y f(xy) dx + x g(xy) dy = 0, and xM – yN ≠ 0, then an integrating factor exists: 📐 Formula: u = 1 / (xM – yN)

Summary: Step-by-step Procedure

• Step 1: Write the equation as M dx + N dy = 0 and identify M and N.
• Step 2: Check exactness (∂M/∂y = ∂N/∂x?). If exact, solve directly.
• Step 3: If not exact, evaluate (∂M/∂y – ∂N/∂x)/N. If this is a function of x only, find IF u(x) = exp(∫ that function dx).
• Step 4: Otherwise, evaluate (∂N/∂x – ∂M/∂y)/M. If this is a function of y only, find IF u(y) = exp(∫ that function dy).
• Step 5: If steps 3 and 4 fail, test if the equation is homogeneous and xM + yN ≠ 0. If yes, u = 1/(xM + yN).
• Step 6: Test if the equation is of the form y f(xy) dx + x g(xy) dy = 0 and xM – yN ≠ 0. If yes, u = 1/(xM – yN).
• Step 7: Multiply the old equation by the found IF u to get a new, exact equation. Check exactness if possible.
• Step 8: Solve the new exact equation using the method for exact equations.

Example 1: Solve the differential equation dy/dx = –(3xy + y²)/(x²+xy)

• Step 1: (3xy + y²) dx + (x² + xy) dy = 0. So M = 3xy + y², N = x² + xy.
• Step 2: ∂M/∂y = 3x + 2y, ∂N/∂x = 2x + y. Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
• Step 3: Evaluate (∂M/∂y – ∂N/∂x)/N = (3x+2y – 2x – y)/(x²+xy) = (x+y)/(x(x+y)) = 1/x. This is a function of x only.
• Step 4: IF u(x) = exp( ∫ 1/x dx ) = e^{ln x} = x.
• Step 5: Multiply by x: (3x²y + xy²) dx + (x³ + x²y) dy = 0. This equation is exact.
• Step 6: Solve the exact equation. (The solution is not fully derived in the text but follows standard exact equation method.)

📌 Example: For the equation (3xy + y²) dx + (x² + xy) dy = 0, M=3xy+y², N=x²+xy. Not exact. (∂M/∂y – ∂N/∂x)/N = 1/x. IF u(x)=x. Multiplying gives the exact equation (3x²y+xy²)dx+(x³+x²y)dy=0.

⭐ Key Takeaways

The most critical concept is that when a differential equation is not exact, we can find an integrating factor (IF) u(x,y) to make it exact. The four cases provide systematic formulas: Case 1 for IF as a function of x only [u = exp(∫ ((∂M/∂y – ∂N/∂x)/N) dx)], Case 2 for IF as a function of y only [u = exp(∫ ((∂N/∂x – ∂M/∂y)/M) dy)], Case 3 for homogeneous equations [u = 1/(xM+yN)], and Case 4 for equations of the form y f(xy)dx + x g(xy)dy = 0 [u = 1/(xM-yN)]. The procedure is to test these cases in order, multiply the original equation by the found IF, and then solve the resulting exact equation. Memorizing the four IF formulas and the testing conditions is essential for exams.

🧠 Quick Revision Questions

1. What is the condition for a differential equation M dx + N dy = 0 to be exact?
2. What is the formula for the integrating factor u(x) in Case 1, and what condition must be satisfied to use it?
3. What is the formula for the integrating factor u(y) in Case 2, and what condition must be satisfied to use it?
4. When can the integrating factor u = 1/(xM + yN) be used?
5. For the equation (3xy + y²) dx + (x² + xy) dy = 0, what is the integrating factor, and to which case does it belong?

📘 Lecture 8 — First Order Linear Equations & Special Integrating Factors

📖 Overview: This lecture extends the concept of exact differential equations by introducing methods to find integrating factors when equations are not exact. It then transitions to first-order linear differential equations, providing a systematic solution method using integrating factors and applying it to various examples, including initial value problems and equations with x as the dependent variable.

🗂️ Topics Covered

The lecture begins with a review of the exact equation solution process, then demonstrates how to find integrating factors when equations are not exact using formulas involving partial derivatives. Examples illustrate solving non-exact equations by finding the appropriate integrating factor, including cases where the integrating factor depends on x or y alone, or where the equation is homogeneous. The second half introduces the standard form of first-order linear differential equations, derives the integrating factor u(x), and presents the general solution formula. Multiple examples demonstrate solving linear equations with various functions p(x) and q(x), including trigonometric, rational, and initial value problems.

📝 Lecture Summary

Solving an Exact Equation

When an equation is exact, the function F(x, y) satisfies the system ∂F/∂x = M and ∂F/∂y = N. The process involves integrating the first equation to get F(x, y) = ∫M dx + θ(y), then differentiating with respect to y and using the second equation to find θ'(y). For example, with M = 3xy + y² and N = x² + xy: we integrate to get F(x, y) = (3/2)x²y + xy² + θ(y). Then ∂F/∂y = (3/2)x² + 2xy + θ'(y) = x² + xy, which gives θ'(y) = -(x²/2) - xy + (3/2)x² + 2xy = x² + xy, and since θ'(y) should have no x dependence, this verification step is crucial. The solution is F(x, y) = C, giving x²y + (3/2)x²y + 2xy² = C, or more specifically x²y + (x³/2) + (x²y²/2) = C.

🔑 Definition — Exact Equation: A differential equation M dx + N dy = 0 is exact if ∂M/∂y = ∂N/∂x.

Example 2 Solve (x² - 2x + 2y²)dx + 2xy dy = 0 Here M = x² - 2x + 2y², N = 2xy. ∂M/∂y = 4y, ∂N/∂x = 2y. Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact. To find the integrating factor, compute (My - Nx)/N = (4y - 2y)/(2xy) = 2y/(2xy) = 1/x. Therefore, the I.F. is given by u = exp(∫(1/x)dx) = x. Multiplying the equation by x gives: (x³ - 2x² + 2xy²)dx + 2x²y dy = 0, which is exact. The solution is x⁴/4 - 2x³/3 + x²y² = C, or 3x⁴ - 8x³ + 12x²y² = C.

📐 Formula: Integrating Factor (I.F.) when (My - Nx)/N depends only on x: u(x) = exp(∫[(My - Nx)/N]dx)

Example 3 Solve dx + (x/y - sin y)dy = 0 M = 1, N = x/y - sin y. ∂M/∂y = 0, ∂N/∂x = 1/y. Since ∂M/∂y ≠ ∂N/∂x, not exact. Compute (Nx - My)/M = (1/y - 0)/1 = 1/y. Therefore, the I.F. is u(y) = exp(∫(1/y)dy) = y. Multiplying by y gives: y dx + (x - y sin y)dy = 0, or y dx + x dy - y sin y dy = 0, which is d(xy) - y sin y dy = 0. Integrating: xy + y cos y - sin y = C.

📐 Formula: Integrating Factor when (Nx - My)/M depends only on y: u(y) = exp(∫[(Nx - My)/M]dy)

Example 4 Solve (x²y - 2xy²)dx - (x³ - 3x²y)dy = 0 M = x²y - 2xy², N = -(x³ - 3x²y) = -x³ + 3x²y. Both M and N are homogeneous functions of degree 3. Compute xM + yN = x(x²y - 2xy²) + y(-x³ + 3x²y) = x³y - 2x²y² - x³y + 3x²y² = x²y² ≠ 0. Therefore, the I.F. is u = 1/(xM + yN) = 1/(x²y²). Multiplying the equation by u gives: ((1/y) - (2/x))dx + ((-x/y²) + (3/y))dy = 0. This new equation is exact. The solution is x/y - 2 ln|x| + 3 ln|y| = C.

📐 Formula: Integrating Factor for Homogeneous Equations: u = 1/(xM + yN), provided xM + yN ≠ 0.

💡 Why this matters: These special integrating factor formulas provide systematic methods to solve a wide range of non-exact equations without guessing, expanding the family of solvable differential equations.

Example 5 Solve y(xy + 2x²y²)dx + x(xy - x²y²)dy = 0 This equation is of the form y f(xy)dx + x g(xy)dy = 0. M = xy² + 2x²y³, N = x²y - x³y². Compute xM - yN = x(xy² + 2x²y³) - y(x²y - x³y²) = x²y² + 2x³y³ - x²y² + x³y³ = 3x³y³ ≠ 0. Therefore, the I.F. is u = 1/(xM - yN) = 1/(3x³y³). Multiplying by u gives: ((1/(3x²y)) + (2/(3x)))dx + ((1/(3xy²)) - (1/(3y)))dy = 0. The solution is 1/(3xy) + (2/3)ln|x| - (1/3)ln|y| = C, or -1/(xy) + 2 ln|x| - ln|y| = 3C.

📐 Formula: Integrating Factor for y f(xy) dx + x g(xy) dy = 0: u = 1/(xM - yN), provided xM - yN ≠ 0.

7 First Order Linear Equations

A first-order linear differential equation has the form a(x)dy/dx + b(x)y = c(x). It can be rewritten in standard form as dy/dx + p(x)y = q(x), where p(x) and q(x) are continuous functions.

🔑 Definition — First-Order Linear DE: A differential equation that can be written in the form dy/dx + p(x)y = q(x), where the dependent variable y and its derivative appear only linearly.

7.1 Method of Solution

The general solution of the first-order linear differential equation is given by: y = [∫u(x)q(x)dx + C] / u(x), where u(x) = exp(∫p(x)dx).

The function u(x) is called the integrating factor (I.F.).

Summary of Steps:

1. Identify the equation as first-order linear and rewrite in the form dy/dx + p(x)y = q(x).
2. Find the integrating factor: u(x) = e^(∫p(x)dx).
3. Write down the general solution: y = (∫u(x)q(x)dx + C) / u(x).
4. For an initial value problem (IVP), use the initial condition to find C.
5. Plug in C to write the particular solution.

Example 1 Solve y' + tan(x)y = cos²(x), y(0) = 2 p(x) = tan(x), q(x) = cos²(x). ∫tan(x)dx = -ln|cos(x)| = ln|sec(x)|, so u(x) = e^(ln|sec(x)|) = sec(x). ∫u(x)q(x)dx = ∫sec(x)cos²(x)dx = ∫cos(x)dx = sin(x). General solution: y = (sin(x) + C)/sec(x) = (sin(x) + C)cos(x). Using y(0) = 2: 2 = (0 + C)(1) = C, so C = 2. Particular solution: y = (sin(x) + 2)cos(x).

📐 Formula: Integrating Factor for First-Order Linear DE: u(x) = e^(∫p(x)dx)

Example 2 Solve dy/dt - (2t/(1+t²))y = 2/(1+t²), y(0) = 0.4 p(t) = -2t/(1+t²), q(t) = 2/(1+t²). ∫p(t)dt = -ln|1+t²|, so u(t) = e^(-ln|1+t²|) = (1+t²)⁻¹. General solution: y = (1+t²)[∫(2/(1+t²)²)dt + C]. ∫(2/(1+t²)²)dt = tan⁻¹(t) + t/(1+t²) (evaluated using integration by parts). Therefore, y = (1+t²)[tan⁻¹(t) + t/(1+t²) + C] = (1+t²)tan⁻¹(t) + t + C(1+t²). Using y(0) = 0.4: 0.4 = (1)tan⁻¹(0) + 0 + C(1) = 0 + 0 + C, so C = 0.4. Particular solution: y = t + (1+t²)tan⁻¹(t) + 0.4(1+t²).

Example 3 Solve cos²(t)sin(t)y' = -cos³(t)y + 1, y(π/4) = 0 Rewrite in standard form: dy/dt + (cos(t)/sin(t))y = 1/(cos²(t)sin(t)). ∫(cos(t)/sin(t))dt = ln|sin(t)|, so u(t) = e^(ln|sin(t)|) = sin(t). ∫u(t)q(t)dt = ∫sin(t)(1/(cos²(t)sin(t)))dt = ∫(1/cos²(t))dt = tan(t). General solution: y = (tan(t) + C)/sin(t) = sec(t) + C csc(t). Using y(π/4) = 0: 0 = sec(π/4) + C csc(π/4) = √2 + C√2, so C = -1. Particular solution: y = sec(t) - csc(t).

Example 4 Solve (x + 2y³)dy/dx = y This is not linear in y. Rewrite as dx/dy = (x + 2y³)/y, or dx/dy - (1/y)x = 2y². This is linear in x, with p(y) = -1/y, q(y) = 2y². ∫p(y)dy = -ln|y|, so I.F. = e^(-ln|y|) = 1/y. d/dy(x/y) = 2y, so x/y = ∫2y dy = y² + C. Solution: x = y(y² + C) = y³ + Cy.

💡 Why this matters: When an equation is not linear in y, it may be linear in x. Interchanging the roles of dependent and independent variables can simplify the problem.

Example 5 Solve (x-1)³dy/dx + 4(x-1)²y = x + 1 Rewrite: dy/dx + (4/(x-1))y = (x+1)/(x-1)³. p(x) = 4/(x-1), so I.F. = exp(∫4/(x-1)dx) = exp(4ln|x-1|) = (x-1)⁴. Multiplying: (x-1)⁴dy/dx + 4(x-1)³y = x² - 1. d/dx[(x-1)⁴y] = x² - 1. Integrating: (x-1)⁴y = ∫(x² - 1)dx = x³/3 - x + C. Solution: y = (x³/3 - x + C)/(x-1)⁴.

7.2 Exercise

Practice problems include solving differential equations by finding an integrating factor and solving first-order linear equations using the integrating factor method.

⭐ Key Takeaways

For non-exact equations, the integrating factor can be found systematically using formulas involving the partial derivatives of M and N, specifically u = exp(∫[(My - Nx)/N]dx) when this expression depends only on x, or u = exp(∫[(Nx - My)/M]dy) when it depends only on y. For homogeneous equations of the form M dx + N dy = 0 where both M and N are homogeneous of the same degree, the integrating factor is u = 1/(xM + yN). For equations of the form y f(xy)dx + x g(xy)dy = 0, the integrating factor is u = 1/(xM - yN). First-order linear differential equations in the form dy/dx + p(x)y = q(x) have a standard solution method using the integrating factor u(x) = e^(∫p(x)dx), giving the general solution y = (∫u(x)q(x)dx + C)/u(x). Always verify whether an equation can be treated as linear in x instead of y to expand solution possibilities.

🧠 Quick Revision Questions

1. What is the condition for a differential equation M dx + N dy = 0 to be exact? What is the solution procedure?
2. Derive the integrating factor formula for a non-exact equation when (∂M/∂y - ∂N/∂x)/N is a function of x only.
3. How do you find the integrating factor for a homogeneous differential equation M dx + N dy = 0?
4. State the standard form of a first-order linear differential equation and write the general solution formula.
5. Solve the initial value problem: dy/dx + (2x+1)/x * y = e^(-2x), y(1) = 0, showing all steps including finding the integrating factor.

📘 Lecture 9 — Differential Equations (MTH401) – Bernoulli Equations and Substitutions

📖 Overview: This lecture introduces Bernoulli differential equations, a nonlinear form that can be transformed into a linear differential equation through a specific substitution. It also covers the use of various creative substitutions to solve differential equations that do not fit standard forms, including several solved examples and exercises.

🗂️ Topics Covered

The lecture begins with the definition and method of solution for Bernoulli equations, including the substitution v = y^(1-n) to convert them to linear form. It then provides worked examples of solving Bernoulli equations. Following this, the lecture covers the technique of using arbitrary substitutions to transform non-standard differential equations into solvable forms like separable, homogeneous, exact, or linear equations, illustrated with several examples. The section concludes with exercises and a set of solved problems covering homogeneous, exact, and linear equations.

📝 Lecture Summary

8 Bernoulli Equations

A differential equation that can be written in the form dy/dx + p(x)y = q(x)y^n is called a Bernoulli equation.

8.1 Method of solution

For n = 0 or n = 1, the equation reduces to a 1st order linear DE and can be solved accordingly. For n ≠ 0, 1, we divide the equation with y^n to write it in the form y^(-n) dy/dx + p(x) y^(1-n) = q(x) and then put v = y^(1-n). Differentiating w.r.t. ‘x’, we obtain v’ = (1-n) y^(-n) y’. Therefore, the equation becomes dv/dx + (1-n) p(x) v = (1-n) q(x). This is a linear equation satisfied by v. Once it is solved, you will obtain the function y = v^(1/(1-n)). If n > 1, then we add the solution y = 0 to the solutions found by the above technique.

Summary

1. Identify the equation as Bernoulli equation. Find n. If n ≠ 0, 1, divide by y^n and substitute v = y^(1-n).
2. Through easy differentiation, find the new equation dv/dx + (1-n) p(x) v = (1-n) q(x).
3. This is a linear equation. Solve the linear equation to find v.
4. Go back to the old function y through the substitution y = v^(1/(1-n)).
5. If n > 1, then include y = 0 in the solution.
6. If you have an IVP, use the initial condition to find the particular solution.

Example 1: Solve the equation dy/dx = y + y^3

1. The given differential can be written as dy/dx - y = y^3, which is a Bernoulli equation with p(x) = -1, q(x) = 1, n = 3. Dividing with y^3, we get y^(-3) dy/dx - y^(-2) = 1. Therefore we substitute v = y^(1-3) = y^(-2).
2. Differentiating w.r.t. ‘x’ we have y^(-3) dy/dx = -(1/2)(dv/dx). So that the equation reduces to dv/dx + 2v = -2.
3. This is a linear equation. To solve this we find the integrating factor u(x) = e^(∫2dx) = e^(2x). The solution of the linear equation is given by v = [∫ u(x)q(x) dx + c] / u(x). Since ∫ e^(2x)(-2)dx = -e^(2x), the solution for v is given by v = (-e^(2x) + c) / e^(2x) = ce^(-2x) - 1.
4. To go back to y we substitute v = y^(-2). Therefore the general solution of the given DE is y^(-2) = ce^(-2x) - 1 or y = ± (ce^(-2x) - 1)^(-1/2).
5. Since n > 1, we include y = 0 in the solutions. Hence, all solutions are y = 0, y = ± (ce^(-2x) - 1)^(-1/2).

🔑 Definition — Bernoulli Equation: A first-order differential equation of the form dy/dx + p(x)y = q(x)y^n, where n is any real number. 📐 Formula: v = y^(1-n) → Substitution to convert a Bernoulli equation into a linear equation in v. 📌 Example: dy/dx - y = y^3. Here n=3, so v = y^(1-3) = y^(-2). The resulting linear equation is dv/dx + 2v = -2, with integrating factor u(x)=e^(2x). The final general solution is y^(-2) = ce^(-2x) - 1.

Example 2: Solve dy/dx + (1/x) y = xy^2 In the given equation we identify p(x) = 1/x, q(x) = x, and n = 2. Thus the substitution w = y^(-1) gives dw/dx - (1/x) w = -x. The integrating factor for this linear equation is u(x) = e^(∫ -1/x dx) = x^(-1). Multiplying gives x^(-1) w = ∫ -x(x^(-1)) dx = ∫ -1 dx = -x + c, or w = -x^2 + cx. Since w = y^(-1), we obtain y = 1 / (cx - x^2). For n > 0 the trivial solution y = 0 is a singular solution of the given equation.

Example 3: Solve dy/dx + (x/(1-x^2)) y = x y^(1/2) Dividing by y^(1/2), the given equation becomes y^(-1/2) dy/dx + (x/(1-x^2)) y^(1/2) = x. Put y^(1/2) = v. Then (1/2) y^(-1/2) dy/dx = dv/dx, so y^(-1/2) dy/dx = 2 dv/dx. The equation reduces to 2 dv/dx + (x/(1-x^2)) v = x, or dv/dx + (x/(2(1-x^2))) v = x/2. This is linear in v. The integrating factor is e^(∫ x/(2(1-x^2)) dx) = (1-x^2)^(-1/4). Multiplying by the I.F. gives d/dx [v(1-x^2)^(-1/4)] = (x/2)(1-x^2)^(-1/4). Integrating, we get v(1-x^2)^(-1/4) = -(1/3)(1-x^2)^(3/4) + c. So v = c(1-x^2)^(1/4) - (1/3)(1-x^2). Since v = y^(1/2), the required solution is y^(1/2) = c(1-x^2)^(1/4) - (1/3)(1-x^2).

8.2 Exercise

Solve the following differential equations:

1. x dy/dx + y = y^2 ln x
2. dy/dx + y = xy^3
3. dy/dx - y = e^x y^2
4. dy/dx = y (xy^3 - 1)
5. x dy/dx - (1+x)y = xy^2
6. x^2 dy/dx + y^2 = xy Solve the initial-value problems:
7. x^2 dy/dx - 2xy = 3y^4, y(1) = 1/2
8. dy/dx = y/x - y^2/x^2, y(-1) = 1
9. 2y^(1/2) dy/dx + y^(3/2) = 1, y(0) = 4
10. y(1+2xy) dx + x(1-2xy) dy = 0

8.3 Substitutions

Sometimes a differential equation can be transformed by means of a substitution into a form that could then be solved by one of the standard methods i.e. Methods used to solve separable, homogeneous, exact, linear, and Bernoulli’s differential equation. An equation may look different from any of those that we have studied in the previous lectures, but through a sensible change of variables perhaps an apparently difficult problem may be readily solved. Although no firm rules can be given on the basis of which these substitution could be selected, a working axiom might be: Try something! It sometimes pays to be clever.

Example 1 The differential equation 2x^2 dy/dx + y^2 = xy is not separable, not homogeneous, not exact, not linear, and not Bernoulli. However, we might try the substitution u = 2xy or y = u/(2x). The equation becomes, after we simplify, 2u^2 dx + (1-u)x du = 0. Separating and integrating gives 2 ln x - u^(-1) - ln u = c. Re-substituting u = 2xy, we obtain ln x = c + 1/(2y) - ln(2xy), which simplifies to x = 2c1 y e^(1/(2xy)), where e^c was replaced by c1. The differential equation also possesses the trivial solution y=0, but this function is not included in the one-parameter family of solution.

💡 Why this matters: When faced with a non-standard differential equation, a creative substitution can transform it into a form we know how to solve, such as a separable equation.

Example 2: Solve x dy/dx + 2y = 3x - 6 The presence of the term 2y dy/dx prompts us to try u = y^2. Since du/dx = 2y dy/dx, the equation becomes x(du/dx)/2 + 2u = 3x - 6, or x du/dx + 4u = 6x - 12, then dividing by x: du/dx + (4/x) u = 6 - 12/x. This equation has the form of a 1st order linear differential equation with P(x) = 4/x and Q(x) = 6 - 12/x. The integrating factor is I.F. = e^(∫ 4/x dx) = x^4. Multiplying gives d/dx [x^4 u] = 6x^4 - 12x^3. Integrating both sides, we obtain x^4 u = (6/5)x^5 - 3x^4 + c, or u = (6/5)x - 3 + c/x^4. Re-substituting u = y^2, the general solution is y^2 = (6/5)x - 3 + c/x^4.

Example 3: Solve x - y = e^(y/x) x^3 dy/dx If we let u = y/x or y = ux, then dy/dx = u + x du/dx. The given differential equation becomes x - ux = e^u x^3 (u + x du/dx), which simplifies to ue^(-u) du = dx. Integrating both sides, we have ∫ u e^(-u) du = ∫ dx. Using integration by parts, we get -ue^(-u) - e^(-u) = x + c, or u + 1 = (c1 - x) e^u, where c1 = -c. We then re-substitute u = y/x and simplify to obtain y + x = x(c1 - x) e^(y/x).

Example 4: Solve dy/dx = 2x (dy/dx)^2 If we let u = dy/dx, then du/dx = d^2y/dx^2. The equation reduces to du/dx = 2x u^2, which is separable. Separating the variables, we obtain du / u^2 = 2x dx. Integrating both sides yields -1/u = x^2 + c2, or u = -1/(x^2 + c2). Since u = dy/dx, we integrate again: ∫ dy = -∫ dx/(x^2 + c2). This gives y + c = -(1/√c2) tan^(-1)(x/√c2).

8.4 Exercise

Solve the differential equations by using an appropriate substitution.

1. y dx + (1 + y e^x) dy = 0
2. (2 + e^(-x/y)) dx + 2(1 - x/y) dy = 0
3. 2x csc(2y) dy/dx = 2x - ln(tan y)
4. dy/dx + 1 = sin x e^(-(x+y))
5. y dy/dx + 2x ln x = x e^y
6. x^2 dy/dx + 2xy = x^4 y^2 + 1
7. x e^y dy/dx - 2e^y dx = x^2

9 Solved Problems

Example 1: y’ = (x^2 + y^2) / (xy) This is a homogeneous equation. Put y = wx, then dy/dx = w + x dw/dx. Substituting, we get w + x dw/dx = (x^2 + w^2 x^2) / (xwx) = (1 + w^2)/w. This simplifies to x dw/dx = 1/w, or w dw = dx/x. Integrating gives w^2/2 = ln|x| + c, or (y/x)^2/2 = ln|x| + c, so y^2 = 2x^2 ln|cx|.

Example 2: dy/dx = (2xy - y) / x This is a homogeneous equation. Put y = wx, then dy/dx = w + x dw/dx. Substituting, we get w + x dw/dx = (2x(wx) - wx)/x = 2w - w = w. This simplifies to x dw/dx = 0, so dw/dx = 0. Integrating gives w = c, so y = cx. Let's re-check: w + x dw/dx = (2x(wx) - wx)/x = (2wx^2 - wx)/x = 2wx - w = w(2x - 1). This is incorrect in the text. The correct simplification: (2√(xy) - y)/x. Let u = √(xy). Actually, the text's solution is: w + x dw/dx = (2√(x(wx)) - wx)/x = (2x√w - wx)/x = 2√w - w. So x dw/dx = 2√w - 2w. Separating: dw/(2√w - 2w) = dx/x. This simplifies to dw/(2√w(1 - √w)) = dx/x. Let t = √w, then dt = dw/(2√w), so dt/(1-t) = dx/x. Integrating: -ln|1-t| = ln|x| + ln|c|, so 1/(1-t) = cx. Therefore t = 1 - 1/(cx), and since t = √w = √(y/x), we get √(y/x) = 1 - 1/(cx).

Example 3: (2y^2 x - 3) dx + (2yx^2 + 4) dy = 0 Check for exactness: M = 2xy^2 - 3, N = 2x^2 y + 4. ∂M/∂y = 4xy, ∂N/∂x = 4xy. The equation is exact. Find f(x,y): ∫ M dx = ∫ (2xy^2 - 3) dx = x^2 y^2 - 3x + h(y). Then ∂f/∂y = 2x^2 y + h'(y) = 2x^2 y + 4 = N, so h'(y) = 4, and h(y) = 4y + c. The solution is x^2 y^2 - 3x + 4y = C.

Example 4: dy/dx = (2xy e^(x/y)^2) / (y^2 + y^2 e^(x/y)^2 + 2x^2 e^(x/y)^2) This requires the substitution x/y = w, which simplifies to a separable equation. After substitution and integration, the solution is y = c(1 + e^(x/y)^2).

Example 5: x dy/dx + y/(ln x) = 3x^2/(ln x) Rewrite as dy/dx + (1/(x ln x)) y = 3x/ln x. This is a linear equation. The integrating factor is I.F. = e^(∫ 1/(x ln x) dx) = e^ln|ln x| = ln x. Multiplying, we get d/dx (y ln x) = 3x. Integrating gives y ln x = 3x^2/2 + c, so y = (3x^2/2 + c) / ln x.

Example 6: (y^2 e^x + 2xy) dx - x^2 dy = 0 Check for exactness: M = y^2 e^x + 2xy, N = -x^2. ∂M/∂y = 2y e^x + 2x, ∂N/∂x = -2x. The equation is not exact. Divide by y^2: (e^x + 2x/y) dx - (x^2/y^2) dy = 0. Now ∂M/∂y = -2x/y^2, ∂N/∂x = -2x/y^2. The equation is exact. Integrate: f(x,y) = ∫ (e^x + 2x/y) dx = e^x + x^2/y + h(y). ∂f/∂y = -x^2/y^2 + h'(y) = -x^2/y^2, so h'(y) = 0, h(y)=c. The solution is e^x + x^2/y = C.

⭐ Key Takeaways

A student must master the identification of Bernoulli equations (dy/dx + p(x)y = q(x)y^n) and the correct substitution method (v = y^(1-n)) to convert them into solvable linear equations, remembering to include the trivial solution y=0 when n > 1. The power of creative substitutions is a critical tool for tackling non-standard differential equations that don't fit the standard forms, but no general rule exists beyond trial and error. Recognizing exact equations and making them exact by multiplying by an appropriate factor is a valuable technique. Finally, practicing the complete solution process—from identification and substitution to solving the transformed equation and back-substituting—is essential, including using initial conditions for particular solutions in Initial Value Problems.

🧠 Quick Revision Questions

1. What is the general form of a Bernoulli differential equation, and what substitution converts it into a linear differential equation?
2. A differential equation is reduced to dv/dx + (2/x) v = x. What is its integrating factor, and what is the general solution for v?
3. In Example 1 of Section 8.3, why is the substitution u = 2xy suggested, and what type of equation does it transform the original equation into?
4. For the initial value problem x^2 dy/dx - 2xy = 3y^4 with y(1) = 1/2, what is the first step and the integrating factor for the resulting linear equation in v?
5. Describe the key difference in the solution process for a Bernoulli equation when n > 1 compared to when n = 0 or n = 1.

📘 Lecture 10 — Differential Equations (MTH401)

📖 Overview: This lecture focuses on solving first-order differential equations using substitution methods to reduce them to standard forms like linear, Bernoulli, or separable equations. It demonstrates how strategic substitutions can simplify complex differential equations into solvable forms, a critical skill for tackling non-standard ODEs.

🗂️ Topics Covered

The lecture presents a series of solved examples (Examples 7 through 17) illustrating various substitution techniques. These include reducing equations to linear form using integrating factors, applying substitutions like ( e^{2y} = u ), (\ln y = u), (\ln \tan y = u), and (x+y=u) to simplify and solve differential equations, along with examples involving Bernoulli's equation and specific function substitutions like (x^3y^3 = u).

📝 Lecture Summary

Example 7: Solve ( x \cos x \frac{dy}{dx} + y(x \sin x + \cos x) = 1 )

The equation is first divided by (x \cos x) to put it in standard linear form: ( \frac{dy}{dx} + y \left[ \frac{x \sin x + \cos x}{x \cos x} \right] = \frac{1}{x \cos x} ). This simplifies to ( \frac{dy}{dx} + y \left[ \tan x + \frac{1}{x} \right] = \frac{1}{x \cos x} ). The integrating factor (I.F.) is calculated as ( \text{I.F.} = \exp\left(\int (\tan x + \frac{1}{x}) dx\right) = x \sec x ). Multiplying through gives ( \frac{d}{dx} [xy \sec x] = \sec^2 x ). Integrating yields the solution: ( xy \sec x = \tan x + c ).

🔑 Definition — Integrating Factor (I.F.): A function used to multiply a non-exact differential equation to make it exact and integrable. 📐 Formula: ( \mu(x) = e^{\int P(x) dx} ) for a linear ODE ( \frac{dy}{dx} + P(x)y = Q(x) ), which simplifies to ( \frac{d}{dx}[\mu y] = \mu Q(x) ). 📌 Example: For the simplified ODE, (P(x) = \tan x + 1/x) and (Q(x) = 1/(x \cos x)). The I.F. is (e^{\int (\tan x + 1/x) dx} = e^{\ln|\sec x| + \ln|x|} = x \sec x). The left side becomes the derivative of (xy\sec x), leading to the final solution.

Example 8: Solve ( x e^{2y} \frac{dy}{dx} + e^{2y} = \frac{\ln x}{x} )

Use the substitution ( e^{2y} = u ). Differentiating gives ( 2e^{2y} \frac{dy}{dx} = \frac{du}{dx} ), so ( e^{2y} \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx} ). Substituting transforms the equation into ( x \left(\frac{1}{2} \frac{du}{dx}\right) + u = \frac{\ln x}{x} ), or ( \frac{du}{dx} + \frac{2}{x} u = \frac{2 \ln x}{x^2} ). This is linear in (u). The I.F. is ( e^{\int (2/x) dx} = x^2 ). Multiplying gives ( \frac{d}{dx}(x^2 u) = 2 \ln x ). Integrating results in ( x^2 u = 2[x \ln x - x] + c ). Substituting back ( u = e^{2y} ) gives the implicit solution: ( x^2 e^{2y} = 2[x \ln x - x] + c ).

Example 9: Solve ( \frac{dy}{dx} + y \ln y = y e^x )

Divide both sides by (y) (assuming (y \neq 0)) to get ( \frac{1}{y} \frac{dy}{dx} + \ln y = e^x ). Use the substitution ( \ln y = u ). Then ( \frac{1}{y} \frac{dy}{dx} = \frac{du}{dx} ). The equation becomes ( \frac{du}{dx} + u = e^x ). This is a linear equation in (u). The I.F. is ( e^{\int 1 dx} = e^x ). Multiplying yields ( \frac{d}{dx}(e^x u) = e^{2x} ). Integrating gives ( e^x u = \frac{e^{2x}}{2} + c ). Substituting back ( u = \ln y ) yields the solution: ( e^x \ln y = \frac{e^{2x}}{2} + c ).

Example 10: Solve ( 2x \csc 2y \frac{dy}{dx} = 2x - \ln \tan y )

Use the substitution ( \ln \tan y = u ). Differentiating gives ( \frac{dy}{dx} = \sin y \cos y \frac{du}{dx} ). The original equation becomes ( 2x \csc 2y (\sin y \cos y \frac{du}{dx}) = 2x - u ). Since ( \csc 2y = 1/\sin 2y ) and ( \sin 2y = 2 \sin y \cos y ), this simplifies to ( x \frac{du}{dx} = 2x - u ). Rearranging gives ( \frac{du}{dx} + \frac{1}{x} u = 2 ). The I.F. is ( e^{\int (1/x) dx} = x ). Multiplying gives ( \frac{d}{dx}(xu) = 2x ). Integrating yields ( xu = x^2 + c ), so ( u = x + c x^{-1} ). Substituting back: ( \ln \tan y = x + c x^{-1} ).

Example 11: Solve ( \frac{dy}{dx} + x + y + 1 = (x + y)^2 e^{3x} )

Use the substitution ( x+y = u ). Then ( dy/dx = du/dx - 1 ). The equation becomes ( (du/dx - 1) + u + 1 = u^2 e^{3x} ), which simplifies to ( \frac{du}{dx} + u = u^2 e^{3x} ). This is Bernoulli's equation. Rewrite as ( u^{-2} \frac{du}{dx} + u^{-1} = e^{3x} ). Let ( w = u^{-1} ). Then ( dw/dx = -u^{-2} du/dx ), so ( -dw/dx + w = e^{3x} ) or ( \frac{dw}{dx} - w = -e^{3x} ). This is linear in (w). The I.F. is ( e^{\int -1 dx} = e^{-x} ). Multiplying: ( d/dx (e^{-x} w) = -e^{2x} ). Integrating: ( e^{-x} w = -e^{2x}/2 + c ), so ( w = -e^{3x}/2 + c e^{x} ). Since ( w = 1/u ), ( 1/u = -e^{3x}/2 + c e^{x} ). Substituting ( u = x+y ) gives ( \frac{1}{x+y} = -\frac{e^{3x}}{2} + c e^{x} ).

🔑 Definition — Bernoulli's Equation: A differential equation of the form ( \frac{dy}{dx} + P(x)y = Q(x) y^n ). 📐 Formula: Solve by dividing by (y^n) and substituting ( w = y^{1-n} ), which transforms it into a linear equation in (w). 📌 Example: Here, ( \frac{du}{dx} + u = u^2 e^{3x} ) is Bernoulli with (n=2). Dividing by (u^2) and substituting (w = u^{-1}) yields the linear ODE ( \frac{dw}{dx} - w = -e^{3x} ).

Example 12: Solve ( \frac{dy}{dx} = (4x + y + 1)^2 )

Use the substitution ( 4x + y + 1 = u ). Then ( dy/dx = du/dx - 4 ). The equation becomes ( du/dx - 4 = u^2 ), or ( du/dx = u^2 + 4 ). This is separable. Separate: ( du/(u^2 + 4) = dx ). Integrate: ( \frac{1}{2} \tan^{-1}(u/2) = x + c ). Multiplying by 2: ( \tan^{-1}(u/2) = 2x + c_1 ). So ( u = 2 \tan(2x + c_1) ). Substituting back: ( 4x + y + 1 = 2 \tan(2x + c_1) ).

Example 13: Solve ( (x + y)^2 \frac{dy}{dx} = a^2 )

Use the substitution ( x+y = u ). Then ( dy/dx = du/dx - 1 ). The equation becomes ( u^2 (du/dx - 1) = a^2 ), or ( u^2 du/dx = a^2 + u^2 ). Rearranging: ( du = (a^2 + u^2)/u^2 dx ), or ( dx = u^2/(a^2 + u^2) du ). Simplify the integrand: ( u^2/(a^2 + u^2) = 1 - a^2/(a^2 + u^2) ). Integrating: ( x + c = \int (1 - a^2/(a^2 + u^2)) du = u - a \tan^{-1}(u/a) ). Substituting back ( u = x+y ): ( x + c = (x+y) - a \tan^{-1}\left(\frac{x+y}{a}\right) ). Rearranged: ( y - a \tan^{-1}\left(\frac{x+y}{a}\right) = c ). (Note: in the text, the final answer is given as ( (x+y) - a \tan^{-1}((x+y)/a) = x + c ), which simplifies to this form.)

Example 14: Solve ( 2y \frac{dy}{dx} + x^2 + y^2 + x = 0 )

Use the substitution ( x^2 + y^2 = u ). Differentiating: ( 2x + 2y \frac{dy}{dx} = \frac{du}{dx} ), or ( 2y \frac{dy}{dx} = \frac{du}{dx} - 2x ). Substituting into the equation: ( (\frac{du}{dx} - 2x) + x^2 + y^2 + x = 0 ). But ( x^2 + y^2 = u ), so ( \frac{du}{dx} - 2x + u + x = 0 ), which simplifies to ( \frac{du}{dx} + u = x ). This is linear in (u). The I.F. is ( e^{\int 1 dx} = e^x ). Multiplying: ( d/dx(e^x u) = x e^x ). Integrating by parts: ( e^x u = x e^x - e^x + c ). Substituting back ( u = x^2 + y^2 ): ( e^x (x^2 + y^2) = x e^x - e^x + c ), or ( x^2 + y^2 = x - 1 + c e^{-x} ).

Example 15: Solve ( y' + 1 = e^{-(x+y)} \sin x )

Use the substitution ( x + y = u ). Then ( y' = du/dx - 1 ). Substituting: ( (du/dx - 1) + 1 = e^{-u} \sin x ), which simplifies to ( du/dx = e^{-u} \sin x ). This is separable. Rearranging: ( e^u du = \sin x dx ). Integrate: ( e^u = -\cos x + c ). Taking the natural log: ( u = \ln |c - \cos x| ). Substituting back ( u = x + y ): ( x + y = \ln |c - \cos x| ).

Example 16: Solve ( x^4 y^2 y' + x^3 y^3 = 2x^3 - 3 )

Use the substitution ( x^3 y^3 = u ). Differentiating: ( 3x^2 y^3 + 3x^3 y^2 y' = \frac{du}{dx} ). Rearranging: ( 3x^3 y^2 y' = \frac{du}{dx} - 3x^2 y^3 ), so ( x^4 y^2 y' = \frac{x}{3} \frac{du}{dx} - x^3 y^3 ). But ( x^3 y^3 = u ), so the left side of the original equation becomes ( \frac{x}{3} \frac{du}{dx} - u ). Substituting into the original ODE: ( \frac{x}{3} \frac{du}{dx} - u + u = 2x^3 - 3 ), which simplifies to ( \frac{x}{3} \frac{du}{dx} = 2x^3 - 3 ). Rearranging: ( \frac{du}{dx} = 6x^2 - \frac{9}{x} ). Integrating: ( u = 2x^3 - 9 \ln x + c ). Substituting back ( u = x^3 y^3 ): ( x^3 y^3 = 2x^3 - 9 \ln x + c ).

Example 17: Solve ( \cos(x + y) dy = dx )

Use the substitution ( x + y = v ). Then ( 1 + dy/dx = dv/dx ), or ( dy/dx = dv/dx - 1 ). The original equation can be written as ( \cos(v) dy = dx ), or ( dy = dx / \cos(v) ). Since ( dy = dv - dx ), we get ( dv - dx = dx / \cos(v) ). Rearranging: ( dv = dx (1 + 1/\cos(v)) = dx ((\cos(v) + 1)/\cos(v)) ). Thus, ( dx = \frac{\cos(v)}{1 + \cos(v)} dv ). Simplify the integrand: ( \frac{\cos v}{1 + \cos v} = 1 - \frac{1}{1 + \cos v} ). Using the half-angle identity, ( 1 + \cos v = 2 \cos^2(v/2) ), so ( 1/(1+\cos v) = \frac{1}{2} \sec^2(v/2) ). Thus, ( dx = [1 - \frac{1}{2} \sec^2(v/2)] dv ). Integrating: ( x + c = v - \tan(v/2) ). Substituting back ( v = x+y ): ( x + c = (x + y) - \tan\left(\frac{x+y}{2}\right) ). Simplifying: ( y = \tan\left(\frac{x+y}{2}\right) + c ). 💡 Why this matters: This example shows how trigonometric substitutions and identities can be combined to solve an implicit ODE.

⭐ Key Takeaways

The most critical skills from this lecture are recognizing when a first-order ODE can be simplified by a clever substitution, such as (x+y), (e^{2y}), (\ln y), or (x^2 + y^2), to transform it into a linear, Bernoulli, or separable equation. Mastery involves identifying the type of substitution based on the structure of the equation (e.g., terms like (x+y) or (\ln y) or (e^{2y}) hint at their use) and then carefully applying the chain rule to rewrite the derivative. For exam success, students must be fluent in computing integrating factors for linear ODEs and solving Bernoulli equations with the standard (y^{1-n}) substitution, and they must practice these techniques to avoid algebraic errors.

🧠 Quick Revision Questions

1. What substitution would you use to solve (\frac{dy}{dx} + y \ln y = y e^x) and why?
2. How does the substitution (x+y=u) transform the term (\frac{dy}{dx}) in an ODE?
3. What is the standard method for solving a Bernoulli equation of the form (\frac{dy}{dx} + P(x)y = Q(x) y^n)?
4. For Example 8, what is the exact substitution used and how does it help eliminate terms involving (e^{2y})?
5. In Example 12, after substituting (4x+y+1=u), the equation becomes separable. Write down the separated form (f(u) du = dx).

📘 Lecture 10 — Applications of First Order Differential Equations

📖 Overview: This lecture introduces how differential equations model physical phenomena, focusing on finding differential equations from families of curves and the concept of orthogonal trajectories. Understanding orthogonal trajectories is crucial for applications in physics, electromagnetism, and fluid dynamics where curves intersect at right angles.

🗂️ Topics Covered

The lecture covers the basic steps of mathematical modeling for physical phenomena, finding differential equations from families of curves by eliminating parameters, defining orthogonal curves and orthogonal trajectories, and presenting a systematic method for finding orthogonal trajectories of a given family of curves through differential equations.

📝 Lecture Summary

Step 1: State Assumptions

We clearly state the assumptions on which the model will be based. These assumptions should describe the relationships among the quantities to be studied.

Step 2: Identify Parameters and Variables

Completely describe the parameters and variables to be used in the model.

Step 3: Derive Equations

Use the assumptions (from Step 1) to derive mathematical equations relating the parameters and variables (from Step 2).

The mathematical models for physical phenomenon often lead to a differential equation or a set of differential equations. The applications of the differential equations discussed in this lecture include:

• Orthogonal Trajectories
• Population dynamics
• Radioactive decay
• Newton’s Law of cooling
• Carbon dating
• Chemical reactions

Finding Differential Equations from Families of Curves

We know that solutions of a 1st order differential equation, e.g. separable equations, may be given by an implicit equation F(x, y, C) = 0 with 1 parameter C, representing a family of curves. Similarly, an nth order DE yields an n-parameter family of curves.

The question arises whether we can turn the problem around: Starting with an n-parameter family of curves, can we find an associated nth order differential equation free of parameters? The answer in most cases is yes.

Procedure for finding DE from a 1-parameter family:

1. Differentiate with respect to x to get an equation involving x, y, dy/dx, and C.
2. Use the original equation to eliminate the parameter C from the new equation.
3. Use algebra to rewrite this equation in explicit form dy/dx = f(x, y).

Example: Find the differential equation satisfied by the family x² + y² = Cx

1. Differentiate with respect to x: 2x + 2y dy/dx = C
2. From the original equation: C = (x² + y²)/x
3. Substitute: 2x + 2y dy/dx = (x² + y²)/x

This gives: dy/dx = (y² - x²)/(2xy)

10.1 Orthogonal Trajectories

Two families of curves example:

• First family: y = mx (all straight lines through origin)
• Second family: x² + y² = C (all circles centered at origin)

When these two families are drawn together, whenever one line intersects one circle, the tangent line to the circle and the line are perpendicular, i.e., orthogonal to each other. The two families of curves are orthogonal at the point of intersection.

10.2 Orthogonal Curves

Any two curves C₁ and C₂ are said to be orthogonal if their tangent lines T₁ and T₂ at their point of intersection are perpendicular. This means that slopes are negative reciprocals of each other, except when T₁ and T₂ are parallel to the coordinate axes.

10.3 Orthogonal Trajectories (OT)

When all curves of a family Γ₁: G(x, y, c₁) = 0 orthogonally intersect all curves of another family Γ₂: H(x, y, c₂) = 0, then each curve of the families is said to be an orthogonal trajectory of the other.

Orthogonal trajectories occur naturally in many areas of physics, fluid dynamics, and the study of electricity and magnetism. For example, the lines of force are perpendicular to the equipotential curves (curves of constant potential).

10.3.1 Method of Finding Orthogonal Trajectories

Consider a family of curves Γ. Assume an associated DE is found: dy/dx = f(x, y)

Since dy/dx gives the slope of the tangent to a curve of the family Γ through (x,y), the slope of the line that is tangent to the orthogonal curve through (x,y) is given by -1/f(x,y). The family of orthogonal curves are solutions to the differential equation: dy/dx = -1/f(x,y)

Summary — Steps for finding Orthogonal Trajectories:

• Step 1: Consider a family of curves Γ and find the associated differential equation.
• Step 2: Rewrite this differential equation in the explicit form dy/dx = f(x,y).
• Step 3: Write down the differential equation associated to the orthogonal family: dy/dx = -1/f(x,y).
• Step 4: Solve the new equation. The solutions are exactly the family of orthogonal curves.
• Step 5: If a specific curve from the orthogonal family is required, solve an IVP.

Example 1

Find the Orthogonal Trajectory to the family of circles x² + y² = C²

Solution: The given equation represents a family of concentric circles centered at the origin.

Step 1: Differentiate w.r.t. x to find the DE: 2x + 2y dy/dx = 0 ⟹ dy/dx = -x/y

Step 2: Rewrite in explicit form: dy/dx = -x/y

Step 3: DE for orthogonal family: dy/dx = y/x

Step 4: This is a separable DE. Using integrating factor for linear equation: u(x) = e^∫(-1/x)dx = 1/x

Solution: y · u(x) = m ⟹ y · (1/x) = m ⟹ y = mx

This represents a family of straight lines through the origin. Hence the family of straight lines y = mx and the family of circles x² + y² = C² are Orthogonal Trajectories.

Example 2

Find the Orthogonal Trajectory to the family of circles x² + y² = 2Cx

Solution:

Step 1: Differentiate: 2x + 2y dy/dx = 2C From original: C = (x² + y²)/(2x)

Substitute: 2x + 2y dy/dx = 2[(x² + y²)/(2x)] = (x² + y²)/x

Step 2: Explicit DE: dy/dx = (y² - x²)/(2xy)

Step 3: DE for orthogonal family: dy/dx = -2xy/(y² - x²)

[The solution continues but is not completed in the provided text.]

⭐ Key Takeaways

The key concept is that orthogonal trajectories are families of curves that intersect at right angles, and their slopes are negative reciprocals at intersection points. To find orthogonal trajectories, first determine the differential equation of the given family, then replace dy/dx with -1/(dy/dx) and solve the resulting equation. This method applies to concentric circles giving straight lines through the origin, and more complex families require similar parameter elimination steps. The process involves differentiation, parameter elimination, forming the orthogonal DE, and solving it.

🧠 Quick Revision Questions

1. What is the condition for two curves to be orthogonal at their point of intersection?
2. What are the four steps to find orthogonal trajectories of a family of curves?
3. What is the orthogonal trajectory family for concentric circles centered at the origin?
4. How do you eliminate the parameter C when finding a DE from a family of curves?
5. Give two physical applications where orthogonal trajectories occur naturally.

📘 Lecture 12 — Exact Lecture Title: Differential Equations (MTH401) — Lecture 12

📖 Overview: This lecture continues the detailed solution of a homogeneous differential equation from the previous lecture, then introduces real-world applications of first-order differential equations. It focuses on population dynamics (exponential growth), radioactive decay, and Newton's Law of Cooling, models that are foundational in biology, physics, and engineering for predicting change over time.

🗂️ Topics Covered

The lecture begins by completing the solution of a homogeneous differential equation from the previous lecture, using a substitution ( y = vx ) to reduce it to a separable form. The remainder of the lecture is dedicated to applying first-order differential equations to three key areas: population dynamics with exponential growth, radioactive decay using the concept of half-life, and Newton's Law of Cooling.

📝 Lecture Summary

4. [Continuation from previous page: Solving the Homogeneous DE]

This DE is a homogeneous equation. To solve it, we substitute ( y = vx ), or equivalently ( v = y/x ). Then we have ( \frac{dy}{dx} = x \frac{dv}{dx} + v ). Therefore the homogeneous differential equation from step 3 becomes ( x \frac{dv}{dx} + v = \frac{2v}{1 - v^2} ). Algebraic manipulations reduce this equation to the separable form: ( \frac{v + v^3}{1 - v^2} dv = \frac{1}{x} dx ).

The constant solutions are given by ( v + v^3 = 0 \Rightarrow v(1 + v^2) = 0 ). The only constant solution is ( v = 0 ). To find the non-constant solutions we separate the variables: ( \int \frac{v + v^3}{1 - v^2} dv = \int \frac{1}{x} dx ). Resolving into partial fractions the integrand on LHS, we obtain: ( \frac{v + v^3}{1 - v^2} = \frac{1}{v} - \frac{2v}{1 + v^2} ). Hence we have: ( \int \left( \frac{1}{v} - \frac{2v}{1 + v^2} \right) dv = \int \frac{1}{x} dx ). Hence the solution of the separable equation becomes: ( \ln |v| - \ln [v^2 + 1] = \ln |x| + \ln C ), where ( C \neq 0 ). This is equivalent to ( \frac{v}{v^2 + 1} = C x ). We go back to ( y ) to get: ( \frac{y/x}{(y/x)^2 + 1} = C x \Rightarrow \frac{y}{x} = C x \left( \frac{y^2}{x^2} + 1 \right) ), and which is equivalent to: The final set of solutions is: ( v = 0 \Rightarrow y = 0 ) (x-axis) and ( \frac{v}{v^2 + 1} = C x \Rightarrow \frac{y/x}{(y/x)^2 + 1} = Cx \Rightarrow x^2 + y^2 = m y ), where ( m = 1/C ). This is a family of circles centered on the y-axis.

10.4 Population Dynamics

Some natural questions related to population problems are the following: What will the population of a certain country be after e.g. ten years? How are we protecting the resources from extinction? The easiest population dynamics model is the exponential model. This model is based on the assumption: The rate of change of the population is proportional to the existing population.

If ( P(t) ) measures the population of a species at any time ( t ), then because of the above-mentioned assumption we can write ( \frac{dP}{dt} = kP ). The rate ( k ) is a constant of proportionality. Clearly, the above equation is linear as well as separable. To solve this equation, we multiply the equation with the integrating factor ( e^{-kt} ) to obtain ( \frac{d}{dt} [P e^{-kt}] = 0 ). Integrating both sides we obtain ( P e^{-kt} = C ) or ( P = C e^{kt} ). Clearly, we must have ( k > 0 ) for growth and ( k < 0 ) for the decay. If ( P_0 ) is the initial population then ( P(0) = P_0 ). So that ( C = P_0 ) and we obtain ( P(t) = P_0 e^{kt} ). 💡 Why this matters: This model, while simple, forms the basis for understanding unrestrained growth (like bacteria in a petri dish) and decay (like drug concentration in the body).

🔑 Definition — Exponential Model: A population model where the rate of change of population is directly proportional to the current population. 📐 Formula: ( P(t) = P_0 e^{kt} ) → The population at time ( t ) equals the initial population multiplied by Euler's number raised to the product of the growth rate and time. 📌 Example: A community's population has doubled in 5 years. Find the growth rate constant ( k ). Solution: ( 2P_0 = P_0 e^{5k} \Rightarrow e^{5k} = 2 \Rightarrow 5k = \ln 2 = 0.69315 \Rightarrow k = 0.69315/5 = 0.13863 ).

🔑 Definition — Half-life: The time it takes for one-half of the atoms in an initial amount ( A_0 ) of a radioactive substance to disintegrate or transmute into atoms of another element. It measures the stability of a radioactive substance; the longer the half-life, the more stable it is. 📐 Formula: ( kT = -\ln 2 ) or ( k = -\frac{\ln 2}{T} ) → The decay constant ( k ) is the negative natural log of 2 divided by the half-life ( T ). The initial amount is given by ( A_0 = 30e^{-30k} ).

📌 Example 1 (Radioactive Decay): A radioactive isotope has a half-life of 16 days. We have 30 g at the end of 30 days. How much radioisotope was initially present?

1. Model: ( \frac{dA}{dt} = kA, A(30) = 30 ). Solution: ( A(t) = A_0 e^{kt} ).
2. Find ( k ) using half-life: ( T=16 ). Formula: ( k = -\frac{\ln 2}{T} = -\frac{\ln 2}{16} ).
3. Use condition ( A(30)=30 ): ( 30 = A_0 e^{30k} = A_0 e^{30 (-\ln 2/16)} ).
4. Solve for initial amount: ( A_0 = 30 e^{-30k} = 30 e^{\frac{30 \ln 2}{16}} \approx 110.04 ) g.

📌 Example 2 (Finding Half-life): A breeder reactor converts uranium into plutonium 239. After 15 years, 0.043% of the initial amount ( A_0 ) of the plutonium has disintegrated. Find the half-life.

1. Model: ( \frac{dA}{dt} = kA, A(0) = A_0, A(15) = (0.99957)A_0 ).
2. Solution: ( A(t) = A_0 e^{kt} ).
3. Find ( k ) using ( A(15) ): ( A_0 e^{15k} = (0.99957)A_0 \Rightarrow e^{15k} = 0.99957 ).
4. Solve for ( k ): ( 15k = \ln(0.99957) \Rightarrow k = \frac{\ln(0.99957)}{15} \approx -0.00002867 ).
5. Find half-life ( T ) using formula: ( T = \frac{\ln 2}{-\ln 2/k} ), so ( T = -\frac{\ln 2}{k} = -\frac{\ln 2}{-0.00002867} \approx 24,180 ) years.

11.1 Newton's Law of Cooling

From experimental observations, it is known that the temperature ( T(t) ) of an object changes at a rate proportional to the difference between the temperature in the body and the temperature ( T_m ) of the surrounding environment. This is what is known as Newton's law of cooling.

The differential equation is ( \frac{dT}{dt} = k(T - T_m) ), where ( k ) is a constant of proportionality. If the initial temperature of the cooling body is ( T_0 ), we obtain the initial value problem ( \frac{dT}{dt} = k(T - T_m), T(0) = T_0 ). The problem is linear as well as separable. Separating the variables and integrating, we obtain: ( \int \frac{dT}{T - T_m} = \int k dt \Rightarrow \ln |T - T_m| = kt + C_1 ). This means that ( T - T_m = e^{kt + C_1} \Rightarrow T(t) = T_m + C e^{kt} ), where ( C = e^{C_1} ). Now applying the initial condition ( T(0) = T_0 ), we see that ( C = T_0 - T_m ). Thus the solution of the initial value problem is given by: ( T(t) = T_m + (T_0 - T_m) e^{kt} ). Hence, if temperatures at times ( t_1 ) and ( t_2 ) are known, then we have: ( T(t_1) - T_m = (T_0 - T_m) e^{kt_1} ) and ( T(t_2) - T_m = (T_0 - T_m) e^{kt_2} ). 💡 Why this matters: This law governs how a cup of coffee cools or how a body's temperature changes in a forensic investigation.

🔑 Definition — Newton's Law of Cooling: The rate of change of an object's temperature is proportional to the difference between its own temperature and the temperature of the surrounding environment. 📐 Formula: ( T(t) = T_m + (T_0 - T_m) e^{kt} ) → The temperature at time ( t ) equals the ambient temperature plus the initial temperature difference multiplied by an exponential decay factor.

⭐ Key Takeaways

A student must remember that the homogeneous differential equation from the previous lecture yields a family of circles and the x-axis as its solution. The exponential model ( P(t) = P_0 e^{kt} ) is the core of population dynamics and radioactive decay, which are both governed by the same differential equation, ( dA/dt = kA ). The sign of ( k ) determines growth (positive) or decay (negative). The half-life ( T ) of a substance is linked to its decay constant by the relation ( kT = -\ln 2 ). Finally, Newton's Law of Cooling, while also an exponential function, has a different formula, ( T(t) = T_m + (T_0 - T_m)e^{kt} ), where the rate of change depends on the temperature difference, not the absolute temperature.

🧠 Quick Revision Questions

1. What is the general solution form for both exponential population growth and radioactive decay?
2. A population doubles in 10 years. What is the exact value of the growth constant ( k )?
3. What is the mathematical relationship between the half-life ( T ) of a radioactive substance and its decay constant ( k )?
4. In Newton's Law of Cooling, what does the constant ( T_m ) physically represent?
5. If the object's initial temperature is lower than ( T_m ), will the constant of proportionality ( k ) be positive or negative in the cooling law equation?

📘 Lecture 13 — Higher Order Linear Differential Equations

📖 Overview: This lecture introduces higher-order linear differential equations, beginning with their general form and classification as homogeneous or non-homogeneous. It then formally defines the initial-value problem for these equations, laying the theoretical groundwork for solving them in subsequent lectures.

🗂️ Topics Covered

The lecture covers the preliminary theory of higher-order linear differential equations, including their general form with variable and constant coefficients, the distinction between homogeneous and non-homogeneous equations, and the formal definition and structure of the initial-value problem for an nth-order equation, with a specific look at the second-order case.

📝 Lecture Summary

13.1 Preliminary theory

A linear differential equation of order n can have variable coefficients and is written in the form: (a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \dots + a_1(x) y' + a_0(x) y = g(x)), where (a_n(x) \neq 0). However, this course will first focus on equations with constant coefficients, where (a_0, a_1, …, a_n) are real constants. When (g(x) \neq 0), the equation is called a non-homogeneous differential equation. If (g(x) = 0), the equation becomes (a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y' + a_0 y = 0), which is known as the associated homogeneous differential equation.

13.2 Initial-Value Problem

For a linear nth-order differential equation, an initial-value problem (IVP) is the problem of solving the differential equation subject to initial conditions at a single point (x_0). These conditions specify the value of the function and its first (n-1) derivatives at (x_0): (y(x_0) = y_0), (y'(x_0) = y'_0), …, (y^{(n-1)}(x_0) = y^{(n-1)}_0), where (y_0, y'_0, …, y^{(n-1)}_0) are arbitrary constants. For the specific case of (n=2), the second-order initial-value problem is to solve: (a_2(x) y'' + a_1(x) y' + a_0(x) y = g(x)), subject to: (y(x_0) = y_0), (y'(x_0) = y'_0).

💡 Why this matters: This formal definition of the IVP is essential because it provides the structure for finding unique solutions to differential equations that model physical systems, where we know the system's state and its rate of change at a starting moment.

⭐ Key Takeaways

A student must remember that a linear nth-order differential equation has a specific form with variable coefficients, but the focus in this course is on equations with constant coefficients. The critical distinction between a non-homogeneous equation and its associated homogeneous equation is fundamental. The initial-value problem (IVP) is a core concept that involves solving a differential equation subject to initial conditions, which specify the function and its first n-1 derivatives at a single point. For an nth-order equation, n initial conditions are required. The specific form of the second-order IVP should be memorized as a common case.

🧠 Quick Revision Questions

1. What is the general form of an nth-order linear differential equation with variable coefficients?
2. What distinguishes a non-homogeneous linear differential equation from its associated homogeneous equation?
3. What are the key components of an initial-value problem for a linear nth-order differential equation?
4. How many initial conditions are required to define an initial-value problem for a second-order differential equation?
5. In the notation for an initial-value problem, what is the variable (x_0) and what do the symbols (y_0) and (y'_0) represent?

📘 Lecture 14 — Boundary-Value Problems and Linear Dependence

📖 Overview: This lecture introduces boundary-value problems (BVPs) as distinct from initial-value problems (IVPs), discussing their unique existence and uniqueness characteristics. It also establishes the fundamental concepts of linear dependence and independence of functions, along with the Wronskian as a tool for determining independence.

🗂️ Topics Covered

The lecture covers the existence and uniqueness of solutions for initial-value problems, the formulation and possible outcomes of boundary-value problems, the definition and examples of linear dependence and independence of functions, the concept of the Wronskian determinant, and the theorem for using it as a criterion for linear independence.

📝 Lecture Summary

13.2.1 Solution of IVP

A function satisfying a differential equation on an interval I whose graph passes through (x₀, y₀) such that the slope of the curve at that point equals y₀′ is called a solution of the initial-value problem.

13.3 Theorem (Existence and Uniqueness of Solutions)

Let aₙ(x), aₙ₋₁(x), …, a₁(x), a₀(x) and g(x) be continuous on an interval I with aₙ(x) ≠ 0 for all x in I. If x₀ is in I, then a solution y(x) of the initial-value problem exists on I and is unique.

🔑 Definition — Initial-Value Problem (IVP): A differential equation together with specified values of the function and its derivatives at a single point x₀.

📌 Example 1: The function y = 3e²ˣ + e⁻²ˣ – 3x solves the IVP y″ – 4y = 12x, y(0)=4, y′(0)=1. Because the equation is linear with constant coefficients, and a₂(x)=1≠0 for all x, this solution is unique.

📌 Example 2: The IVP 3y‴ + 5y″ – y′ + 7y = 0, y(1)=0, y′(1)=0, y″(1)=0 has only the trivial solution y=0, which is unique.

💡 Why this matters: If the leading coefficient aₙ(x)=0 at some point in I, the solution of an IVP may not be unique or may not even exist.

📌 Example 4: The function y = cx² + x + 3 satisfies the IVP x²y″ – 2xy′ + 2y = 6, y(0)=3, y′(0)=1 for any choice of c. The leading coefficient a₂(x)=x²=0 at x=0, causing non-uniqueness.

13.4 Boundary-value problem (BVP)

For a 2ⁿᵈ-order linear differential equation, the problem:

• Solve: a₂(x) d²y/dx² + a₁(x) dy/dx + a₀(x) y = g(x)
• Subject to: y(a)=y₀, y(b)=y₁ is called a boundary-value problem (BVP). The specified values y(a)=y₀ and y(b)=y₁ are called boundary conditions.

13.4.1 Solution of BVP

A solution of the BVP is a function satisfying the differential equation on an interval I containing a and b, whose graph passes through the two points (a, y₀) and (b, y₁).

📌 Example 5: The function y = 3x² – 6x + 3 satisfies the BVP x²y″ – 2xy′ + 2y = 6, y(1)=0, y(2)=3, as verified by substitution.

13.4.2 Possible Boundary Conditions

For a 2ⁿᵈ-order linear differential equation, the possible pairs of boundary conditions include: y(a) = y₀, y(b) = y₁; y′(a) = y₀′, y(b) = y₁; y(a) = y₀, y′(b) = y₁′; y′(a) = y₀′, y′(b) = y₁′. These are special cases of the general form: α₁y(a) + β₁y′(a) = γ₁, α₂y(b) + β₂y′(b) = γ₂, where α₁, α₂, β₁, β₂ ∈ {0,1}.

💡 Why this matters: A BVP can have several solutions, a unique solution, or no solution at all.

📌 Example 1: The BVP y″ + 16y = 0, y(0)=0, y(π/2)=0 has the one-parameter family of solutions y = c₂ sin 4x, meaning infinite solutions. The condition y(π/2)=0 is satisfied for any c₂ because sin(2π)=0.

📌 Example 2: The BVP y″ + 16y = 0, y(0)=0, y(π/8)=0 has only the trivial solution y = 0, because c₁ = 0 from y(0)=0 and c₂ = 0 from y(π/8)=0.

📌 Example 3: The BVP y″ + 16y = 0, y(0)=0, y(π/2)=1 has no solution. Applying y(π/2)=1 to y = c₂ sin 4x yields 1 = c₂·0, a contradiction.

13.5 Linear Dependence

A set of functions {f₁(x), f₂(x), …, fₙ(x)} is linearly dependent on an interval I if there exist constants c₁, c₂, …, cₙ, not all zero, such that c₁ f₁(x) + c₂ f₂(x) + … + cₙ fₙ(x) = 0 for all x in I.

13.6 Linear Independence

A set of functions {f₁(x), f₂(x), …, fₙ(x)} is linearly independent on an interval I if the equation c₁ f₁(x) + c₂ f₂(x) + … + cₙ fₙ(x) = 0 for all x in I implies c₁ = c₂ = … = cₙ = 0.

13.6.1 Case of two functions

Two functions f₁(x) and f₂(x) are linearly dependent on an interval I if and only if one is a constant multiple of the other. They are linearly independent when neither is a constant multiple of the other. In general, a set of n functions is linearly dependent if at least one can be expressed as a linear combination of the rest.

📌 Example 1: f₁(x) = sin 2x and f₂(x) = sin x cos x are linearly dependent because sin 2x = 2 sin x cos x, so (1/2) sin 2x – sin x cos x = 0 with non-zero coefficients (c₁=1/2, c₂=-1).

📌 Example 2: f₁(x) = cos²x, f₂(x) = sin²x, f₃(x) = sec²x, f₄(x) = tan²x are linearly dependent because sec²x = cos²x + sin²x + tan²x, so the combination with c₁=1, c₂=1, c₃=-1, c₄=1 gives zero.

📌 Example 3: f₁(x) = 1 + x, f₂(x) = x, f₃(x) = x² are linearly independent because the equation c₁(1+x) + c₂x + c₃x² = 0 forces c₁=0, c₂=0, c₃=0 when coefficients of like powers are equated.

13.7 Wronskian

If functions f₁, f₂, …, fₙ possess at least n–1 derivatives, the Wronskian, denoted W(f₁, f₂, …, fₙ), is the determinant:

| f₁      f₂      …  fₙ      |
| f₁′     f₂′     …  fₙ′     |
|  ⁝       ⁝      ⋱   ⁝      |
| f₁⁽ⁿ⁻¹⁾ f₂⁽ⁿ⁻¹⁾ … fₙ⁽ⁿ⁻¹⁾ |

🔑 Definition — Wronskian: A determinant used to test for linear independence of a set of functions.

📐 Formula: For two functions f₁ and f₂, W = f₁ f₂′ – f₂ f₁′.

13.8 Theorem (Criterion for Linearly Independent Functions)

If functions f₁, f₂, …, fₙ possess at least n–1 derivatives on an interval I and W(f₁, f₂, …, fₙ) ≠ 0 for at least one point in I, then the functions are linearly independent on I. If the functions are linearly dependent on I, then W = 0 for all x in I (the converse is not necessarily true).

📌 Example 1: f₁(x) = sin²x and f₂(x) = 1 – cos 2x are linearly dependent, and their Wronskian W = 0 for all x. This confirms the vanishing Wronskian for dependent functions.

📌 Example 2: f₁(x) = e^(m₁x) and f₂(x) = e^(m₂x) with m₁ ≠ m₂ are linearly independent (neither is a constant multiple of the other). Their Wronskian W = (m₂ – m₁)e^((m₁+m₂)x) ≠ 0.

📌 Example 3: The functions y₁ = e^(αx) cos βx and y₂ = e^(αx) sin βx are linearly independent for β ≠ 0 on any interval, with W = βe^(2αx) ≠ 0.

📌 Example 4: The functions f₁(x) = eˣ, f₂(x) = xeˣ, f₃(x) = x²eˣ are linearly independent on any interval, with W = 2e^(3x) ≠ 0.

⭐ Key Takeaways

You must understand the fundamental difference between initial-value problems (which have a unique solution when the leading coefficient never vanishes) and boundary-value problems (which may have zero, one, or infinitely many solutions). The concept of linear dependence is that one function can be written as a combination of others, while independence requires that only the trivial combination yields zero. The Wronskian provides a practical computational tool: if it is non-zero at any point, the functions are linearly independent. Remember, a zero Wronskian does not guarantee dependence—it is only a sufficient condition for independence.

🧠 Quick Revision Questions

1. State the conditions for a unique solution to an initial-value problem.
2. What is a boundary-value problem, and how does it differ from an initial-value problem?
3. What are the three possible outcomes when solving a boundary-value problem?
4. Define linear dependence and linear independence for a set of functions.
5. If the Wronskian of two functions is non-zero for all x, what can you conclude?

📘 Lecture 14 — Solutions of Higher Order Linear Equations

📖 Overview: This lecture introduces the theory for solving higher-order linear differential equations, beginning with homogeneous equations. It covers the superposition principle, linear independence, the Wronskian, fundamental sets of solutions, and the structure of general solutions for both homogeneous and non-homogeneous equations.

🗂️ Topics Covered

This lecture covers the preliminary theory for solving nth-order linear differential equations, starting with the homogeneous case. Key concepts include the superposition principle for solutions, the Wronskian's role in determining linear independence, and the definition of a fundamental set of solutions. It also explains the general solution structure for homogeneous equations and introduces the concepts of particular solutions and complementary functions for non-homogeneous equations.

📝 Lecture Summary

14.1 Preliminary Theory

To solve an nth-order non-homogeneous linear differential equation, we first solve the associated homogeneous differential equation. The focus is on the theory and methods for solving homogeneous linear equations. A function that satisfies the homogeneous equation is called a solution.

14.2 Superposition Principle

If ( y_1, y_2, ..., y_n ) are solutions on an interval ( I ) of the homogeneous linear differential equation, then the linear combination ( y = c_1 y_1(x) + c_2 y_2(x) + ... + c_n y_n(x) ), where ( c_1, c_2, ..., c_n ) are arbitrary constants, is also a solution. A constant multiple of a solution is also a solution. Homogeneous linear equations always have the trivial solution ( y = 0 ). This principle does not hold for non-linear equations.

💡 Why this matters: The Superposition Principle is the foundation for constructing general solutions from a set of individual solutions.

🔑 Definition — Superposition Principle: If ( y_1, y_2, ..., y_n ) are solutions to a homogeneous linear DE, then any linear combination of them is also a solution.

📌 Example 1: The functions ( y = e^x, y = e^{2x}, y = e^{3x} ) all satisfy ( y''' - 6y'' + 11y' - 6y = 0 ). Substituting the linear combination ( y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} ) into the DE, the result is 0, confirming it is also a solution.

📌 Example 2: The function ( y = x^2 ) is a solution of ( x^2 y'' - 3xy' + 4y = 0 ) on ( (0, \infty) ). Substituting ( y = cx^2 ) (a scalar multiple) also satisfies the equation, confirming it is a solution.

The Wronskian

For two solutions ( y_1, y_2 ) of a second-order homogeneous linear DE on interval ( I ), the Wronskian ( W(y_1, y_2) ) is either identically zero or never zero on ( I ). This is derived by showing the Wronskian satisfies the linear first-order DE ( \frac{dW}{dx} + PW = 0 ), leading to ( W = ce^{-\int P dx} ). If ( c \neq 0 ), ( W \neq 0 ); if ( c = 0 ), ( W = 0 ). This property extends to n solutions of an nth-order homogeneous linear DE.

🔑 Definition — Wronskian (for two functions): ( W(y_1, y_2) = y_1 y_2' - y_1' y_2 )

📐 Formula: ( W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} ) → The value of this 2x2 determinant determines if the functions are linearly dependent or independent.

14.3 Linear Independence of Solutions

A set of n solutions ( y_1, y_2, ..., y_n ) of a homogeneous linear nth-order DE is linearly independent on interval ( I ) if and only if ( W(y_1, y_2, ..., y_n) \neq 0 ). Conversely, the solutions are linearly dependent if and only if ( W(y_1, y_2, ..., y_n) = 0 ) for all ( x \in I ).

14.4 Fundamental Set of Solutions

A set ( { y_1, y_2, ..., y_n } ) of n linearly independent solutions of a homogeneous linear nth-order DE on interval ( I ) constitutes a fundamental set of solutions.

14.4.1 Existence of a Fundamental Set

A fundamental set of solutions always exists for a linear nth-order homogeneous differential equation on any interval ( I ).

14.5 General Solution-Homogeneous Equations

If ( { y_1, y_2, ..., y_n } ) is a fundamental set of solutions on interval ( I ) for a homogeneous linear nth-order DE, then the general solution on ( I ) is defined as ( y = c_1 y_1(x) + c_2 y_2(x) + ... + c_n y_n(x) ), where ( c_1, c_2, ..., c_n ) are arbitrary constants.

📌 Example 1: For ( y'' - 9y = 0 ), the functions ( y_1 = e^{3x} ) and ( y_2 = e^{-3x} ) are solutions. Since ( W(e^{3x}, e^{-3x}) = -6 \neq 0 ), they form a fundamental set. The general solution is ( y = c_1 e^{3x} + c_2 e^{-3x} ).

📌 Example 2: The function ( y = 4 \sinh 3x - 5e^{-3x} ) satisfies ( y'' - 9y = 0 ) and is a particular solution. Choosing ( c_1 = 2, c_2 = -7 ) from the general solution gives ( y = 2e^{3x} - 7e^{-3x} ), which is equal to ( 4 \sinh 3x - 5e^{-3x} ). This demonstrates that a particular solution can be obtained from the general solution.

📌 Example 3: For ( y''' - 6y'' + 11y' - 6y = 0 ), the functions ( y_1 = e^x, y_2 = e^{2x}, y_3 = e^{3x} ) are all solutions. The Wronskian ( W(e^x, e^{2x}, e^{3x}) = 2e^{6x} \neq 0 ), so they form a fundamental set. The general solution is ( y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} ).

14.6 Non-Homogeneous Equations

A function ( y_p ) that satisfies the non-homogeneous differential equation and contains no arbitrary parameters is called a particular solution.

📌 Example 1: ( y_p = 3 ) is a particular solution of ( y'' + 9y = 27 ).

📌 Example 2: ( y_p = x^3 - x ) is a particular solution of ( x^2 y'' + 2xy' - 8y = 4x^3 + 6x ).

14.7 Complementary Function

The general solution of the associated homogeneous equation, ( y_c = c_1 y_1 + c_2 y_2 + ... + c_n y_n ), is called the complementary function for the non-homogeneous equation.

⭐ Key Takeaways

The Superposition Principle is crucial for building general solutions of homogeneous linear DEs from individual solutions. The Wronskian is the definitive test for linear independence of solutions: a non-zero Wronskian indicates a fundamental set, which is necessary to form the general solution. The general solution of a homogeneous linear DE is a linear combination of a fundamental set of solutions. For non-homogeneous equations, the general solution includes the complementary function (solution to the homogeneous part) plus any particular solution.

🧠 Quick Revision Questions

1. State the Superposition Principle for homogeneous linear differential equations.
2. What is the Wronskian used for in the context of solutions to differential equations?
3. What condition must a set of solutions satisfy to be considered a fundamental set?
4. What is the relationship between the general solution of a homogeneous equation and the general solution of the corresponding non-homogeneous equation?
5. Why does the Superposition Principle not apply to non-linear differential equations?

📘 Lecture 14 — General Solution of Non-Homogeneous Equations

📖 Overview: This lecture explains how to construct the general solution of non-homogeneous linear differential equations by combining the complementary function and a particular solution. It also introduces the superposition principle for non-homogeneous equations, which allows us to break down complicated forcing functions into simpler components. Understanding this structure is essential for solving real-world problems where external forces or inputs are present.

🗂️ Topics Covered

The lecture begins by defining the general solution of non-homogeneous equations as the sum of the complementary function and a particular solution, illustrated with a worked example. It then presents the superposition principle, which states that the sum of particular solutions for individual forcing terms is a particular solution for the sum of those terms. Finally, the lecture provides a set of exercises for verifying fundamental solution sets and general solutions.

📝 Lecture Summary

14.8 General Solution of Non-Homogeneous Equations

Suppose that:

• The particular solution of the non-homogeneous equation aₙ(x) dⁿy/dxⁿ + aₙ₋₁(x) dⁿ⁻¹y/dxⁿ⁻¹ + … + a₁(x) dy/dx + a₀(x) y = g(x) is yₚ.
• The complementary function of the associated homogeneous equation aₙ(x) dⁿy/dxⁿ + aₙ₋₁(x) dⁿ⁻¹y/dxⁿ⁻¹ + … + a₁(x) dy/dx + a₀(x) y = 0 is y_c = c₁y₁ + c₂y₂ + … + cₙyₙ.

Then the general solution of the non-homogeneous equation on the interval I is given by: y = y_c + yₚ or y = c₁y₁(x) + c₂y₂(x) + … + cₙyₙ(x) + yₚ(x) = y_c(x) + yₚ(x)

Hence, General Solution = Complementary solution + any particular solution.

🔑 Definition — Complementary Function: The general solution of the associated homogeneous differential equation (where the right-hand side is zero).

📌 Example: Suppose yₚ = -11/12 - (1/2)x. Then yₚ' = -1/2, yₚ'' = 0. Substituting into d³y/dx³ - 6 d²y/dx² + 11 dy/dx - 6y gives 0 - 0 - 11/2 + 11/2 + 3x = 3x. Hence yₚ = -11/12 - (1/2)x is a particular solution of d³y/dx³ - 6 d²y/dx² + 11 dy/dx - 6y = 3x.

Now consider y_c = c₁eˣ + c₂e²ˣ + c₃e³ˣ. Computing y_c', y_c'', y_c''' and substituting into the homogeneous equation d³y/dx³ - 6 d²y/dx² + 11 dy/dx - 6y = 0 yields 0, confirming y_c is the complementary function. Therefore, the general solution of the non-homogeneous equation is: y = y_c + yₚ = c₁eˣ + c₂e²ˣ + c₃e³ˣ - 11/12 - (1/2)x.

14.9 Superposition Principle for Non-homogeneous Equations

Suppose that yₚ₁, yₚ₂, …, yₚₖ denote the particular solutions of the k differential equations aₙ(x)y⁽ⁿ⁾ + aₙ₋₁(x)y⁽ⁿ⁻¹⁾ + … + a₁(x)y' + a₀(x)y = gᵢ(x), for i = 1, 2, …, k, on an interval I. Then yₚ = yₚ₁(x) + yₚ₂(x) + … + yₚₖ(x) is a particular solution of aₙ(x)y⁽ⁿ⁾ + aₙ₋₁(x)y⁽ⁿ⁻¹⁾ + … + a₁(x)y' + a₀(x)y = g₁(x) + g₂(x) + … + gₖ(x).

🔑 Definition — Superposition Principle for Non-homogeneous Equations: The sum of particular solutions corresponding to individual forcing functions is a particular solution of the equation with the sum of those forcing functions.

📌 Example: Consider y'' - 3y' + 4y = -16x² + 24x - 8 + 2e²ˣ + 2xeˣ - eˣ. Suppose yₚ₁ = -4x², yₚ₂ = e²ˣ, yₚ₃ = xeˣ. Then yₚ₁'' - 3yₚ₁' + 4yₚ₁ = -8 + 24x - 16x², so yₚ₁ is a particular solution of y'' - 3y' + 4y = -16x² + 24x - 8. Similarly, yₚ₂ = e²ˣ is a particular solution of y'' - 3y' + 4y = 2e²ˣ, and yₚ₃ = xeˣ is a particular solution of y'' - 3y' + 4y = 2xeˣ - eˣ. Hence, by the superposition principle, yₚ = yₚ₁ + yₚ₂ + yₚ₃ = -4x² + e²ˣ + xeˣ is a particular solution of the original equation.

14.10 Exercise

The lecture provides exercises for verifying that given functions form a fundamental set of solutions on an indicated interval and for forming the general solution. Examples include:

• y'' - y' - 12y = 0; e⁻³ˣ, e⁴ˣ, (-∞, ∞)
• y'' - 2y' + 5y = 0; eˣ cos 2x, eˣ sin 2x, (-∞, ∞)
• x²y'' + xy' + y = 0; cos(ln x), sin(ln x), (0, ∞)
• 4y'' - 4y' + y = 0; eˣ/², xeˣ/², (-∞, ∞)
• x²y'' - 6xy' + 12y = 0; x³, x⁴, (0, ∞)
• y'' - 4y = 0; cosh 2x, sinh 2x, (-∞, ∞)

Additionally, exercises ask to verify that given two-parameter families are the general solution of non-homogeneous differential equations on indicated intervals.

⭐ Key Takeaways

The central concept of this lecture is that the general solution of any non-homogeneous linear differential equation is the sum of the complementary function (the general solution of the associated homogeneous equation) and any particular solution of the non-homogeneous equation. The superposition principle is a powerful tool that allows us to find a particular solution for a complicated forcing function by breaking it into simpler parts, finding a particular solution for each part, and summing them. For exam preparation, you must be able to identify the complementary function from the homogeneous equation, verify if a given function is a particular solution, apply the superposition principle, and write the general solution. The exercises reinforce the method of verifying fundamental solution sets and general solutions for both homogeneous and non-homogeneous equations.

🧠 Quick Revision Questions

1. What is the general solution of a non-homogeneous linear differential equation, and what two components make it up?
2. State the superposition principle for non-homogeneous differential equations in your own words.
3. If yₚ₁ is a particular solution for g₁(x) and yₚ₂ is a particular solution for g₂(x), what is a particular solution for g₁(x) + g₂(x)?
4. Given a non-homogeneous equation, how do you find its complementary function?
5. In the example, why is y_c = c₁eˣ + c₂e²ˣ + c₃e³ˣ the complementary function for the equation d³y/dx³ - 6 d²y/dx² + 11 dy/dx - 6y = 3x?

📘 Lecture 17 — Method of Undetermined Coefficients (Superposition Approach)

📖 Overview: This lecture introduces a method for finding particular solutions of non-homogeneous linear differential equations with constant coefficients. The method of undetermined coefficients provides a systematic way to guess the form of a particular solution based on the input function ( g(x) ), and its importance lies in being a straightforward technique for equations where ( g(x) ) has a specific form.

🗂️ Topics Covered

This lecture begins by recalling the structure of non-homogeneous linear differential equations and the need for both a complementary function and a particular integral. It then discusses the limitations and specific forms of the input function ( g(x) ) that make the method of undetermined coefficients applicable. The core of the lecture details the step-by-step solution process, emphasizing how to assume the form of the particular solution ( y_p ) and the critical caution about avoiding duplication with terms in the complementary function ( y_c ).

📝 Lecture Summary

17 Method of Undetermined Coefficients (Superposition Approach)

This section introduces the context for solving non-homogeneous linear differential equations. It recalls that the general solution is the sum of the complementary function ( y_c ) (the general solution of the associated homogeneous equation) and any particular solution ( y_p ) of the non-homogeneous equation. The lecture focuses on non-homogeneous equations with constant coefficients. The three main methods for finding a particular integral are listed: the method of undetermined coefficients (superposition approach), the annihilator operator approach, and the method of variation of parameters.

💡 Why this matters: Understanding that the general solution is ( y = y_c + y_p ) is fundamental to solving all non-homogeneous linear differential equations.

The Method of Undetermined Coefficient

The method is limited to non-homogeneous linear differential equations with constant coefficients.

17.1 The form of Input function ( g(x) )

The input function ( g(x) ) must be of a specific form for the method to be applicable. These forms include:

• A constant function ( k )
• A polynomial function
• An exponential function ( e^{\alpha x} )
• The trigonometric functions ( \sin(\beta x) ) or ( \cos(\beta x) )
• Finite sums and products of these functions.

If ( g(x) ) does not have one of these forms, the method cannot be applied.

17.2 Solution Steps

The solution process involves a systematic series of steps:

• Step 1: Determine the form of the input function ( g(x) ).
• Step 2: Assume the general form of ( y_p ) according to the form of ( g(x) ).
• Step 3: Substitute the assumed ( y_p ) and its derivatives into the given non-homogeneous differential equation.
• Step 4: Simplify and equate coefficients of like terms from both sides of the equation.
• Step 5: Solve the resulting system of equations to find the unknown coefficients.
• Step 6: Substitute the calculated values of the coefficients back into the assumed form for ( y_p ).

17.2.1 Restriction on Input function ( g )

The restriction on the form of ( g(x) ) is necessary because the derivatives of sums and products of polynomials, exponentials, and trigonometric functions are again sums and products of similar kinds of functions. Therefore, the expression ( a y_p'' + b y_p' + c y_p ) must be able to identically equal the input function ( g(x) ).

💡 Why this matters: This restriction ensures that the assumed form for ( y_p ) is an educated guess that can be made to match ( g(x) ).

Caution! A critical point to remember is that the educated guess for ( y_p ) must take into consideration the functions that make up the complementary function ( y_c ). No function in the assumed ( y_p ) must be a solution of the associated homogeneous differential equation. This means the assumed ( y_p ) should not contain terms that duplicate terms in ( y_c ).

🔑 Definition — Complementary function ( y_c ): The general solution of the associated homogeneous linear differential equation. 🔑 Definition — Particular solution ( y_p ): Any solution of the non-homogeneous linear differential equation. 🔑 Definition — Method of Undetermined Coefficients: A technique for finding a particular solution ( y_p ) of a non-homogeneous linear differential equation with constant coefficients, where the input function ( g(x) ) has a specific form (polynomial, exponential, sine/cosine, or sums/products of these). It involves assuming a general form for ( y_p ) with unknown coefficients and then solving for those coefficients.

⭐ Key Takeaways

The method of undetermined coefficients is a direct approach for finding a particular solution to non-homogeneous linear differential equations, but it is restricted to equations with constant coefficients and input functions ( g(x) ) that are constants, polynomials, exponentials, sines, cosines, or sums/products thereof. The core process involves assuming a general form for ( y_p ) that matches the form of ( g(x) ), substituting it into the differential equation, and solving for the unknown coefficients. A crucial caution is that the assumed ( y_p ) must not contain any terms that are already present in the complementary function ( y_c ), as this would lead to an incorrect guess and failure to find a valid solution. This method is a specific case of the superposition approach, where the general solution is constructed as the sum of ( y_c ) and ( y_p ).

🧠 Quick Revision Questions

1. What is the general solution of a non-homogeneous linear differential equation?
2. List the specific forms that the input function ( g(x) ) must have for the method of undetermined coefficients to be applicable.
3. What is the critical caution to consider when assuming the form of the particular solution ( y_p )?
4. Describe the six steps involved in the method of undetermined coefficients.
5. What does it mean if a term in the assumed ( y_p ) is "duplicated" in the complementary function ( y_c )?

📘 Lecture 18 — Differential Equations (MTH401)

📖 Overview: This lecture introduces the Method of Undetermined Coefficients for solving non-homogeneous linear differential equations with constant coefficients. It explains how to select trial particular solutions based on the form of the input function, including when the input is a sum of functions and when duplication occurs between the particular solution and the complementary function.

🗂️ Topics Covered

This lecture covers trial particular solutions for various input functions including polynomials, exponentials, and trigonometric functions. It discusses how to handle input functions that are sums of terms, how to address duplication between yₚ and y꜀ by multiplying by powers of x, and applies these concepts to second-order and higher-order equations with examples including initial value problems.

📝 Lecture Summary

17.3 Trial particular solutions

This section provides a table of assumed particular solutions yₚ for common input functions g(x). For any constant input, assume yₚ = A. For a linear function (5x + 7), assume yₚ = Ax + B. For a quadratic (3x² − 2), assume yₚ = Ax² + Bx + c. For a cubic (x³ − x + 1), assume yₚ = Ax³ + Bx² + Cx + D. For sin 4x or cos 4x, assume yₚ = A cos 4x + B sin 4x. For e⁵ˣ, assume yₚ = Ae⁵ˣ. For (9x − 2)e⁵ˣ, assume yₚ = (Ax + B)e⁵ˣ. For x²e⁵ˣ, assume yₚ = (Ax² + Bx + C)e⁵ˣ. For e³ˣ sin 4x, assume yₚ = A e³ˣ cos 4x + B e³ˣ sin 4x. For 5x² sin 4x, assume a more complex form involving both cosine and sine terms with quadratic coefficients.

17.4 Input function g(x) as a sum

When the input function g(x) is a sum of m terms of the kind listed in the table, the trial particular solution yₚ is the sum of the trial forms corresponding to each term. This means yₚ is a linear combination of all linearly independent functions generated by repeated differentiation of g(x).

🔑 Definition — Superposition Principle: If g(x) = g₁(x) + g₂(x) + … + gₘ(x) and yₚ₁, yₚ₂, …, yₚₘ are the trial forms, then the particular solution is yₚ = yₚ₁ + yₚ₂ + … + yₚₘ.

📐 Formula: yₚ = yₚ₁ + yₚ₂ + … + yₚₘ (for input function g(x) = g₁(x) + g₂(x) + … + gₘ(x))

📌 Example 1: Solve y″ + 4y′ − 2y = 2x² − 3x + 6 The auxiliary equation m² + 4m − 2 = 0 gives roots m = −2 ± √6, so y꜀ = c₁e^(−2−√6)x + c₂e^(−2+√6)x. For g(x) = 2x² − 3x + 6 (quadratic), assume yₚ = Ax² + Bx + C. After substitution and equating coefficients: −2A = 2, 8A − 2B = −3, 2A + 4B − 2C = 6. Solving gives A = −1, B = −5/2, C = −9, so yₚ = −x² − (5/2)x − 9. General solution: y = y꜀ + yₚ = c₁e^(−2−√6)x + c₂e^(−2+√6)x − x² − (5/2)x − 9.

📌 Example 2: Solve y″ − y′ + y = 2 sin 3x Auxiliary equation m² − m + 1 = 0 gives m = (1 ± i√3)/2, so y꜀ = e^(x/2)[c₁ cos(√3 x/2) + c₂ sin(√3 x/2)]. For g(x) = sin 3x, assume yₚ = A cos 3x + B sin 3x. Substituting: (−8A − 3B) cos 3x + (3A − 8B) sin 3x = 0·cos 3x + 2 sin 3x. Solving −8A − 3B = 0 and 3A − 8B = 2 gives A = 6/73, B = −16/73. General solution: y = e^(x/2)[c₁ cos(√3 x/2) + c₂ sin(√3 x/2)] + (6/73) cos 3x − (16/73) sin 3x.

📌 Example 3: Solve y″ − 2y′ − 3y = 4x − 5 + 6xe²ˣ Auxiliary equation m² − 2m − 3 = 0 gives m = −1, 3, so y꜀ = c₁e⁻ˣ + c₂e³ˣ. g(x) = (4x − 5) + 6xe²ˣ, so assume yₚ₁ = Ax + B for the polynomial part and yₚ₂ = (Cx + D)e²ˣ for the exponential part. After substitution and equating coefficients: −3A = 4, −2A − 3B = −5, −3C = 6, 2C − 3D = 0. Solving gives A = −4/3, B = 23/9, C = −2, D = −4/3.

17.5 Duplication between yₚ and y꜀

If a function in the assumed yₚ is also present in y꜀, then that function is a solution of the associated homogeneous equation, making the assumption incorrect. To resolve this duplication, multiply the duplicate term(s) by xⁿ, where n is the smallest positive integer that eliminates the duplication.

🔑 Definition — Duplication Rule: If yₚᵢ contains terms that duplicate terms in y꜀, multiply yₚᵢ by xⁿ where n is the least positive integer that eliminates the duplication.

💡 Why this matters: Without addressing duplication, the assumed form would lead to a contradiction (like 0 = 8eˣ), making it impossible to find the particular solution.

📌 Example 4: Find particular solution of y″ − 5y′ + 4y = 8eˣ y꜀ = c₁eˣ + c₂e⁴ˣ. Assuming yₚ = Aeˣ leads to 0 = 8eˣ (wrong). Since eˣ duplicates y꜀, multiply by x: yₚ = Axeˣ. Substituting: Axeˣ + 2Aeˣ − 5Axeˣ − 5Aeˣ + 4Axeˣ = 8eˣ, giving −3Aeˣ = 8eˣ, so A = −8/3.

📌 Example 5: Determine form of particular solution for y″ − 8y′ + 25y = 5x³e⁻ˣ − 7e⁻ˣ y꜀ = e⁴ˣ(c₁ cos 3x + c₂ sin 3x). g(x) = (5x³ − 7)e⁻ˣ, assume yₚ = (Ax³ + Bx² + Cx + D)e⁻ˣ. No duplication since y꜀ contains e⁴ˣ terms, not e⁻ˣ.

📌 Example 6: Determine form of particular solution for y″ − y′ + y = 3x² − 5 sin 2x + 7xe⁶ˣ y꜀ = e^(x/2)[c₁ cos(√3 x/2) + c₂ sin(√3 x/2)]. For g₁(x) = 3x², yₚ₁ = Ax² + Bx + C. For g₂(x) = −5 sin 2x, yₚ₂ = D cos 2x + E sin 2x. For g₃(x) = 7xe⁶ˣ, yₚ₃ = (Fx + G)e⁶ˣ. No duplication with y꜀.

📌 Example 7: Find particular solution of y″ − 2y′ + y = eˣ y꜀ = c₁eˣ + c₂xeˣ (repeated root m = 1). Assuming yₚ = Aeˣ duplicates y꜀. Multiply by x: yₚ = Axeˣ still duplicates (xeˣ is in y꜀). Multiply by x²: yₚ = Ax²eˣ has no duplication. Substituting gives yₚ = (1/2)x²eˣ.

📌 Example 8: Solve initial value problem y″ + y = 4x + 10 sin x, y(π) = 0, y′(π) = 2 y꜀ = c₁ cos x + c₂ sin x. For g₁(x) = 4x, yₚ₁ = Ax + B. For g₂(x) = 10 sin x, assuming yₚ₂ = C cos x + D sin x duplicates y꜀. Multiply by x: yₚ₂ = Cx cos x + Dx sin x. Final yₚ = Ax + B + Cx cos x + Dx sin x. Substituting: Ax + B − 2C sin x + 2Dx cos x = 4x + 10 sin x. Solving: B = 0, A = 4, −2C = 10 gives C = −5, 2D = 0 gives D = 0. So yₚ = 4x − 5x cos x. General: y = c₁ cos x + c₂ sin x + 4x − 5x cos x. Using y(π) = 0: c₁(−1) + c₂(0) + 4π − 5π(−1) = 0 → −c₁ + 4π + 5π = 0 → c₁ = 9π. Using y′(π) = 2: −9π sin π + c₂ cos π + 4 + 5π sin π − 5 cos π = 2 → 0 − c₂ + 4 − 0 + 5 = 2 → −c₂ + 9 = 2 → c₂ = 7. Solution: y = 9π cos x + 7 sin x + 4x − 5x cos x.

📌 Example 9: Solve y″ − 6y′ + 9y = 6x² + 2 − 12e³ˣ y꜀ = c₁e³ˣ + c₂xe³ˣ (repeated root m = 3). For g₁(x) = 6x² + 2, yₚ₁ = Ax² + Bx + C. For g₂(x) = −12e³ˣ, yₚ₂ = De³ˣ duplicates y꜀; multiplying by x gives Dxe³ˣ still duplicates; multiplying by x² gives Dx²e³ˣ with no duplication. Final yₚ = Ax² + Bx + C + Dx²e³ˣ. After substitution: 9Ax² + (−12A + 9B)x + (2A − 6B + 9C) + 2De³ˣ = 6x² + 2 − 12e³ˣ. Solving: 9A = 6 → A = 2/3; −12A + 9B = 0 → −8 + 9B = 0 → B = 8/9; 2A − 6B + 9C = 2 → 4/3 − 16/3 + 9C = 2 → −4 + 9C = 2 → C = 2/3; 2D = −12 → D = −6.

📌 Example 10: Solve y‴ + y″ = eˣ cos x Auxiliary equation m³ + m² = 0 → m²(m + 1) = 0 → m = 0, 0, −1, so y꜀ = c₁ + c₂x + c₃e⁻ˣ. For g(x) = eˣ cos x, assume yₚ = Aeˣ cos x + Beˣ sin x. No duplication (y꜀ has e⁻ˣ and constants, not eˣ cos x). After substitution and equating coefficients: (−2A + 4B)eˣ cos x + (−4A − 2B)eˣ sin x = eˣ cos x. Solving −2A + 4B = 1 and −4A − 2B = 0 gives A = −1/10, B = 1/5.

⭐ Key Takeaways

The method of undetermined coefficients requires matching the form of the particular solution to the input function using the provided table of trial forms. When the input function is a sum, the particular solution is the sum of individual trial solutions. Crucially, if any term in the assumed yₚ duplicates a term in y꜀, it must be multiplied by xⁿ until the duplication is eliminated, starting with n = 1 and increasing as needed. For repeated roots, higher powers of x may be required. This method applies to higher-order linear equations with constant coefficients as long as the input function consists of polynomials, exponentials, sines, cosines, or products thereof.

🧠 Quick Revision Questions

1. What trial particular solution would you assume for g(x) = 3x² − 5x + 2?
2. How do you handle the input function g(x) = 4x + 3e²ˣ + sin x?
3. What happens if your assumed yₚ contains a term that is already in y꜀?
4. Find the particular solution for y″ − 2y′ + y = eˣ if the complementary function is y꜀ = c₁eˣ + c₂xeˣ.
5. What is the form of yₚ for y‴ + y″ = eˣ cos x?

📘 Lecture 19 — Undetermined Coefficient (Annihilator Operator Approach)

📖 Overview: This lecture introduces the annihilator operator approach for finding particular solutions to non-homogeneous linear differential equations with constant coefficients. It presents differential operators as a compact notation and shows how to construct operators that "annihilate" specific functions, enabling a systematic method for determining particular integrals.

🗂️ Topics Covered

The lecture begins by reviewing non-homogeneous differential equations and the need for complementary functions and particular integrals. It then introduces differential operators and notation using D. The concept of annihilator operators is developed, showing how operators like D^n, (D-α)^n, and (D^2-2αD+(α²+β²))^n annihilate polynomial, exponential, and trigonometric functions. Examples demonstrate finding annihilators for various functions and combinations of functions.

📝 Lecture Summary

18 Undetermined Coefficient (Annihilator Operator Approach)

A non-homogeneous linear differential equation of order n has the form aₙ(dⁿy/dxⁿ) + aₙ₋₁(dⁿ⁻¹y/dxⁿ⁻¹) + ... + a₁(dy/dx) + a₀y = g(x). The associated homogeneous equation is aₙ(dⁿy/dxⁿ) + aₙ₋₁(dⁿ⁻¹y/dxⁿ⁻¹) + ... + a₁(dy/dx) + a₀y = 0. To obtain the general solution, we must find: the complementary function y_c (general solution of the associated homogeneous equation) and any particular solution y_p of the non-homogeneous equation. The general solution equals Complementary Function + Particular Integral. In this lecture, we learn to find particular integrals using the concept of differential annihilator operators.

18.1 Differential Operators

The symbol D denotes the differential operator d/dx, so dy/dx = Dy. D is a linear differential operator because D{af(x) + bg(x)} = aDf(x) + bDg(x). Higher order derivatives are expressed as D²y = d²y/dx², D³y = d³y/dx³, and Dⁿy = dⁿy/dxⁿ. A polynomial expression aₙDⁿ + aₙ₋₁Dⁿ⁻¹ + ... + a₁D + a₀ is also a linear differential operator.

🔑 Definition — Differential Operator: The symbol D represents the operation of differentiation with respect to x, so Dy = dy/dx. It is a linear operator.

📐 Formula: Dⁿ(f) = dⁿf/dxⁿ → the nth derivative of f with respect to x

📌 Example: D(e⁴ˣ) = 4e⁴ˣ, D(5x³ - 6x²) = 15x² - 12x, D(cos 2x) = -2 sin 2x

18.2 Differential Equation in Terms of D

Any linear differential equation can be expressed using D notation. For a 2nd order equation ay″ + by′ + cy = g(x), since dy/dx = Dy and d²y/dx² = D²y, we can write (aD² + bD + c)y = g(x). Defining L = aD² + bD + c, the equation becomes L(y) = g(x). L is a second-order linear differential operator with constant coefficients.

Factorization of a differential operator: An nth-order linear differential operator L = aₙDⁿ + aₙ₋₁Dⁿ⁻¹ + ... + a₁D + a₀ with constant coefficients can be factorized whenever the characteristic polynomial can be factorized. The factors of a linear differential operator with constant coefficients commute.

📐 Formula: (D² + 5D + 6) = (D + 2)(D + 3) → factorization treating D as an algebraic quantity

📌 Example: (D² + 5D + 6)y = (D + 2)(D + 3)y = (D + 3)(D + 2)y, since (D+2)(D+3)y = y″ + 5y′ + 6y and (D+3)(D+2)y = y″ + 5y′ + 6y

18.3 Annihilator Operator

Suppose L is a linear differential operator with constant coefficients and y = f(x) is a sufficiently differentiable function. If L(y) = 0, then L is called an annihilator operator of the function f.

🔑 Definition — Annihilator Operator: A linear differential operator L such that L(f) = 0 for a given function f

The differential operator Dⁿ annihilates each of the functions 1, x, x², ..., xⁿ⁻¹. The polynomial function c₀ + c₁x + ... + cₙ₋₁xⁿ⁻¹ can be annihilated by finding an operator that annihilates the highest power of x. The differential operator (D - α)ⁿ annihilates each of the functions e^(αx), xe^(αx), x²e^(αx), ..., xⁿ⁻¹e^(αx). The differential operator (D² - 2αD + (α² + β²))ⁿ is the annihilator operator of the functions e^(αx)cos(βx), xe^(αx)cos(βx), ..., xⁿ⁻¹e^(αx)cos(βx) and e^(αx)sin(βx), xe^(αx)sin(βx), ..., xⁿ⁻¹e^(αx)sin(βx).

💡 Why this matters: The annihilator approach provides a systematic algebraic method for finding particular solutions. Instead of guessing the form of y_p, we apply an annihilator to both sides of the differential equation, converting the non-homogeneous equation into a homogeneous one that can be solved using standard methods.

📌 Example 5: Since Dx = 0, D²x = 0, D³x² = 0, D⁴x³ = 0, the operators D, D², D³, D⁴ are annihilators of constant, x, x², x³ respectively.

📌 Example 6: For y = 1 - 5x² + 8x³, since D⁴x³ = 0, then D⁴ is the annihilator operator.

📌 Example 7: The differential equation (D - α)ⁿy = 0 has general solution y = c₁e^(αx) + c₂xe^(αx) + ... + cₙxⁿ⁻¹e^(αx), so (D - α)ⁿ annihilates these functions.

📌 Example 8a: For f(x) = e^(5x), (D - 5)e^(5x) = 5e⁵ˣ - 5e⁵ˣ = 0, so annihilator is L = D - 5 (α = 5, n = 1).

📌 Example 8b: For g(x) = 4e^(2x) - 6xe^(2x), (D - 2)²(4e^(2x) - 6xe^(2x)) = 0, so annihilator is L = (D - 2)² (α = 2, n = 2).

📌 Example 10: (D² + 2D + 5)e^(-x)cos(2x) = 0 and (D² + 2D + 5)e^(-x)sin(2x) = 0, so D² + 2D + 5 annihilates these functions (α = -1, β = 2, n = 1).

📌 Example 11: (D² + 1)² = D⁴ + 2D² + 1 annihilates cos x, sin x, x cos x, x sin x (α = 0, β = 1, n = 2).

📌 Example 13: For f(x) = 7 - x + 6 sin(3x), D² annihilates y₁(x) = 7 - x, and (D² + 9) annihilates y₂(x) = 6 sin(3x), so D²(D² + 9) annihilates f(x).

📌 Example 14: For f(x) = e^(-3x) + xe^x, (D + 3) annihilates e^(-3x) and (D - 1)² annihilates xe^x, so (D + 3)(D - 1)² annihilates f(x).

Important properties: If L annihilates y₁ and y₂, then L annihilates their linear combination c₁y₁ + c₂y₂. If L₁ annihilates y₁ and L₂ annihilates y₂, but L₁(y₂) ≠ 0 and L₂(y₁) ≠ 0, then the product L₁L₂ annihilates the sum y₁ + y₂. The annihilator of a function is not unique; we seek the operator of lowest possible order.

⭐ Key Takeaways

The D operator notation transforms differential equations into algebraic form, where D represents differentiation. An annihilator operator is a linear differential operator that yields zero when applied to a specific function or class of functions; for polynomials, Dⁿ is the annihilator; for exponentials e^(αx), (D-α)ⁿ is the annihilator; for sine/cosine combinations, (D² - 2αD + (α²+β²))ⁿ is the annihilator. When finding annihilators for sums of functions, we take the product of the individual annihilator operators, always seeking the lowest possible order operator. The annihilator approach provides a systematic, algebraic method for finding particular solutions to non-homogeneous differential equations by converting them into higher-order homogeneous equations.

🧠 Quick Revision Questions

1. What is the annihilator operator for the function f(x) = 5x³ - 2x + 7?
2. Find the annihilator operator of lowest order for f(x) = e^(4x) + 3xe^(4x).
3. What annihilator operator corresponds to the function f(x) = e^(-2x)cos(3x)?
4. If L₁ annihilates y₁ and L₂ annihilates y₂, under what condition does L₁L₂ annihilate y₁ + y₂?
5. Write the differential equation y″ + 4y′ - 5y = 3x² - 1 in operator notation using D.

📘 Lecture 20 — Undetermined Coefficients (Annihilator Operator Approach)

📖 Overview: This lecture presents the annihilator operator approach for solving non-homogeneous linear differential equations with constant coefficients. It is a systematic method for finding particular solutions when the input function has specific forms like polynomials, exponentials, sines, cosines, and their sums/products.

🗂️ Topics Covered

The lecture covers the method of undetermined coefficients using the annihilator operator, including: the types of input functions suitable for this method, the 8-step solution procedure (from writing the equation to forming the general solution), and several fully worked examples involving polynomial, exponential, and trigonometric forcing functions. Examples range from simple quadratic inputs to complex combinations like ( x \cos x - \cos x ) and ( 10 e^{-2x} \cos x ). The lecture concludes with an exercise set for independent practice.

📝 Lecture Summary

Undetermined Coefficients (Annihilator Operator Approach)

The method of undetermined coefficients that uses the annihilator operator approach is limited to non-homogeneous linear differential equations that have constant coefficients AND where the function ( g(x) ) has a specific form.

🔑 Definition — Input function forms: The function ( g(x) ) must be one of these types: a constant ( k ), a polynomial, an exponential ( e^{\alpha x} ), trigonometric functions ( \sin(\beta x), \cos(\beta x) ), or finite sums and products of these functions. Otherwise, the method cannot be applied.

19.1 Solution Method

Consider a non-homogeneous linear differential equation with constant coefficients of order ( n ). If ( L ) denotes the differential operator ( L = a_n D^n + a_{n-1} D^{n-1} + \cdots + a_1 D + a_0 ), then the equation is written as ( L(y) = g(x) ).

The method consists of the following 8 steps:

1. Step 1: Write the equation as ( L(y) = g(x) ).
2. Step 2: Find the complementary solution ( y_c ) by solving the associated homogeneous equation ( L(y) = 0 ).
3. Step 3: Operate on both sides with a differential operator ( L_1 ) that annihilates ( g(x) ).
4. Step 4: Find the general solution of the higher-order homogeneous equation ( L_1 L(y) = 0 ).
5. Step 5: Delete all terms from the solution in Step 4 that are duplicated in ( y_c ).
6. Step 6: Form a linear combination ( y_p ) of the remaining terms — this is the form of the particular solution.
7. Step 7: Substitute ( y_p ) into ( L(y) = g(x) ), match coefficients, and solve for the unknown constants.
8. Step 8: Form the general solution as ( y = y_c + y_p ).

💡 Why this matters: The annihilator operator provides a systematic, algebraic way to "destroy" the forcing function, turning a non-homogeneous problem into a homogeneous one of higher order, from which the form of the particular solution can be extracted.

🔑 Definition — Annihilator operator: A differential operator that, when applied to a function, yields zero. For example, ( D^3 ) annihilates ( x^2 ) because the third derivative of ( x^2 ) is zero. ( (D-3) ) annihilates ( e^{3x} ). ( (D^2+1) ) annihilates ( \sin x ) and ( \cos x ).

📐 Formula: Annihilator for ( x^n ) is ( D^{n+1} ). Annihilator for ( e^{\alpha x} ) is ( D - \alpha ). Annihilator for ( \sin(\beta x) ) or ( \cos(\beta x) ) is ( D^2 + \beta^2 ). For products like ( x^2 e^{2x} ), the annihilator is ( (D-2)^3 ).

Example 1 — Solve ( y'' + 3y' + 2y = 4x^2 )

Step 1: Write as ( (D^2 + 3D + 2)y = 4x^2 ). Step 2: Auxiliary equation: ( m^2 + 3m + 2 = (m+1)(m+2) = 0 \Rightarrow m = -1, -2 ). So ( y_c = c_1 e^{-x} + c_2 e^{-2x} ). Step 3: Since ( D^3(4x^2) = 0 ), apply ( D^3 ): ( D^3(D^2 + 3D + 2)y = 0 ). Step 4: Auxiliary equation: ( m^3(m+1)(m+2) = 0 \Rightarrow m = 0,0,0,-1,-2 ). General solution: ( y = c_1 + c_2 x + c_3 x^2 + c_4 e^{-x} + c_5 e^{-2x} ). Step 5: Remove ( c_4 e^{-x} + c_5 e^{-2x} ) (duplicated in ( y_c )). Remaining: ( c_1 + c_2 x + c_3 x^2 ). Step 6: Form ( y_p = A + Bx + Cx^2 ). Step 7: ( y_p' = B + 2Cx ), ( y_p'' = 2C ). Substitute: ( 2C + 3(B + 2Cx) + 2(A + Bx + Cx^2) = 4x^2 ). Simplify: ( 2Cx^2 + (2B + 6C)x + (2A + 3B + 2C) = 4x^2 + 0x + 0 ). Equate coefficients: ( 2C = 4 ), ( 2B + 6C = 0 ), ( 2A + 3B + 2C = 0 ). Solve: ( C = 2 ), ( B = -6 ), ( A = 7 ). So ( y_p = 7 - 6x + 2x^2 ). Step 8: General solution: ( y = c_1 e^{-x} + c_2 e^{-2x} + 7 - 6x + 2x^2 ).

📌 Example: For ( y = c_1 e^{-x} + c_2 e^{-2x} + 7 - 6x + 2x^2 ), the particular solution ( y_p = 7 - 6x + 2x^2 ) was found by assuming a quadratic form and solving the coefficient equations.

Example 2 — Solve ( y'' - 3y' = 8e^{3x} + 4\sin x )

Step 1: Write as ( (D^2 - 3D)y = 8e^{3x} + 4\sin x ). Step 2: Auxiliary equation: ( m(m-3) = 0 \Rightarrow m = 0, 3 ). So ( y_c = c_1 + c_2 e^{3x} ). Step 3: ( (D-3) ) annihilates ( 8e^{3x} ), ( (D^2+1) ) annihilates ( 4\sin x ). Apply ( (D-3)(D^2+1) ): ( (D-3)(D^2+1)(D^2-3D)y = 0 ). Step 4: Auxiliary: ( m(m-3)^2(m^2+1) = 0 \Rightarrow m = 0, 3, 3, \pm i ). General solution: ( y = c_1 + c_2 e^{3x} + c_3 x e^{3x} + c_4 \cos x + c_5 \sin x ). Step 5: Remove ( c_1 + c_2 e^{3x} ). Remaining: ( c_3 x e^{3x} + c_4 \cos x + c_5 \sin x ). Step 6: Form ( y_p = A x e^{3x} + B \cos x + C \sin x ). Step 7: Substitute into ( y'' - 3y' ). Obtain: ( 3A e^{3x} + (-B - 3C) \cos x + (3B - C) \sin x = 8e^{3x} + 4\sin x ). Equate: ( 3A = 8 ), ( -B - 3C = 0 ), ( 3B - C = 4 ). Solve: ( A = \frac{8}{3} ), ( B = \frac{6}{5} ), ( C = -\frac{2}{5} ). So ( y_p = \frac{8}{3} x e^{3x} + \frac{6}{5} \cos x - \frac{2}{5} \sin x ). Step 8: General solution: ( y = c_1 + c_2 e^{3x} + \frac{8}{3} x e^{3x} + \frac{6}{5} \cos x - \frac{2}{5} \sin x ).

Example 3 — Solve ( y'' + 8y = 5x + 2e^{-x} )

Step 1: Write as ( (D^2 + 8)y = 5x + 2e^{-x} ). Step 2: Auxiliary: ( m^2 + 8 = 0 \Rightarrow m = \pm 2\sqrt{2} i ). So ( y_c = c_1 \cos(2\sqrt{2}x) + c_2 \sin(2\sqrt{2}x) ). Step 3: ( D^2 ) annihilates ( 5x ), ( (D+1) ) annihilates ( 2e^{-x} ). Apply ( D^2(D+1) ): ( D^2(D+1)(D^2+8)y = 0 ). Step 4: Auxiliary: ( m^2(m+1)(m^2+8) = 0 \Rightarrow m = 0,0,-1,\pm 2\sqrt{2} i ). General solution: ( y = c_1 \cos(2\sqrt{2}x) + c_2 \sin(2\sqrt{2}x) + c_3 + c_4 x + c_5 e^{-x} ). Step 5: Remove ( c_1 \cos(2\sqrt{2}x) + c_2 \sin(2\sqrt{2}x) ). Remaining: ( c_3 + c_4 x + c_5 e^{-x} ). Step 6: Form ( y_p = A + Bx + C e^{-x} ). Step 7: Substitute: ( y_p'' + 8y_p = 8A + 8Bx + 9C e^{-x} = 5x + 2e^{-x} ). Equate: ( 8A = 0 ), ( 8B = 5 ), ( 9C = 2 ). Solve: ( A = 0 ), ( B = \frac{5}{8} ), ( C = \frac{2}{9} ). So ( y_p = \frac{5}{8}x + \frac{2}{9} e^{-x} ). Step 8: General solution: ( y = c_1 \cos(2\sqrt{2}x) + c_2 \sin(2\sqrt{2}x) + \frac{5}{8}x + \frac{2}{9} e^{-x} ).

Example 4 — Solve ( y'' + y = x \cos x - \cos x )

Step 1: Write as ( (D^2 + 1)y = x \cos x - \cos x ). Step 2: Auxiliary: ( m^2 + 1 = 0 \Rightarrow m = \pm i ). So ( y_c = c_1 \cos x + c_2 \sin x ). Step 3: ( (D^2+1)^2 ) annihilates ( x \cos x ) and ( \cos x ). Apply ( (D^2+1)^2 ): ( (D^2+1)^3 y = 0 ). Step 4: Auxiliary: ( (m^2+1)^3 = 0 \Rightarrow m = i, i, i, -i, -i, -i ) (each multiplicity 3). General solution: ( y = c_1 \cos x + c_2 \sin x + c_3 x \cos x + c_4 x \sin x + c_5 x^2 \cos x + c_6 x^2 \sin x ). Step 5: Remove ( c_1 \cos x + c_2 \sin x ). Remaining: ( c_3 x \cos x + c_4 x \sin x + c_5 x^2 \cos x + c_6 x^2 \sin x ). Step 6: Form ( y_p = A x \cos x + B x \sin x + C x^2 \cos x + E x^2 \sin x ). Step 7: Substitute into ( y'' + y ): ( y_p'' + y_p = 4Ex \cos x - 4Cx \sin x + (2B + 2C) \cos x + (-2A + 2E) \sin x = x \cos x - \cos x ). Equate: ( 4E = 1 ), ( -4C = 0 ), ( 2B + 2C = -1 ), ( -2A + 2E = 0 ). Solve: ( A = \frac{1}{4} ), ( B = -\frac{1}{2} ), ( C = 0 ), ( E = \frac{1}{4} ). So ( y_p = \frac{1}{4} x \cos x - \frac{1}{2} x \sin x + \frac{1}{4} x^2 \sin x ). Step 8: General solution: ( y = c_1 \cos x + c_2 \sin x + \frac{1}{4} x \cos x - \frac{1}{2} x \sin x + \frac{1}{4} x^2 \sin x ).

Example 5 — Determine the form of a particular solution for ( y'' - 2y' + y = 10e^{-2x} \cos x )

Step 1: Write as ( (D^2 - 2D + 1)y = 10e^{-2x} \cos x ). Step 2: Auxiliary: ( m^2 - 2m + 1 = (m-1)^2 = 0 \Rightarrow m = 1,1 ). So ( y_c = c_1 e^x + c_2 x e^x ). Step 3: ( (D^2 + 4D + 5) ) annihilates ( e^{-2x} \cos x ). Apply: ( (D^2 + 4D + 5)(D^2 - 2D + 1)y = 0 ). Step 4: Auxiliary: ( (m^2 + 4m + 5)(m-1)^2 = 0 \Rightarrow m = -2 \pm i, 1, 1 ). General solution: ( y = c_1 e^x + c_2 x e^x + c_3 e^{-2x} \cos x + c_4 e^{-2x} \sin x ). Step 5: Remove ( c_1 e^x + c_2 x e^x ). Remaining: ( c_3 e^{-2x} \cos x + c_4 e^{-2x} \sin x ). Step 6: Form ( y_p = A e^{-2x} \cos x + B e^{-2x} \sin x ). (Note: Steps 7 and 8 not needed — only the form was requested.)

Example 6 — Determine the form of a particular solution for ( y''' - 4y'' + 4y' = 5x^2 - 6x + 4x^2 e^{2x} + 3e^{5x} )

Step 1: Write as ( (D^3 - 4D^2 + 4D)y = 5x^2 - 6x + 4x^2 e^{2x} + 3e^{5x} ). Step 2: Auxiliary: ( m^3 - 4m^2 + 4m = m(m-2)^2 = 0 \Rightarrow m = 0, 2, 2 ). So ( y_c = c_1 + c_2 e^{2x} + c_3 x e^{2x} ). Step 3: ( D^3 ) annihilates ( 5x^2 - 6x ), ( (D-2)^3 ) annihilates ( x^2 e^{2x} ), ( (D-5) ) annihilates ( 3e^{5x} ). Apply ( D^3(D-2)^3(D-5) ): ( D^4 (D-2)^5 (D-5) y = 0 ). Step 4: Auxiliary: ( m^4 (m-2)^5 (m-5) = 0 \Rightarrow m = 0,0,0,0, 2,2,2,2,2, 5 ). General solution: ( y = c_1 + c_2 x + c_3 x^2 + c_4 x^3 + c_5 e^{2x} + c_6 x e^{2x} + c_7 x^2 e^{2x} + c_8 x^3 e^{2x} + c_9 x^4 e^{2x} + c_{10} e^{5x} ). Step 5: Remove ( c_1 + c_5 e^{2x} + c_6 x e^{2x} ). Remaining: ( c_2 x + c_3 x^2 + c_4 x^3 + c_7 x^2 e^{2x} + c_8 x^3 e^{2x} + c_9 x^4 e^{2x} + c_{10} e^{5x} ). Step 6: Form ( y_p = A x + B x^2 + C x^3 + E x^2 e^{2x} + F x^3 e^{2x} + G x^4 e^{2x} + H e^{5x} ).

⭐ Key Takeaways

The annihilator operator approach is a powerful, step-by-step method for finding particular solutions of non-homogeneous linear differential equations with constant coefficients, provided the forcing function ( g(x) ) is of a specific algebraic or trigonometric type. The core idea is to find a differential operator that "annihilates" ( g(x) ), turning the non-homogeneous equation into a higher-order homogeneous one. The general solution of that higher-order equation contains both the complementary solution and the form of the particular solution. By removing terms duplicated in ( y_c ), the correct form of ( y_p ) is obtained, and its unknown coefficients are found by substituting back into the original equation and matching terms. Mastery of this method requires knowing the annihilators for common functions and carefully handling cases where the forcing function includes terms that already appear in the homogeneous solution.

🧠 Quick Revision Questions

1. What types of functions ( g(x) ) are suitable for the annihilator operator approach?
2. What is the annihilator operator for the function ( x^3 )?
3. In Example 1, why was the term ( c_4 e^{-x} + c_5 e^{-2x} ) removed from the general solution of the higher-order equation?
4. In Example 2, which operators annihilate ( 8e^{3x} ) and ( 4\sin x ), respectively?
5. In Example 4, why is the annihilator for ( x \cos x ) equal to ( (D^2+1)^2 )?

📘 Lecture 21 — Variation of Parameters

📖 Overview: This lecture introduces the method of variation of parameters for finding particular solutions of non-homogeneous linear differential equations. Unlike the method of undetermined coefficients, variation of parameters is more general and can be applied to equations with variable coefficients and non-polynomial forcing functions. The method extends from first-order to second-order equations systematically.

🗂️ Topics Covered

The lecture begins by recalling the structure of non-homogeneous linear differential equations and the general solution format. It then derives the variation of parameters method for first-order equations, showing how it reproduces the familiar integrating factor solution. The core of the lecture develops the method for second-order equations, including the derivation of the two conditions on the unknown functions, the use of Cramer's rule with the Wronskian, and a detailed 8-step summary procedure. Three worked examples demonstrate the complete solution process, including handling repeated roots and trigonometric forcing functions.

📝 Lecture Summary

Recall

A non-homogeneous linear differential equation with constant coefficients is an equation of the form ( a_n \frac{d^n y}{dx^n} + a_{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \cdots + a_1 \frac{dy}{dx} + a_0 y = g(x) ). The general solution is given by General Solution = Complementary Function + Particular Integral. For a first-order linear differential equation ( \frac{dy}{dx} + P(x)y = f(x) ), the general solution is ( y = e^{-\int P dx} \cdot \int e^{\int P dx} f(x) dx + c_1 e^{-\int P dx} ). Here, the second term ( y_c = c_1 e^{-\int P dx} ) is the solution of the associated homogeneous equation, and the first term ( y_p = e^{-\int P dx} \cdot \int e^{\int P dx} f(x) dx ) is a particular solution.

The Variation of Parameters

20.1 First order equation

The method of variation of parameters provides an alternative derivation of the particular solution for first-order equations. Since ( y_1 = e^{-\int P dx} ) solves the homogeneous equation ( \frac{dy}{dx} + P(x)y = 0 ), the general solution is ( y = c_1 y_1(x) ). The variation of parameters consists of finding a function ( u_1(x) ) such that ( y_p = u_1(x) y_1(x) ) is a particular solution of the non-homogeneous equation.

🔑 Definition — Variation of Parameters (first-order): Replace the constant parameter ( c_1 ) in the complementary function with a variable function ( u_1(x) ) and determine ( u_1 ) by substituting into the non-homogeneous equation.

Substituting ( y_p = u_1 y_1 ) into ( \frac{dy}{dx} + P(x)y = f(x) ) gives ( u_1 \left[ \frac{dy_1}{dx} + P(x)y_1 \right] + y_1 \frac{du_1}{dx} = f(x) ). Since ( y_1 ) solves the homogeneous equation, the bracket term is zero, yielding ( y_1 \frac{du_1}{dx} = f(x) ).

📐 Formula for ( u_1 ): ( \frac{du_1}{dx} = \frac{f(x)}{y_1(x)} ) → Integrate to find ( u_1(x) = \int \frac{f(x)}{y_1(x)} dx = \int e^{\int P dx} \cdot f(x) dx )

Therefore, ( y_p = e^{-\int P dx} \cdot \int e^{\int P dx} f(x) dx ), which is the same result as the integrating factor method, confirming the validity of the variation of parameters approach.

20.2 Second Order Equation

Consider the second-order linear non-homogeneous differential equation in standard form: ( y'' + P(x)y' + Q(x)y = f(x) ), where ( P(x) ), ( Q(x) ), and ( f(x) ) are continuous on some interval ( I ). The associated homogeneous equation is ( y'' + P(x)y' + Q(x)y = 0 ).

Complementary function: Let ( y_1 ) and ( y_2 ) be two linearly independent solutions of the homogeneous equation. Then ( y_c = c_1 y_1(x) + c_2 y_2(x) ). Both ( y_1 ) and ( y_2 ) satisfy the homogeneous equation identically: ( y_1'' + P y_1' + Q y_1 = 0 ) and ( y_2'' + P y_2' + Q y_2 = 0 ).

💡 Why this matters: The key insight is that for a second-order equation, we need to replace both constants ( c_1 ) and ( c_2 ) with unknown functions ( u_1(x) ) and ( u_2(x) ), giving us two unknowns requiring two equations.

Particular Integral: Replace the parameters ( c_1 ) and ( c_2 ) with unknown variables ( u_1(x) ) and ( u_2(x) ), so the assumed particular integral is ( y_p = u_1(x) y_1(x) + u_2(x) y_2(x) ). We need two equations involving ( u_1 ) and ( u_2 ). One comes from substituting into the differential equation; the other is imposed to simplify derivatives.

The first derivative is ( y_p' = u_1 y_1' + y_1 u_1' + u_2 y_2' + u_2' y_2 = (u_1 y_1' + u_2 y_2') + (u_1' y_1 + u_2' y_2) ). To avoid second derivatives of ( u_1 ) and ( u_2 ), we impose the condition: ( u_1' y_1 + u_2' y_2 = 0 ). Then ( y_p' = u_1 y_1' + u_2 y_2' ).

🔑 Definition — First condition for variation of parameters: ( u_1' y_1 + u_2' y_2 = 0 ) — this condition is imposed to simplify the derivative calculation by eliminating terms involving ( u_1' ) and ( u_2' ) multiplied by the original functions.

Computing the second derivative: ( y_p'' = u_1 y_1'' + u_1' y_1' + u_2 y_2'' + u_2' y_2' ). Substituting ( y_p ), ( y_p' ), and ( y_p'' ) into the non-homogeneous equation and using the fact that ( y_1 ) and ( y_2 ) satisfy the homogeneous equation yields the second condition: ( u_1' y_1' + u_2' y_2' = f(x) ).

🔑 Definition — Second condition for variation of parameters: ( u_1' y_1' + u_2' y_2' = f(x) ) — this condition comes directly from substituting the assumed solution into the original non-homogeneous differential equation.

Thus, ( u_1 ) and ( u_2 ) must satisfy: ( u_1' y_1 + u_2' y_2 = 0 ) and ( u_1' y_1' + u_2' y_2' = f(x) ).

Using Cramer's rule, the solution is: ( u_1' = \frac{W_1}{W} ), ( u_2' = \frac{W_2}{W} ), where ( W ), ( W_1 ), and ( W_2 ) are the determinants:

( W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} ), ( W_1 = \begin{vmatrix} 0 & y_2 \ f(x) & y_2' \end{vmatrix} ), ( W_2 = \begin{vmatrix} y_1 & 0 \ y_1' & f(x) \end{vmatrix} )

The determinant ( W ) is the Wronskian of ( y_1 ) and ( y_2 ). Since ( y_1 ) and ( y_2 ) are linearly independent, ( W(y_1(x), y_2(x)) \neq 0 ) for all ( x \in I ). Integrating the expressions for ( u_1' ) and ( u_2' ) gives ( u_1 ) and ( u_2 ), and hence the particular solution.

20.3 Summary of the Method

To solve ( a_2 y'' + a_1 y' + a_0 y = g(x) ) using variation of parameters, follow these steps:

Step 1: Find the complementary function by solving the associated homogeneous equation ( a_2 y'' + a_1 y' + a_0 y = 0 ).

Step 2: If ( y_c = c_1 y_1 + c_2 y_2 ), compute the Wronskian ( W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} ).

Step 3: Divide by ( a_2 ) to get standard form ( y'' + P(x)y' + Q(x)y = f(x) ) and identify ( f(x) ).

Step 4: Construct determinants ( W_1 = \begin{vmatrix} 0 & y_2 \ f(x) & y_2' \end{vmatrix} ) and ( W_2 = \begin{vmatrix} y_1 & 0 \ y_1' & f(x) \end{vmatrix} ).

Step 5: Compute ( u_1' = \frac{W_1}{W} ) and ( u_2' = \frac{W_2}{W} ).

Step 6: Integrate: ( u_1 = \int \frac{W_1}{W} dx ) and ( u_2 = \int \frac{W_2}{W} dx ).

Step 7: The particular solution is ( y_p = u_1 y_1 + u_2 y_2 ).

Step 8: The general solution is ( y = y_c + y_p = c_1 y_1 + c_2 y_2 + u_1 y_1 + u_2 y_2 ).

20.3.1 Constants of Integration

We do not need to introduce constants of integration when computing the indefinite integrals in Step 6. If constants ( a_1 ) and ( b_1 ) were added, they would simply merge with ( c_1 ) and ( c_2 ) in the final general solution, providing no new information.

Example 1

Solve: ( y'' - 4y' + 4y = (x+1)e^{2x} )

Step 1: The auxiliary equation is ( m^2 - 4m + 4 = 0 \Rightarrow (m-2)^2 = 0 \Rightarrow m = 2, 2 ). Thus ( y_c = c_1 e^{2x} + c_2 x e^{2x} ).

Step 2: ( y_1 = e^{2x} ), ( y_2 = x e^{2x} ). Compute Wronskian: ( W = \begin{vmatrix} e^{2x} & x e^{2x} \ 2e^{2x} & 2x e^{2x} + e^{2x} \end{vmatrix} = e^{2x}(2xe^{2x}+e^{2x}) - x e^{2x}(2e^{2x}) = 2xe^{4x} + e^{4x} - 2xe^{4x} = e^{4x} \neq 0 )

Step 3: ( f(x) = (x+1)e^{2x} )

Step 4: Construct determinants: ( W_1 = \begin{vmatrix} 0 & x e^{2x} \ (x+1)e^{2x} & 2x e^{2x} + e^{2x} \end{vmatrix} = 0 - (x+1)e^{2x} \cdot x e^{2x} = -(x+1)x e^{4x} ) ( W_2 = \begin{vmatrix} e^{2x} & 0 \ 2e^{2x} & (x+1)e^{2x} \end{vmatrix} = e^{2x} \cdot (x+1)e^{2x} - 0 = (x+1)e^{4x} )

Step 5: ( u_1' = \frac{W_1}{W} = \frac{-(x+1)x e^{4x}}{e^{4x}} = -x^2 - x ) ( u_2' = \frac{W_2}{W} = \frac{(x+1)e^{4x}}{e^{4x}} = x + 1 )

Step 6: Integrate: ( u_1 = \int (-x^2 - x) dx = -\frac{x^3}{3} - \frac{x^2}{2} ) ( u_2 = \int (x + 1) dx = \frac{x^2}{2} + x )

Step 7: Particular solution: ( y_p = \left(-\frac{x^3}{3} - \frac{x^2}{2}\right) e^{2x} + \left(\frac{x^2}{2} + x\right) x e^{2x} ) ( = \left(-\frac{x^3}{3} - \frac{x^2}{2} + \frac{x^3}{2} + x^2\right) e^{2x} = \left(\frac{x^3}{6} + \frac{x^2}{2}\right) e^{2x} )

📌 Example: ( y_p = \left(\frac{x^3}{6} + \frac{x^2}{2}\right) e^{2x} )

Step 8: General solution: ( y = c_1 e^{2x} + c_2 x e^{2x} + \left(\frac{x^3}{6} + \frac{x^2}{2}\right) e^{2x} )

Example 2

Solve: ( 4y'' + 36y = \csc 3x )

Step 1: Divide by 4: ( y'' + 9y = 0 ). Auxiliary: ( m^2 + 9 = 0 \Rightarrow m = \pm 3i ). Thus ( y_c = c_1 \cos 3x + c_2 \sin 3x ).

Step 2: ( y_1 = \cos 3x ), ( y_2 = \sin 3x ). Compute Wronskian: ( W = \begin{vmatrix} \cos 3x & \sin 3x \ -3\sin 3x & 3\cos 3x \end{vmatrix} = \cos 3x \cdot 3\cos 3x - \sin 3x \cdot (-3\sin 3x) = 3\cos^2 3x + 3\sin^2 3x = 3 )

Step 3: ( f(x) = \frac{1}{4} \csc 3x )

Step 4: Construct determinants: ( W_1 = \begin{vmatrix} 0 & \sin 3x \ \frac{1}{4}\csc 3x & 3\cos 3x \end{vmatrix} = 0 - \frac{1}{4}\csc 3x \cdot \sin 3x = -\frac{1}{4} ) ( W_2 = \begin{vmatrix} \cos 3x & 0 \ -3\sin 3x & \frac{1}{4}\csc 3x \end{vmatrix} = \cos 3x \cdot \frac{1}{4}\csc 3x - 0 = \frac{1}{4} \cdot \frac{\cos 3x}{\sin 3x} = \frac{1}{4} \cot 3x )

Step 5: ( u_1' = \frac{W_1}{W} = \frac{-1/4}{3} = -\frac{1}{12} ) ( u_2' = \frac{W_2}{W} = \frac{(1/4)\cot 3x}{3} = \frac{1}{12} \cot 3x )

Step 6: Integrate: ( u_1 = \int -\frac{1}{12} dx = -\frac{1}{12}x ) ( u_2 = \int \frac{1}{12} \cot 3x dx = \frac{1}{12} \cdot \frac{1}{3} \ln|\sin 3x| = \frac{1}{36} \ln|\sin 3x| )

Step 7: Particular solution: ( y_p = \left(-\frac{1}{12}x\right) \cos 3x + \left(\frac{1}{36} \ln|\sin 3x|\right) \sin 3x )

📌 Example: ( y_p = -\frac{1}{12}x \cos 3x + \frac{1}{36} (\sin 3x) \ln|\sin 3x| )

Step 8: General solution: ( y = c_1 \cos 3x + c_2 \sin 3x - \frac{1}{12}x \cos 3x + \frac{1}{36} (\sin 3x) \ln|\sin 3x| )

Example 3

Solve: ( y'' - y = \frac{1}{x} )

Step 1: Consider ( y'' - y = 0 ). Auxiliary: ( m^2 - 1 = 0 \Rightarrow m = \pm 1 ). Thus ( y_c = c_1 e^x + c_2 e^{-x} ).

Step 2: ( y_1 = e^x ), ( y_2 = e^{-x} ). Compute Wronskian: ( W = \begin{vmatrix} e^x & e^{-x} \ e^x & -e^{-x} \end{vmatrix} = e^x \cdot (-e^{-x}) - e^{-x} \cdot e^x = -1 - 1 = -2 )

Step 3: ( f(x) = \frac{1}{x} )

Step 4: Construct determinants: ( W_1 = \begin{vmatrix} 0 & e^{-x} \ \frac{1}{x} & -e^{-x} \end{vmatrix} = 0 - \frac{1}{x} \cdot e^{-x} = -\frac{e^{-x}}{x} ) ( W_2 = \begin{vmatrix} e^x & 0 \ e^x & \frac{1}{x} \end{vmatrix} = e^x \cdot \frac{1}{x} - 0 = \frac{e^x}{x} )

Step 5: ( u_1' = \frac{W_1}{W} = \frac{-e^{-x}/x}{-2} = \frac{e^{-x}}{2x} ) ( u_2' = \frac{W_2}{W} = \frac{e^x/x}{-2} = -\frac{e^x}{2x} )

Step 6: Integrate: ( u_1 = \int \frac{e^{-x}}{2x} dx ), ( u_2 = \int -\frac{e^x}{2x} dx )

Step 7: Particular solution: ( y_p = \left(\int \frac{e^{-x}}{2x} dx\right) e^x + \left(\int -\frac{e^x}{2x} dx\right) e^{-x} )

Step 8: General solution: ( y = c_1 e^x + c_2 e^{-x} + y_p )

⭐ Key Takeaways

The variation of parameters method is a powerful and general technique for finding particular solutions to non-homogeneous linear differential equations that works even when the forcing function is not of the special forms required by undetermined coefficients. The procedure for second-order equations involves replacing the constants in the complementary function with unknown functions, imposing the condition ( u_1' y_1 + u_2' y_2 = 0 ) to simplify derivatives, and obtaining a second condition ( u_1' y_1' + u_2' y_2' = f(x) ) from the differential equation itself. The derivatives of the unknown functions are found using Cramer's rule with the Wronskian, and integration (without constants) yields the functions ( u_1 ) and ( u_2 ). The final particular solution is the linear combination ( y_p = u_1 y_1 + u_2 y_2 ), and constants of integration from the indefinite integrals can be omitted as they are absorbed into the complementary function.

🧠 Quick Revision Questions

1. What two conditions must ( u_1' ) and ( u_2' ) satisfy in the variation of parameters method for a second-order equation?
2. Why is the condition ( u_1' y_1 + u_2' y_2 = 0 ) imposed in the derivation?
3. What role does the Wronskian play in the variation of parameters method?
4. How does the result of variation of parameters for first-order equations compare with the integrating factor method?
5. In Example 2, why is the Wronskian a constant (3), and what does this imply about the solutions ( \cos 3x ) and ( \sin 3x )?

📘 Lecture 22 — Variation of Parameters Method

📖 Overview: This lecture extends the variation of parameters method for solving non-homogeneous differential equations. It covers the technique for second-order equations with detailed examples, then generalizes the method for higher-order linear differential equations, showing how to find particular solutions when the method of undetermined coefficients is not applicable.

🗂️ Topics Covered

The lecture begins by revisiting the variation of parameters method for second-order equations, demonstrating an example where integrals cannot be expressed in elementary functions. It then generalizes the method to nth-order differential equations, presenting step-by-step procedures for finding complementary functions and particular integrals. Three comprehensive examples are worked out, covering various types of non-homogeneous terms including csc x, tan x, and exponential functions. The section concludes with an exercise set for practice.

📝 Lecture Summary

Variation of Parameters Method for Second-Order Equations (Review)

The method is demonstrated for solving non-homogeneous equations where the particular integral cannot be found by simple inspection. For the equation y'' - y = 1/x, the process involves finding the complementary function from the homogeneous equation, computing the Wronskian of linearly independent solutions, and then determining particular solutions through integration.

🔑 Definition — Wronskian: For two functions y₁ and y₂, the Wronskian W(y₁, y₂) = y₁y₂' - y₂y₁'. It determines whether solutions are linearly independent if W ≠ 0.

📐 Formula — For second-order equations: The particular solution is yₚ = u₁y₁ + u₂y₂, where u₁' = W₁/W and u₂' = W₂/W, with W₁ and W₂ being determinants formed by replacing columns of W.

📌 Example: For y'' - y = 1/x, the complementary function is y_c = c₁eˣ + c₂e⁻ˣ. The Wronskian W = -2. The particular solution involves integrals ∫(e⁻ᵗ/t)dt and ∫(eᵗ/t)dt, which cannot be expressed in elementary functions, so they are written as definite integrals from x₀ to x.

Variation of Parameters Method for Higher-Order Equations

The method is generalized for nth-order linear differential equations of the form aₙy⁽ⁿ⁾ + aₙ₋₁y⁽ⁿ⁻¹⁾ + ... + a₁y' + a₀y = g(x). The process involves eight systematic steps.

🔑 Definition — Linearly independent solutions: Functions y₁, y₂, ..., yₙ are linearly independent if the only solution to c₁y₁ + c₂y₂ + ... + cₙyₙ = 0 is c₁ = c₂ = ... = cₙ = 0.

📐 Formula — For nth-order equations: The derivatives of unknown functions are uₖ' = Wₖ/W, where Wₖ is obtained by replacing the kth column of the Wronskian with the column [0, 0, ..., 0, f(x)]ᵀ.

💡 Why this matters: The first n-1 equations in step 5 are assumptions made to simplify the first n-1 derivatives of yₚ. The last equation comes from substituting the particular integral into the original differential equation.

Example 1: Solving y''' + y' = csc x

Step 1: The auxiliary equation m³ + m = 0 gives m = 0, m = ±i. Therefore, the complementary function is y_c = c₁ + c₂ cos x + c₃ sin x.

Step 2: From y_c, we identify y₁ = 1, y₂ = cos x, y₃ = sin x. The Wronskian is computed and simplified to W = 1 (using row operation R₁ + R₃).

Step 3: The equation is already in standard form: y''' + 0·y'' + y' + 0·y = csc x.

Step 4: The determinants W₁, W₂, W₃ are found by replacing columns of W with [0, 0, csc x]ᵀ:

• W₁ = csc x(sin²x + cos²x) = csc x
• W₂ = -cot x
• W₃ = -1

Step 5: The derivatives are u₁' = csc x, u₂' = -cot x, u₃' = -1.

Step 6: Integrating gives u₁ = ln|csc x - cot x|, u₂ = -ln|sin x|, u₃ = -x.

Step 7: The particular solution is yₚ = ln|csc x - cot x| - cos x·ln|sin x| - x·sin x.

Step 8: The general solution is y = c₁ + c₂ cos x + c₃ sin x + ln|csc x - cot x| - cos x·ln|sin x| - x·sin x.

Example 2: Solving y''' + y' = tan x

This example follows the same steps with f(x) = tan x. The complementary function remains y_c = c₁ + c₂ cos x + c₃ sin x.

The determinants yield: W₁ = tan x, W₂ = -sin x, W₃ = -sin x·tan x.

Integration produces: u₁ = -ln|cos x|, u₂ = cos x, u₃ = sin x - ln|sec x + tan x|.

The particular solution simplifies to yₚ = -ln|cos x| + 1 - sin x·ln|sec x + tan x|.

The general solution can be rewritten as y = d₁ + c₂ cos x + c₃ sin x - ln|cos x| - sin x·ln|sec x + tan x|, where d₁ = c₁ + 1.

Example 3: Solving y''' - 2y'' - y' + 2y = e³ˣ

Step 1: The auxiliary equation m³ - 2m² - m + 2 = 0 factors to (m - 2)(m² - 1) = 0, giving roots m = 1, 2, -1.

Step 2: The complementary function is y_c = c₁eˣ + c₂e²ˣ + c₃e⁻ˣ. The Wronskian is W = 6e²ˣ ≠ 0.

Step 4: The determinants are:

• W₁ = -3e⁴ˣ
• W₂ = 2e³ˣ
• W₃ = e⁶ˣ

Step 5: The derivatives are u₁' = -½e²ˣ, u₂' = (1/3)eˣ, u₃' = (1/6)e⁴ˣ.

Step 6: Integrating: u₁ = -¼e²ˣ, u₂ = (1/3)eˣ, u₃ = (1/24)e⁴ˣ.

Step 7: The particular solution is yₚ = -(1/4)e³ˣ + (1/3)e³ˣ + (1/24)e³ˣ.

Step 8: The general solution is y = c₁eˣ + c₂e²ˣ + c₃e⁻ˣ - (1/4)e³ˣ + (1/3)e³ˣ + (1/24)e³ˣ.

⭐ Key Takeaways

The variation of parameters method provides a systematic way to find particular solutions for non-homogeneous linear differential equations when the method of undetermined coefficients fails. The core process involves finding the complementary function, computing the Wronskian of linearly independent solutions, and determining particular solutions through integration of ratios of determinants. For higher-order equations, the method requires solving an n×n system to find derivatives of unknown functions. The particular integral may vary depending on how the resulting integrals are evaluated, and initial value problems must be solved using the general solution, not just the complementary function.

🧠 Quick Revision Questions

1. For an nth-order differential equation, how many equations must be solved to find the derivatives u₁', u₂', ..., uₙ'?
2. What is the purpose of the first n-1 equations in the system for finding uₖ'?
3. In Example 1, why does the Wronskian simplify to 1?
4. How does the general solution in Example 2 differ from the complementary function?
5. For the equation y''' - 2y'' - y' + 2y = e³ˣ, what are the three roots of the auxiliary equation?