MTH401 — Final Term Summary (Lectures 23–44)

📘 Lecture 23 — Damped Motion

📖 Overview: This lecture extends the analysis of spring-mass systems to include realistic damping forces (e.g., friction, air resistance). It derives the differential equation for free damped motion and classifies the three types of damping (over-damped, critically damped, and under-damped) based on the system parameters, analyzing the resulting motion in each case.

🗂️ Topics Covered

This lecture introduces the concept of a damping force proportional to velocity, derives the differential equation for free damped motion from Newton's Second Law, solves this equation using the auxiliary equation, and then distinguishes between the three possible cases of motion—over-damped, critically damped, and under-damped—based on the sign of the discriminant λ² - ω².

📝 Lecture Summary

23.1 Damping Force

In realistic systems, a resisting force (damping) always acts on a moving mass, for instance, due to a viscous medium. The damping force is modeled as being proportional to a power of the instantaneous velocity. In this lecture, we assume it is proportional to the first power of velocity.

🔑 Definition — Damping force: A retarding force acting opposite to the direction of motion, given by Damping force = -β (dx/dt), where β is a positive damping constant.

23.2 The Differential Equation

When a mass m attached to a spring is displaced, three forces act: weight (mg), the restoring force (-k(s+x)), and the damping force. Applying Newton's Second Law, F = ma, and using the equilibrium condition mg - ks = 0 leads to the differential equation of free damped motion.

For algebraic convenience, we define 2λ = β/m and ω² = k/m, where λ is the damping coefficient and ω is the natural frequency. The standard form of the equation is: [ \frac{d^2x}{dt^2} + 2\lambda \frac{dx}{dt} + \omega^2 x = 0 ]

23.2.1 Solution of the Differential Equation

To solve this second-order linear homogeneous ODE, we assume a solution of the form x(t) = e^{mt}, leading to the auxiliary equation: m² + 2λm + ω² = 0. The roots are found using the quadratic formula: [ m = -\lambda \pm \sqrt{\lambda^2 - \omega^2} ] The nature of the motion is entirely determined by the sign of the discriminant λ² - ω².

🔑 Definition — Over-damped motion: Occurs when λ² - ω² > 0 (or β > k). The system returns to equilibrium slowly without oscillating. 📐 Formula: x(t) = c₁e^{m₁t} + c₂e^{m₂t} or x(t) = e^{-λt}[c₁e^{√(λ²-ω²)t} + c₂e^{-√(λ²-ω²)t}] 📌 Example: A mass in a very thick, viscous fluid.

🔑 Definition — Critically damped motion: Occurs when λ² - ω² = 0 (or β = k). This represents the smallest amount of damping for which the system returns to equilibrium without oscillating. 📐 Formula: x(t) = c₁e^{m₁t} + c₂ t e^{m₁t} or x(t) = e^{-λt}(c₁ + c₂t) 📌 Example: The "perfect" shock absorber.

🔑 Definition — Under-damped motion: Occurs when λ² - ω² < 0 (or β < k). The system oscillates with a diminishing amplitude over time. 📐 Formula: x(t) = e^{-λt}[c₁ cos(√(ω²-λ²)t) + c₂ sin(√(ω²-λ²)t)] 📌 Example: A mass oscillating in water.

💡 Why this matters: In each case, the damping factor e^{-λt} (with λ > 0) causes the amplitude of motion to approach zero as time t → ∞, reflecting that the mass effectively comes to rest at its equilibrium position.

⭐ Key Takeaways

The lecture formalizes how to model resistance in a spring-mass system by introducing a damping force proportional to velocity. The resulting second-order ODE, x'' + 2λx' + ω²x = 0, has solutions that describe three distinct behaviors based on the discriminant λ² - ω². In over-damped and critically damped systems, the mass returns to equilibrium without oscillating, while in an under-damped system, the mass undergoes oscillations of decreasing amplitude. Critically damped motion is the boundary case that represents the fastest return to equilibrium without oscillation, which is crucial for practical applications like shock absorbers.

🧠 Quick Revision Questions

1. What is the mathematical expression for the damping force considered in this lecture, and why is the sign negative?
2. Derive the differential equation for the free damped motion of a spring-mass system, starting from Newton's Second Law.
3. What are the three possible types of motion for a damped free system, and what condition on the discriminant λ² - ω² defines each one?
4. Explain the physical meaning of the term e^{-λt} in the solution for the under-damped case. What happens to the amplitude of oscillation as time increases?
5. A critically damped system is achieved when λ² = ω². In terms of the spring constant k and mass m, what must be the value of the damping coefficient β?

📘 Lecture 24 — Forced Motion

📖 Overview: This lecture examines vibrational systems where an external driving force acts on the mass-spring-damper system, leading to non-homogeneous differential equations. It covers forced motion with damping, the solution process using undetermined coefficients, and the analogous electrical circuit (LRC series circuit). This material is essential for understanding how systems respond to periodic external inputs.

🗂️ Topics Covered

The lecture introduces forced motion with damping, deriving the non-homogeneous differential equation governing the system. It then solves an example problem of a forced damped spring-mass system with an initial value problem, followed by a discussion of the analogous electrical circuit consisting of an inductor, resistor, and capacitor in series, and its similarities to the mechanical system.

📝 Lecture Summary

24.1 Forced motion with damping

When an external force f(t) acts on the system, the total forces include the weight mg, the restoring force -k(s+x), the damping force -β(dx/dt), and the external force f(t). Applying Newton's second law of motion and using the equilibrium condition mg - ks = 0, we derive the non-homogeneous differential equation for forced motion with damping:

d²x/dt² + 2λ (dx/dt) + ω² x = F(t)

where F(t) = f(t)/m, 2λ = β/m, and ω² = k/m. To solve this equation, we can use either the method of undetermined coefficients or the variation of parameters.

Example 1

Interpret and solve the initial value problem: (1/5) d²x/dt² + 1.2 dx/dt + 2x = 5 cos 4t, with x(0) = 1/2, x'(0) = 0.

Interpretation The problem represents a vibrational system with mass m = 1/5 slugs (or kg), spring constant k = 2 lb/ft (or N/m), damping constant β = 1.2, and external periodic force f(t) = 5 cos 4t with period T = π/2. The mass is released from rest 1/2 ft (or m) below the equilibrium position.

Solution First, rewrite the differential equation by multiplying both sides by 5: d²x/dt² + 6 dx/dt + 10x = 25 cos 4t

Consider the associated homogeneous equation: d²x/dt² + 6 dx/dt + 10x = 0

Put x = e^(mt), so dx/dt = me^(mt) and d²x/dt² = m²e^(mt). The auxiliary equation is: m² + 6m + 10 = 0

📌 Note: The solution of this auxiliary equation is not fully computed in the lecture text, but would proceed using the quadratic formula to find m = -3 ± i. The homogeneous solution would then be x_h(t) = e^(-3t) (c₁ cos t + c₂ sin t). The particular solution and final solution would involve applying the method of undetermined coefficients to the non-homogeneous equation d²x/dt² + 6 dx/dt + 10x = 25 cos 4t, then using the initial conditions x(0) = 1/2 and x'(0) = 0 to determine the constants.

💡 Why this matters: The solution of this forced damped system will exhibit both a transient component (from the homogeneous solution that decays over time) and a steady-state component (from the particular solution that persists as long as the driving force is applied).

24.2 Forced Motion without Damping

The lecture text does not contain a section under this heading. The lecture moves directly to discussing the analogous electrical circuit.

24.3 Electrical Circuit Analogy: Series LRC Circuit

The flow of current in an electrical circuit consisting of an inductor (L), resistor (R), and capacitor (C) connected in series is analogous to the forced motion of a mechanical system. This similarity allows us to apply the same mathematical techniques to analyze electrical circuits.

⭐ Key Takeaways

The key concepts from this lecture are: (1) Forced motion introduces an external driving force f(t) into the damped spring-mass system, resulting in a non-homogeneous second-order linear differential equation. (2) The governing equation for forced motion with damping is d²x/dt² + 2λ (dx/dt) + ω² x = F(t), where F(t) = f(t)/m. (3) This non-homogeneous equation is solved by finding the complementary solution (from the homogeneous part) and a particular solution using methods like undetermined coefficients. (4) The solution consists of a transient part (decaying with time due to damping) and a steady-state part (driven by the external force). (5) The series LRC circuit provides a direct electrical analogy to the mechanical forced damped system.

🧠 Quick Revision Questions

1. What are the four forces acting on a mass in a forced damped spring-mass system, and how do they combine to form the differential equation?
2. What is the standard form of the non-homogeneous differential equation governing forced motion with damping, and what do the parameters λ, ω, and F(t) represent?
3. Explain the difference between the transient and steady-state components of the solution to a forced damped system.
4. What two methods can be used to find the particular solution of the non-homogeneous differential equation for forced motion?
5. Which physical components in an electrical circuit correspond to the mass, damping, and spring constant in a mechanical forced system?

📘 Lecture 25 — Differential Equations (MTH401) — Forced Oscillations

📖 Overview: This lecture explores forced oscillations in spring-mass systems where an external periodic force is applied. It demonstrates how to solve non-homogeneous second-order linear differential equations with constant coefficients, and introduces the critical concepts of transient and steady-state solutions through detailed examples.

🗂️ Topics Covered

The lecture covers solving initial value problems for forced damped oscillations using the method of undetermined coefficients, analyzing transient and steady-state terms in solutions with damping, examining motion without damping including the special case of pure resonance when the forcing frequency equals the natural frequency, and applying L’Hôpital’s rule to find limiting solutions.

📝 Lecture Summary

24.2 Transient and Steady-State Terms

When a damped system is subjected to a periodic forcing function, the general solution consists of two distinct parts. The complementary function ( x_c(t) ) contains the factor ( e^{-\lambda t} ) which causes it to approach zero as ( t \to \infty ). This decaying part is called the transient term or transient solution. The particular solution ( x_p(t) ), which persists as time increases, is called the steady-state solution.

🔑 Definition — Transient Term: The part of the solution that approaches zero as ( t \to \infty ), typically containing exponential decay factors like ( e^{-\lambda t} ).

🔑 Definition — Steady-State Solution: The particular solution that remains after the transient has decayed, representing the long-term behavior of the system.

💡 Why this matters: In physical systems, the transient represents the initial disturbances that die out, while the steady-state represents the ongoing forced response that actually matters for most practical applications.

Example 2 — Solve the initial value problem: [ \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 2x = 4\cos t + 2\sin t, \quad x(0) = 0, ; x'(0) = 3 ]

Step 1: Find complementary function. The auxiliary equation for the homogeneous part is: [ m^2 + 2m + 2 = 0 ] Using the quadratic formula: [ m = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i ] Thus: [ x_c = e^{-t}(c_1\cos t + c_2\sin t) ]

Step 2: Find particular integral using undetermined coefficients. Assume: [ x_p = A\cos t + B\sin t ] [ x'_p = -A\sin t + B\cos t ] [ x''_p = -A\cos t - B\sin t ]

Substituting into the differential equation: [ (-A\cos t - B\sin t) + 2(-A\sin t + B\cos t) + 2(A\cos t + B\sin t) ] [ = (-A + 2B + 2A)\cos t + (-B - 2A + 2B)\sin t ] [ = (A + 2B)\cos t + (-2A + B)\sin t = 4\cos t + 2\sin t ]

Equating coefficients: [ A + 2B = 4 ] [ -2A + B = 2 ]

Solving: From the second equation, ( B = 2 + 2A ). Substituting into the first: [ A + 2(2 + 2A) = 4 ] [ A + 4 + 4A = 4 ] [ 5A = 0 \implies A = 0, ; B = 2 ]

Thus ( x_p = 2\sin t ).

Step 3: General solution: [ x(t) = e^{-t}(c_1\cos t + c_2\sin t) + 2\sin t ]

Step 4: Apply initial conditions. [ x'(t) = -e^{-t}(c_1\cos t + c_2\sin t) + e^{-t}(-c_1\sin t + c_2\cos t) + 2\cos t ]

From ( x(0) = 0 ): ( c_1 \cdot 1 + 0 + 0 = 0 \implies c_1 = 0 )

From ( x'(0) = 3 ): ( -c_1 \cdot 1 + c_2 \cdot 1 + 2 = 3 \implies c_2 = 1 )

Final solution: [ x(t) = e^{-t}\sin t + 2\sin t ]

The transient term is ( e^{-t}\sin t ) (decays to 0), and the steady-state is ( 2\sin t ). The transient becomes negligible for ( t > 2\pi ).

24.3 Motion without Damping

When a system is subjected to a periodic force but has no damping, there is no transient term in the solution. The system oscillates indefinitely with both the natural frequency and the forcing frequency present.

Example 3 — Solve the initial value problem: [ \frac{d^2x}{dt^2} + \omega^2 x = F_0 \sin \gamma t, \quad x(0) = 0, ; x'(0) = 0 ]

Step 1: Find complementary function. The auxiliary equation is: [ m^2 + \omega^2 = 0 \implies m = \pm \omega i ] Thus: [ x_c(t) = c_1\cos \omega t + c_2\sin \omega t ]

Step 2: Find particular integral. Assume ( x_p = A\cos \gamma t + B\sin \gamma t ). [ x'_p = -A\gamma\sin \gamma t + B\gamma\cos \gamma t ] [ x''_p = -A\gamma^2\cos \gamma t - B\gamma^2\sin \gamma t ]

Substituting: [ x''_p + \omega^2 x_p = (-A\gamma^2\cos \gamma t - B\gamma^2\sin \gamma t) + \omega^2(A\cos \gamma t + B\sin \gamma t) ] [ = A(\omega^2 - \gamma^2)\cos \gamma t + B(\omega^2 - \gamma^2)\sin \gamma t = F_0\sin \gamma t ]

Equating coefficients: [ A(\omega^2 - \gamma^2) = 0 \implies A = 0 ] [ B(\omega^2 - \gamma^2) = F_0 \implies B = \frac{F_0}{\omega^2 - \gamma^2} \quad (\gamma \neq \omega) ]

Thus: [ x_p(t) = \frac{F_0}{\omega^2 - \gamma^2}\sin \gamma t ]

Step 3: General solution: [ x(t) = c_1\cos \omega t + c_2\sin \omega t + \frac{F_0}{\omega^2 - \gamma^2}\sin \gamma t ] [ x'(t) = -c_1\omega\sin \omega t + c_2\omega\cos \omega t + \frac{F_0\gamma}{\omega^2 - \gamma^2}\cos \gamma t ]

Step 4: Apply initial conditions. From ( x(0) = 0 ): ( c_1 = 0 ) From ( x'(0) = 0 ): [ c_2\omega + \frac{F_0\gamma}{\omega^2 - \gamma^2} = 0 \implies c_2 = \frac{-F_0\gamma}{\omega(\omega^2 - \gamma^2)} ]

Final solution: [ x(t) = \frac{F_0}{\omega(\omega^2 - \gamma^2)}(-\gamma\sin \omega t + \omega\sin \gamma t), \quad \gamma \neq \omega ]

Important: This solution is not defined when ( \gamma = \omega ) (resonance). The limit as ( \gamma \to \omega ) can be found using L’Hôpital’s rule.

🔑 Definition — Resonance: The condition that occurs when the forcing frequency ( \gamma ) equals the natural frequency ( \omega ), causing the amplitude of oscillation to grow without bound in undamped systems.

📐 Formula: ( \lim_{\gamma \to \omega} x(t) = \frac{F_0}{2\omega^2}(\sin \omega t - \omega t \cos \omega t) )

⭐ Key Takeaways

The general solution of a forced damped system always consists of a transient part (complementary function, decaying to zero) and a steady-state part (particular solution, persisting indefinitely). For damped systems, the transient becomes negligible after approximately ( t > 2\pi ). In undamped systems, there is no transient, and the solution contains both natural and forcing frequency components. When the forcing frequency equals the natural frequency in an undamped system, resonance occurs producing unbounded oscillations. L’Hôpital’s rule is essential for finding the limiting solution at resonance when the standard formula becomes undefined.

🧠 Quick Revision Questions

1. What distinguishes a transient term from a steady-state term in the solution of a forced damped oscillator?
2. In Example 2, why does the assumed particular solution ( x_p = A\cos t + B\sin t ) work even though the complementary function also contains ( \cos t ) and ( \sin t )?
3. What happens to the transient term in Example 2 when ( t > 2\pi ) and why?
4. Why is the solution ( x(t) = \frac{F_0}{\omega(\omega^2 - \gamma^2)}(-\gamma\sin \omega t + \omega\sin \gamma t) ) not valid when ( \gamma = \omega )?
5. In Example 3, what mathematical technique is used to find the limiting solution as ( \gamma \to \omega )?

📘 Lecture 26 — Forced Motion (Examples)

📖 Overview: This lecture examines the behavior of LRC series circuits by applying the mathematical framework of second-order linear differential equations developed for forced mechanical oscillations. It demonstrates how the charge on a capacitor and current in the circuit can be determined under various conditions, including the special case of resonance when damping is absent.

🗂️ Topics Covered

This lecture begins by deriving the response of an undamped system at resonance, showing that the amplitude grows without bound. It then introduces the LRC series circuit as an electrical analog of the forced damped harmonic oscillator, defining the roles of the resistor, inductor, and capacitor. Kirchhoff’s Voltage Law is applied to derive the governing differential equation for the charge, and the nature of the solution is classified based on the discriminant (overdamped, critically damped, underdamped). Finally, two detailed examples are solved: an LC circuit with no resistance and an LRC circuit with given parameters.

📝 Lecture Summary

24.4 Electric Circuits

Many physical systems, including electrical circuits, are governed by a second-order linear differential equation identical in form to that of forced mechanical motion: m d²x/dt² + β dx/dt + kx = f(t). The LRC series circuit is a direct electrical analog, allowing insights from one system to be applied to the other.

24.5 The LRC Series Circuits

An LRC series circuit consists of an inductor (L), a resistor (R), and a capacitor (C) connected in series with a time-varying voltage source E(t).

🔑 Definition — Resistor: An electrical component that opposes the flow of current. Its opposition is called resistance (R). Ohm’s Law states V = IR, where V is voltage and I is current. Current is inversely proportional to resistance.

🔑 Definition — Inductor: A passive component that stores energy in a magnetic field. It opposes changes in current, producing a counter emf. The voltage across an inductor is V = L dI/dt, where inductance (L) is measured in Henry (H).

🔑 Definition — Capacitor: A passive component that stores charge and opposes voltage changes. Its ability to store charge is capacitance (C). The voltage across a capacitor is V = q/C, where q is the charge, and capacitance is measured in farad (F).

24.6 Kirchhoff’s Voltage Law

Kirchhoff’s 2nd law states that the sum of voltage drops around any closed loop equals the sum of voltage rises.

24.6.1 The Differential Equation

For the LRC circuit, the voltage drops across the inductor (V_L = L dI/dt), resistor (V_R = RI), and capacitor (V_C = q/C) sum to the source voltage E(t). Substituting I = dq/dt (current is the rate of flow of charge) gives the governing differential equation: 📐 Formula: L d²q/dt² + R dq/dt + q/C = E(t) 💡 Why this matters: This equation is mathematically identical to the driven, damped harmonic oscillator. If E(t)=0 and R≠0, the circuit exhibits free damped oscillations. If E(t)=0 and R=0, it exhibits free undamped oscillations.

24.6.2 Solution of the differential equation

The general solution consists of a complementary function (solution to the homogeneous equation) and a particular integral. Substituting q = e^(mt) into the homogeneous equation (L d²q/dt² + R dq/dt + q/C = 0) yields the auxiliary equation: 📐 Formula: Lm² + Rm + 1/C = 0 The nature of the roots, and thus the system’s behavior, depends on the discriminant Disc = R² - 4L/C:

• Overdamped: Disc > 0 → real and distinct roots.
• Critically damped: Disc = 0 → real and equal roots.
• Underdamped: Disc < 0 → complex roots.

In all three cases where R ≠ 0, the solution contains the factor e^(-Rt/2L), meaning q(t) → 0 as t → ∞. In the underdamped case, the charge oscillates as it decays. If R=0 and E(t)=0, the electrical vibrations are Simple Harmonic and do not decay.

25 Forced Motion (Examples)

Example 1: LC Series Circuit

An LC circuit has no resistor, so R dq/dt = 0. The source voltage E(t) = 0. The governing equation becomes: 📐 Formula: L d²q/dt² + q/C = 0 The initial conditions are q(0) = q₀ and I(0) = 0 (which implies q'(0) = 0). The auxiliary equation is Lm² + 1/C = 0, giving m = ± i/√(LC). The solution is: 📌 Example: q(t) = c₁ cos(t/√(LC)) + c₂ sin(t/√(LC)) Applying q(0) = q₀ gives c₁ = q₀. Applying q'(0) = 0 gives c₂ = 0. Therefore, the charge is q(t) = q₀ cos(t/√(LC)). The current is I(t) = dq/dt = -(q₀/√(LC)) sin(t/√(LC)).

Example 2: LRC Series Circuit

Find the charge q(t) on the capacitor in an LRC series circuit with L = 0.25 H (= 1/4), R = 10 Ω, C = 0.001 F (= 1/1000), E(t) = 0, q(0) = q₀, and I(0) = 0. 📌 Example: The governing equation is L d²q/dt² + R dq/dt + q/C = 0. Substituting values: (1/4) d²q/dt² + 10 dq/dt + 1000q = 0. Multiplying by 4: d²q/dt² + 40 dq/dt + 4000q = 0. The auxiliary equation is m² + 40m + 4000 = 0. The discriminant is 40² - 4*1*4000 = 1600 - 16000 = -14400 < 0, so the circuit is underdamped with complex roots: m = [-40 ± √(-14400)]/2 = [-40 ± 120i]/2 = -20 ± 60i.

⭐ Key Takeaways

The LRC series circuit is an exact electrical analog of the forced, damped harmonic oscillator, governed by the equation L q'' + R q' + q/C = E(t). The system’s behavior—overdamped, critically damped, or underdamped—is determined solely by the discriminant R² - 4L/C of the auxiliary equation. When R=0 and E(t)=0, the circuit produces simple harmonic oscillations described by q(t) = q₀ cos(t/√(LC)). The charge q(t) always decays to zero as t→∞ when R≠0. A key skill is translating circuit parameters into the differential equation and applying initial conditions to find the specific solution for charge and current.

🧠 Quick Revision Questions

1. Write the differential equation governing an LRC series circuit and identify the electrical analog of each term in the mechanical equation m x'' + β x' + kx = f(t).
2. What three types of damping behavior are possible in an LRC circuit, and what condition on the discriminant defines each?
3. In a damped LRC circuit (R≠0), what happens to the charge on the capacitor as t → ∞?
4. Derive the expression for the charge q(t) on the capacitor in an LC series circuit with E(t)=0, initial charge q₀, and no initial current.
5. For an LRC circuit with L=0.25 H, R=10 Ω, C=0.001 F, and E(t)=0, determine if the circuit is overdamped, critically damped, or underdamped.

📘 Lecture 27 — Differential Equations with Variable Coefficients

📖 Overview: This lecture introduces differential equations with variable (non-constant) coefficients, which arise frequently in applications such as potential theory, temperature distributions, and vibration phenomena. The focus is on the Cauchy-Euler equation, a special type of variable-coefficient ODE that can be solved using a power substitution method analogous to constant-coefficient equations. The lecture covers three cases for the roots of the auxiliary equation—distinct real, repeated real, and complex conjugate—and provides examples for each.

🗂️ Topics Covered

The lecture begins by motivating the study of variable-coefficient equations with examples from physics and engineering. It then defines the Cauchy-Euler equation and introduces the method of solution by assuming (y = x^m). The solution procedure is developed for the second-order homogeneous case, and three distinct cases are analyzed depending on the nature of the roots of the auxiliary equation: distinct real roots, repeated real roots, and complex conjugate roots. For the repeated root case, the method of reduction of order is used to find a second linearly independent solution involving (\ln x). The complex root case is handled using Euler's formula to express the solution in terms of real functions involving (\cos(\beta \ln x)) and (\sin(\beta \ln x)). The lecture also notes that the method extends to higher-order equations and non-homogeneous equations.

📝 Lecture Summary

Differential Equations with Variable Coefficients

So far, we have solved linear differential equations with constant coefficients. Now, we discuss equations with non-constant (variable) coefficients. These equations arise in applications such as temperature or potential (u) in the region between two concentric spheres, leading to equations like (r \frac{d^2 u}{dr^2} + 2 \frac{du}{dr} = 0) where (r > 0). Other important examples include Bessel's equation (x^2 y'' + x y' + (x^2 - v^2)y = 0), Legendre's equation ((1-x^2)y'' - 2xy' + n(n+1)y = 0), and Hermite's equation (y'' - 2xy' + 2ny = 0). These equations cannot be solved as easily as constant-coefficient ones.

26.1 Cauchy-Euler Equation

Any linear differential equation of the form: [ a_n x^n \frac{d^n y}{dx^n} + a_{n-1} x^{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \cdots + a_1 x \frac{dy}{dx} + a_0 y = g(x) ] where (a_n, a_{n-1}, \ldots, a_0) are constants, is called a Cauchy-Euler equation or equi-dimensional equation. The degree of each monomial coefficient matches the order of differentiation. For convenience, we focus on the homogeneous second-order case: [ ax^2 \frac{d^2 y}{dx^2} + bx \frac{dy}{dx} + cy = 0, \quad x \neq 0 ] Solutions for higher-order equations follow analogously. Non-homogeneous equations can be solved by variation of parameters after finding the complementary function (y_c(x)).

26.1.1 Method of Solution

We try a solution of the form (y = x^m), where (m) is to be determined. The first and second derivatives are: [ \frac{dy}{dx} = m x^{m-1}, \quad \frac{d^2 y}{dx^2} = m(m-1) x^{m-2} ] Substituting into the differential equation yields: [ ax^2 \cdot m(m-1)x^{m-2} + bx \cdot m x^{m-1} + c x^m = x^m \left( am(m-1) + bm + c \right) = 0 ] Thus, (y = x^m) is a solution whenever (m) satisfies the auxiliary equation: [ am(m-1) + bm + c = 0 \quad \text{or} \quad am^2 + (b-a)m + c = 0 ] The solution depends on the nature of the roots of this equation.

26.1.2 Case I (Distinct Real Roots)

Let (m_1) and (m_2) be real roots such that (m_1 \neq m_2). Then (y = x^{m_1}) and (y = x^{m_2}) form a fundamental set of solutions. The general solution is: [ y = c_1 x^{m_1} + c_2 x^{m_2} ]

📌 Example 1: Solve (x^2 \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} - 4y = 0).

Assume (y = x^m, \frac{dy}{dx} = m x^{m-1}, \frac{d^2 y}{dx^2} = m(m-1) x^{m-2}). Substituting: [ x^2 \cdot m(m-1)x^{m-2} - 2x \cdot m x^{m-1} - 4x^m = x^m (m(m-1) - 2m - 4) = x^m (m^2 - 3m - 4) = 0 ] The auxiliary equation is (m^2 - 3m - 4 = 0), which gives (m_1 = -1, m_2 = 4). Since the roots are real and distinct, the general solution is: [ y = c_1 x^{-1} + c_2 x^4 ]

26.1.3 Case II (Repeated Real Roots)

If the roots are equal ((m_1 = m_2)), we obtain only one solution (y_1 = x^{m_1}). To find a second linearly independent solution, we use the method of reduction of order. Writing the Cauchy-Euler equation in standard form (\frac{d^2 y}{dx^2} + P(x) \frac{dy}{dx} + Q(x) y = 0), we identify (P(x) = \frac{b}{ax}). The second solution is given by: [ y_2 = y_1 \int \frac{e^{-\int P(x) dx}}{(y_1)^2} dx = x^{m_1} \int \frac{e^{-(b/a) \ln x}}{x^{2m_1}} dx = x^{m_1} \int x^{-b/a} \cdot x^{-2m_1} dx ] Since the roots are equal, (m_1 = -\frac{b-a}{2a}), so (-2m_1 = \frac{b-a}{a}). This gives: [ y_2 = x^{m_1} \int x^{-b/a} \cdot x^{(b-a)/a} dx = x^{m_1} \int x^{-1} dx = x^{m_1} \ln x ] The general solution is: [ y = c_1 x^{m_1} + c_2 x^{m_1} \ln x ] For higher-order equations, if (m_1) is a root of multiplicity (k), the (k) linearly independent solutions are (x^{m_1}, x^{m_1} \ln x, x^{m_1} (\ln x)^2, \ldots, x^{m_1} (\ln x)^{k-1}).

📌 Example 2: Solve (4x^2 \frac{d^2 y}{dx^2} + 8x \frac{dy}{dx} + y = 0).

Assume (y = x^m). Substituting: [ 4x^2 \cdot m(m-1)x^{m-2} + 8x \cdot m x^{m-1} + x^m = x^m (4m(m-1) + 8m + 1) = x^m (4m^2 + 4m + 1) = 0 ] The auxiliary equation is (4m^2 + 4m + 1 = 0) or ((2m+1)^2 = 0), so the repeated root is (m = -1/2). The general solution is: [ y = c_1 x^{-1/2} + c_2 x^{-1/2} \ln x ]

26.1.4 Case III (Conjugate Complex Roots)

If the roots are a conjugate pair (m_1 = \alpha + i\beta, m_2 = \alpha - i\beta) with (\alpha, \beta > 0) real, the solution is: [ y = c_1 x^{\alpha + i\beta} + c_2 x^{\alpha - i\beta} ] To express this in terms of real functions, we use the identity: [ x^{i\beta} = (e^{\ln x})^{i\beta} = e^{i\beta \ln x} = \cos(\beta \ln x) + i \sin(\beta \ln x) ] Similarly, (x^{-i\beta} = \cos(\beta \ln x) - i \sin(\beta \ln x)). Adding and subtracting gives: [ x^{i\beta} + x^{-i\beta} = 2 \cos(\beta \ln x), \quad x^{i\beta} - x^{-i\beta} = 2i \sin(\beta \ln x) ] Choosing appropriate constants ((c_1 = c_2 = 1) and (c_1 = 1, c_2 = -1)), we obtain two real linearly independent solutions: [ y_1 = x^\alpha \cos(\beta \ln x), \quad y_2 = x^\alpha \sin(\beta \ln x) ] The general solution in real form is: [ y = x^\alpha \left[ c_1 \cos(\beta \ln x) + c_2 \sin(\beta \ln x) \right] ]

🔑 Definition — Auxiliary Equation for Cauchy-Euler: The equation (am^2 + (b-a)m + c = 0) obtained by substituting (y = x^m) into the Cauchy-Euler equation (ax^2 y'' + bx y' + cy = 0). 📐 Formula (General Solution for Complex Roots): (y = x^\alpha [c_1 \cos(\beta \ln x) + c_2 \sin(\beta \ln x)]), where (\alpha \pm i\beta) are the complex conjugate roots of the auxiliary equation. 💡 Why this matters: The use of (\ln x) in the solution (rather than (x) as in constant-coefficient cases) reflects the variable coefficient structure and is crucial for correctly modeling physical systems where the independent variable is a radial distance or other non-uniform coordinate.

⭐ Key Takeaways

The Cauchy-Euler equation is a special type of variable-coefficient ODE that can be solved by assuming a solution of the form (y = x^m), leading to an auxiliary equation. The nature of the roots—distinct real, repeated real, or complex conjugate—determines the form of the general solution. For distinct real roots (m_1) and (m_2), the general solution is a linear combination of (x^{m_1}) and (x^{m_2}). For repeated roots, a second solution containing (\ln x) is obtained via reduction of order. For complex conjugate roots (\alpha \pm i\beta), the real-valued solution involves (x^\alpha) multiplied by (\cos(\beta \ln x)) and (\sin(\beta \ln x)). This method extends naturally to higher-order equations and to non-homogeneous problems.

🧠 Quick Revision Questions

1. What is the standard form of a second-order homogeneous Cauchy-Euler equation?
2. How do you find the auxiliary equation for a Cauchy-Euler equation?
3. What is the general solution when the auxiliary equation has two distinct real roots?
4. How do you construct a second linearly independent solution when the auxiliary equation has a repeated real root?
5. Write the general solution in real form when the auxiliary equation has complex conjugate roots (\alpha \pm i\beta).

📘 Lecture 27 — Cauchy-Euler Equation (Alternative Method of Solution)

📖 Overview: This lecture introduces an alternative method for solving Cauchy-Euler differential equations by reducing them to differential equations with constant coefficients through the substitution (x = e^t) or (t = \ln x). This approach simplifies the solving process, especially for higher-order equations, by transforming variable coefficients into constant coefficients, allowing standard solution techniques to be applied.

🗂️ Topics Covered

This lecture covers the alternative method of solving Cauchy-Euler equations using the substitution (x = e^t), which transforms the equation into one with constant coefficients. It introduces new differential operators (\Delta = d/dt) and the operational relationships (xD = \Delta), (x^2 D^2 = \Delta(\Delta - 1)), etc. Several examples are solved, including second-order homogeneous equations with distinct real, repeated, and complex roots, as well as a third-order equation and a non-homogeneous equation solved via variation of parameters after transformation.

📝 Lecture Summary

The Substitution Method

The Cauchy-Euler equation can be reduced to a differential equation with constant coefficients through the substitution: [ x = e^t \quad \text{or} \quad t = \ln x ]

Using the chain rule, the derivatives transform as: [ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \cdot \frac{dy}{dt} ] [ \frac{d^2 y}{dx^2} = \frac{1}{x^2} \cdot \frac{d^2 y}{dt^2} - \frac{1}{x^2} \cdot \frac{dy}{dt} ]

Therefore: [ x\frac{dy}{dx} = \frac{dy}{dt} ] [ x^2\frac{d^2 y}{dx^2} = \frac{d^2 y}{dt^2} - \frac{dy}{dt} ]

Introducing the notation: [ D = \frac{d}{dx}, \quad D^2 = \frac{d^2}{dx^2}, \quad \Delta = \frac{d}{dt}, \quad \Delta^2 = \frac{d^2}{dt^2} ]

We obtain the operational relationships: [ xD = \Delta ] [ x^2 D^2 = \Delta^2 - \Delta = \Delta(\Delta - 1) ] [ x^3 D^3 = \Delta(\Delta - 1)(\Delta - 2) ] [ x^4 D^4 = \Delta(\Delta - 1)(\Delta - 2)(\Delta - 3) ]

💡 Why this matters: This substitution transforms any Cauchy-Euler equation into a constant-coefficient differential equation in terms of (t), which can then be solved using standard methods with (y = e^{mt}).

Example 1: Distinct Real Roots

Solve: (x^2 \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} - 4y = 0)

Solution: Write the equation as ((x^2 D^2 - 2xD - 4)y = 0).

With substitution (x = e^t), using (xD = \Delta) and (x^2 D^2 = \Delta(\Delta - 1)): [ [\Delta(\Delta - 1) - 2\Delta - 4]y = 0 ] [ (\Delta^2 - 3\Delta - 4)y = 0 ] [ \frac{d^2 y}{dt^2} - 3\frac{dy}{dt} - 4y = 0 ]

Substitute (y = e^{mt}): [ (m^2 - 3m - 4)e^{mt} = 0 ] [ m^2 - 3m - 4 = 0 ] [ (m + 1)(m - 4) = 0 \implies m = -1, 4 ]

🔑 Definition — Auxiliary equation: (m^2 - 3m - 4 = 0) with distinct real roots.

The solution in terms of (t) is: [ y = c_1 e^{-t} + c_2 e^{4t} ]

Since (x = e^t), the solution in terms of (x) is: [ y = c_1 x^{-1} + c_2 x^{4} ]

Example 2: Repeated Real Roots

Solve: (4x^2 \frac{d^2 y}{dx^2} + 8x \frac{dy}{dx} + y = 0)

Solution: Write as ((4x^2 D^2 + 8xD + 1)y = 0).

With (x = e^t), (xD = \Delta), (x^2 D^2 = \Delta(\Delta - 1)): [ (4\Delta(\Delta - 1) + 8\Delta + 1)y = 0 ] [ (4\Delta^2 + 4\Delta + 1)y = 0 ] [ 4\frac{d^2 y}{dt^2} + 4\frac{dy}{dt} + y = 0 ]

Substitute (y = e^{mt}): [ (4m^2 + 4m + 1)e^{mt} = 0 ] [ 4m^2 + 4m + 1 = 0 ] [ (2m + 1)^2 = 0 \implies m = -\frac{1}{2}, -\frac{1}{2} \text{ (repeated)} ]

📐 Formula: For repeated roots, the solution is (y = (c_1 + c_2 t)e^{mt}).

Solution in terms of (t): [ y = (c_1 + c_2 t)e^{-\frac{1}{2}t} ]

Since (t = \ln x): [ y = (c_1 + c_2 \ln x)x^{-\frac{1}{2}} ] [ y = c_1 x^{-\frac{1}{2}} + c_2 x^{-\frac{1}{2}} \ln x ]

Example 3: Complex Roots (Initial Value Problem)

Solve: (x^2 \frac{d^2 y}{dx^2} + 3x \frac{dy}{dx} + 3y = 0), (y(1) = 1), (y'(1) = -5)

Solution: Write as ((x^2 D^2 + 3xD + 3)y = 0).

With (x = e^t), (xD = \Delta), (x^2 D^2 = \Delta(\Delta - 1)): [ (\Delta(\Delta - 1) + 3\Delta + 3)y = 0 ] [ (\Delta^2 + 2\Delta + 3)y = 0 ] [ \frac{d^2 y}{dt^2} + 2\frac{dy}{dt} + 3y = 0 ]

Substitute (y = e^{mt}): [ m^2 + 2m + 3 = 0 ] [ m = \frac{-2 \pm \sqrt{4 - 12}}{2} = \frac{-2 \pm \sqrt{-8}}{2} = -1 \pm i\sqrt{2} ]

📌 Example: For complex roots (\alpha \pm i\beta), the solution is (y = e^{\alpha t}(c_1 \cos \beta t + c_2 \sin \beta t)).

Here (\alpha = -1), (\beta = \sqrt{2}), so: [ y = e^{-t}(c_1 \cos \sqrt{2}t + c_2 \sin \sqrt{2}t) ] [ y = x^{-1}(c_1 \cos(\sqrt{2}\ln x) + c_2 \sin(\sqrt{2}\ln x)) ]

Apply initial conditions:

• (y(1) = 1): (1 = c_1 \cos 0 + c_2 \sin 0 \implies c_1 = 1)
• Find (y'): [ y' = -x^{-2}(c_1 \cos(\sqrt{2}\ln x) + c_2 \sin(\sqrt{2}\ln x)) + x^{-2}(-\sqrt{2}c_1 \sin(\sqrt{2}\ln x) + \sqrt{2}c_2 \cos(\sqrt{2}\ln x)) ]
• (y'(1) = -5): (-5 = -(1) + \sqrt{2}c_2 \implies \sqrt{2}c_2 = -4 \implies c_2 = -2\sqrt{2})

Solution: (y = x^{-1}[\cos(\sqrt{2}\ln x) - 2\sqrt{2}\sin(\sqrt{2}\ln x)])

Example 4: Third-Order Cauchy-Euler Equation

Solve: (x^3 \frac{d^3 y}{dx^3} + 5x^2 \frac{d^2 y}{dx^2} + 7x \frac{dy}{dx} + 8y = 0)

Solution: Write as ((x^3 D^3 + 5x^2 D^2 + 7xD + 8)y = 0).

With (x = e^t), using: [ xD = \Delta, \quad x^2 D^2 = \Delta(\Delta - 1), \quad x^3 D^3 = \Delta(\Delta - 1)(\Delta - 2) ]

[ [\Delta(\Delta - 1)(\Delta - 2) + 5\Delta(\Delta - 1) + 7\Delta + 8]y = 0 ] [ (\Delta^3 - 3\Delta^2 + 2\Delta + 5\Delta^2 - 5\Delta + 7\Delta + 8)y = 0 ] [ (\Delta^3 + 2\Delta^2 + 4\Delta + 8)y = 0 ] [ \frac{d^3 y}{dt^3} + 2\frac{d^2 y}{dt^2} + 4\frac{dy}{dt} + 8y = 0 ]

Substitute (y = e^{mt}): [ m^3 + 2m^2 + 4m + 8 = 0 ] [ (m^2 + 4)(m + 2) = 0 \implies m = -2, \pm 2i ]

Solution: [ y = c_1 e^{-2t} + c_2 \cos 2t + c_3 \sin 2t ] [ y = c_1 x^{-2} + c_2 \cos(2\ln x) + c_3 \sin(2\ln x) ]

Example 5: Non-Homogeneous Equation (Using Alternative Method)

Solve: (x^2 y'' - 3xy' + 3y = 2x^4 e^x)

Solution: First solve the homogeneous part: (x^2 y'' - 3xy' + 3y = 0).

Write as ((x^2 D^2 - 3xD + 3)y = 0).

With (x = e^t), (xD = \Delta), (x^2 D^2 = \Delta(\Delta - 1)): [ [\Delta(\Delta - 1) - 3\Delta + 3]y = 0 ] [ (\Delta^2 - 4\Delta + 3)y = 0 ] [ \frac{d^2 y}{dt^2} - 4\frac{dy}{dt} + 3y = 0 ]

Substitute (y = e^{mt}): [ m^2 - 4m + 3 = 0 \implies (m - 3)(m - 1) = 0 \implies m = 1, 3 ]

Complementary solution: [ y_c = c_1 e^t + c_2 e^{3t} = c_1 x + c_2 x^3 ]

For the particular solution, use variation of parameters. First write the original equation in standard form: [ y'' - \frac{3}{x}y' + \frac{3}{x^2}y = 2x^2 e^x ]

Here (f(x) = 2x^2 e^x), (y_1 = x), (y_2 = x^3).

Wronskian: [ W = \begin{vmatrix} x & x^3 \ 1 & 3x^2 \end{vmatrix} = 3x^3 - x^3 = 2x^3 ]

[ W_1 = \begin{vmatrix} 0 & x^3 \ 2x^2 e^x & 3x^2 \end{vmatrix} = 0 - 2x^5 e^x = -2x^5 e^x ]

[ W_2 = \begin{vmatrix} x & 0 \ 1 & 2x^2 e^x \end{vmatrix} = 2x^3 e^x - 0 = 2x^3 e^x ]

Then: [ u_1' = \frac{W_1}{W} = \frac{-2x^5 e^x}{2x^3} = -x^2 e^x ] [ u_2' = \frac{W_2}{W} = \frac{2x^3 e^x}{2x^3} = e^x ]

Integrate: [ u_1 = \int -x^2 e^x dx = -(x^2 - 2x + 2)e^x ] [ u_2 = \int e^x dx = e^x ]

Particular solution: [ y_p = u_1 y_1 + u_2 y_2 = [-(x^2 - 2x + 2)e^x]x + e^x \cdot x^3 ] [ y_p = -x^3 e^x + 2x^2 e^x - 2xe^x + x^3 e^x = 2x^2 e^x - 2xe^x ]

General solution: [ y = y_c + y_p = c_1 x + c_2 x^3 + 2x^2 e^x - 2x e^x ]

⭐ Key Takeaways

The alternative method using the substitution (x = e^t) or (t = \ln x) transforms any Cauchy-Euler equation into a constant-coefficient differential equation, which can be solved using standard techniques. The key operational relationships are (xD = \Delta), (x^2 D^2 = \Delta(\Delta - 1)), (x^3 D^3 = \Delta(\Delta - 1)(\Delta - 2)), and so on. After solving the transformed equation in terms of (t), substitute back using (x = e^t) to obtain the solution in terms of (x). This method works for homogeneous equations with all types of roots (distinct real, repeated real, and complex) as well as non-homogeneous equations when combined with variation of parameters.

🧠 Quick Revision Questions

1. What substitution transforms a Cauchy-Euler equation into a constant-coefficient equation?
2. What are the operational relationships for (xD), (x^2 D^2), and (x^3 D^3) in terms of (\Delta)?
3. How do you handle repeated roots when using this alternative method?
4. In Example 3, how were the constants (c_1) and (c_2) determined from the initial conditions?
5. After finding the solution in terms of (t), how do you convert it back to the solution in terms of (x)?

📘 Lecture 28 — Power Series (An Introduction)

📖 Overview: This lecture introduces power series as a standard technique for solving linear differential equations with variable coefficients. It covers the fundamental concepts of power series, including convergence, divergence, the Ratio Test, and how to determine the interval and radius of convergence—essential tools for finding series solutions to differential equations.

🗂️ Topics Covered

The lecture begins by defining power series in (x - a) and their components. It then explains convergence and divergence of power series, introduces the Ratio Test for determining convergence, and describes the interval of convergence and radius of convergence. Finally, it discusses how to investigate convergence at the endpoints of the interval.

📝 Lecture Summary

28 Power Series (An Introduction)

A standard technique for solving linear differential equations with variable coefficients is to find a solution as an infinite series, often in the form of a power series. This lecture discusses important facts about power series, but an in-depth review of infinite series concepts should be consulted in a standard calculus text.

28.1 Power Series

A power series in ((x - a)) is an infinite series of the form:

[ \sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1 (x - a) + c_2 (x - a)^2 + \ldots ]

The coefficients (c_0, c_1, c_2, \ldots) and (a) are constants, and (x) represents a variable. In this discussion, we consider only real numbers for the coefficients, (x), and (a). The number (a) is known as the centre of the power series.

📌 Example 1: The infinite series (\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} x^n = x - \frac{x^2}{2^2} + \frac{x^3}{3^2} - \ldots) is a power series in (x) centered at zero.

28.2 Convergence and Divergence

If we choose a specified value of the variable (x), the power series becomes an infinite series of constants. If, for the given (x), the sum of terms of the power series equals a finite real number, the series is said to be convergent at (x). A power series that is not convergent is said to be divergent, meaning the sum of its terms is not a finite real number.

📌 Example 2: (a) The series (\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots) converges at (x=1) to the number (e). (b) The series (\sum_{n=0}^{\infty} n!(x+2)^n = 1 + (x+2) + 2!(x+2)^2 + 3!(x+2)^3 + \ldots) diverges for all (x) except (x=-2).

28.2.1 The Ratio Test

To determine for which values of (x) a power series converges, one can often use the Ratio Test. For a power series (\sum_{n=0}^{\infty} a_n = \sum_{n=0}^{\infty} c_n (x - a)^n), if: [ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} |x - a| \cdot \left| \frac{c_{n+1}}{c_n} \right| = L ] Then:

• The power series converges absolutely for values of (x) where (L < 1).
• The power series diverges for values of (x) where (L > 1) or (L = \infty).
• The test is inconclusive for values of (x) where (L = 1).

28.2.2 Interval of Convergence

The set of all real values of (x) for which a power series (\sum_{n=0}^{\infty} c_n (x - a)^n) converges is known as the interval of convergence of the power series.

28.2.3 Radius of Convergence

Consider a power series (\sum_{n=0}^{\infty} c_n (x - a)^n). Exactly one of three possibilities is true:

• The series converges only at its center (x = a).
• The series converges for all values of (x).
• There is a number (R > 0) such that the series converges absolutely for all (x) satisfying (|x - a| < R) and diverges for (|x - a| > R). The series converges for (x \in (a - R, a + R)) and diverges outside this interval.

The number (R) is called the radius of convergence. If the first possibility holds, then (R = 0). In the second case, we write (R = \infty). From the Ratio Test, the radius of convergence is given by: 🔑 Definition — Radius of Convergence (R): [ R = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| ] provided the limit exists.

28.2.4 Convergence at an Endpoint

If the radius of convergence is (R > 0), the interval of convergence is one of the following: ((a - R, a + R)), ((a - R, a + R]), ([a - R, a + R)), or ([a - R, a + R]). To determine which interval applies, separate investigations are needed for the numbers (x = a - R) and (x = a + R).

📌 Example 3: Consider the power series (\sum_{n=1}^{\infty} \frac{1}{n} x^n). Using the Ratio Test: [ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| = \lim_{n \to \infty} \frac{n}{n+1} |x| = |x| ] The series converges absolutely for (|x| < 1), i.e., the interval ((-1, 1)), and diverges for (|x| > 1). To check the endpoints:

• At (x = 1): (\sum_{n=1}^{\infty} \frac{1}{n}) (the harmonic series), which diverges.
• At (x = -1): The series would be (\sum_{n=1}^{\infty} \frac{(-1)^n}{n}), an alternating harmonic series, which converges conditionally.

⭐ Key Takeaways

The most critical concept from this lecture is understanding the definition of a power series and how to determine its convergence behavior. You must be able to apply the Ratio Test to find the radius of convergence (R) and the open interval of convergence. Remember that the ratio test is inconclusive at the endpoints, so each endpoint must be tested separately by substitution. The ultimate goal is to find the interval of convergence, which is the set of (x) values for which the series converges.

🧠 Quick Revision Questions

1. What is the general form of a power series centered at (a)?
2. What does the Ratio Test tell us about the convergence of a power series?
3. How do you calculate the radius of convergence (R) from the coefficients of a power series?
4. If the radius of convergence is (R = 2) and the center is (a = 0), what is the open interval of convergence?
5. Why must the endpoints of the interval of convergence be tested separately?

📘 Lecture 30 — Power Series Solution of a Differential Equation

📖 Overview: This lecture focuses on the concept of power series solutions for differential equations, covering the interval and radius of convergence, representation of functions as power series, and the arithmetic operations that can be performed on them. It culminates with the direct method of finding a power series solution to a first-order linear differential equation, a technique analogous to the method of undetermined coefficients.

🗂️ Topics Covered

The lecture begins with finding the interval of convergence for power series using the ratio test, including endpoint analysis. It then explains absolute convergence within that interval and how a power series defines a function. Key theorems on the term-by-term differentiation and integration of power series are stated. The concept of a function being analytic at a point is introduced, followed by arithmetic operations (addition, multiplication, division) on power series. The section concludes with techniques for shifting summation indices to combine series. Finally, a direct method for solving differential equations using power series is demonstrated.

📝 Lecture Summary

Interval and Radius of Convergence

To find the interval of convergence for a power series, we first apply the Ratio Test to determine the radius of convergence (R). The series converges absolutely for (|x-a| < R). The convergence at the endpoints (x = a \pm R) must be checked separately by substituting these values into the original series to get a series of constants, which is then tested using standard convergence tests like the Alternating Series Test or the p-series test.

🔑 Definition — Interval of Convergence: The set of all x for which a power series converges. It is an interval centered at (a) with radius (R), which may be open, closed, or half-open at its endpoints. 📐 Formula: (R = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right|) or (R = \lim_{n \to \infty} \sqrt[n]{|c_n|}) → This formula gives the radius of the interval where the power series converges absolutely. 📌 Example 4: Find interval of convergence for (\sum_{n=1}^{\infty} \frac{(x-3)^n}{2^n n}).

• Step 1: Find radius. (R = \lim_{n \to \infty} \left| \frac{(x-3)^{n} / (2^n n)}{(x-3)^{n+1} / (2^{n+1} (n+1))} \right| = \lim_{n \to \infty} \frac{2^{n+1}(n+1)}{2^n n} = 2).
• Step 2: Test interval. The series converges for (|x-3| < 2 \Rightarrow 1 < x < 5).
• Step 3: Test endpoints. At (x=1), the series is (\sum \frac{(-2)^n}{2^n n} = \sum \frac{(-1)^n}{n}), which converges by the Alternating Series Test. At (x=5), the series is (\sum \frac{(2)^n}{2^n n} = \sum \frac{1}{n}), the harmonic series, which diverges.
• Result: The interval of convergence is ([1, 5)).

Absolute Convergence

Within its interval of convergence (and excluding the endpoints), a power series converges absolutely. This means that the series formed by taking the absolute values of its terms also converges for all x in the open interval ((a-R, a+R)).

Power Series Representation of Functions

A power series (\sum_{n=0}^{\infty} c_n (x-a)^n) defines a function (f(x)) on its interval of convergence. We say the power series is a representation of the function on that domain.

Theorem (Term-by-Term Differentiation and Integration)

If a power series (\sum c_n (x-a)^n) has a radius of convergence (R>0) and defines a function (f) on ((a-R, a+R)), then (f) is continuous, differentiable, and integrable on that interval. Its derivative and integral can be found by differentiating and integrating the series term-by-term.

• Derivative: (f'(x) = \sum_{n=1}^{\infty} n c_n (x-a)^{n-1})
• Integral: (\int f(x) dx = C + \sum_{n=0}^{\infty} \frac{c_n (x-a)^{n+1}}{n+1})

The new series have the same radius of convergence (R), but their convergence at the endpoints may differ from the original series.

📌 Example 5: Find the function represented by (1 - x + x^2 - x^3 + \dots).

• This is a geometric series with first term (a=1) and common ratio (r=-x).
• It converges for (|x| < 1), and its sum is (S = \frac{a}{1-r} = \frac{1}{1-(-x)} = \frac{1}{1+x}).
• Therefore, (f(x) = \frac{1}{1+x}). This is the power series representation for this function.

Series that are Identically Zero

If a power series (\sum_{n=0}^{\infty} c_n (x-a)^n = 0) for all x in some interval with (R>0), then all its coefficients must be zero: (c_n = 0) for all (n).

Analytic at a Point

A function (f) is said to be analytic at a point (a) if it can be represented by a power series in ((x-a)) with a positive radius of convergence. 💡 Why this matters: This concept is crucial for finding power series solutions to differential equations. If a function is analytic at a point, we can assume a solution exists as a power series centered at that point. 📌 Example 6: The functions (e^x), (\cos x), and (\ln(1+x)) are all analytic at (x=0) because each can be represented by a convergent Maclaurin series (a power series in (x)).

Arithmetic of Power Series

Power series can be combined through addition, multiplication, and division, similar to polynomials, as long as they have a common interval of convergence. This involves adding coefficients of like powers, using the distributive law, or performing long division. 📌 Example 7: If (f(x) = \sum c_n x^n) and (g(x) = \sum b_n x^n) converge for (|x| < R), then:

• (f(x) + g(x) = \sum (c_n + b_n) x^n)
• (f(x) \cdot g(x) = c_0 b_0 + (c_0 b_1 + c_1 b_0)x + (c_0 b_2 + c_1 b_1 + c_2 b_0)x^2 + \dots)

📌 Example 8: Find first four terms of power series for (e^x \cos x).

• Use known Maclaurin series: (e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \dots) and (\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots).
• Multiply and collect like terms:
  • Constant: (1 \cdot 1 = 1)
  • (x^1): (1 \cdot x = x)
  • (x^2): (1 \cdot (-\frac{x^2}{2}) + \frac{x^2}{2} \cdot 1 = 0)
  • (x^3): (x \cdot (-\frac{x^2}{2}) + \frac{x^3}{6} \cdot 1 = -\frac{x^3}{2} + \frac{x^3}{6} = -\frac{x^3}{3})
  • (x^4): (1 \cdot \frac{x^4}{24} + \frac{x^2}{2} \cdot (-\frac{x^2}{2}) + \frac{x^4}{24} \cdot 1 = \frac{x^4}{24} - \frac{x^4}{4} + \frac{x^4}{24} = -\frac{x^4}{6})
• Result: (e^x \cos x = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + \dots)

📌 Example 9: Find first four terms of power series for (\sec x).

• Use series for (\cos x) and perform long division: (\sec x = 1 / \cos x).
• Dividing (1) by ( (1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots) ) yields: (1 + \frac{x^2}{2} + \frac{5x^4}{24} + \frac{61x^6}{720} + \dots)

Shifting the Summation Index

When adding two or more power series to form a single series, the indices must be the same, and the powers of (x) must be "in phase". This is achieved by re-indexing the series. A common method is to let (k = n + \text{constant}). 📌 Example 10: Write (\sum_{n=1}^{\infty} 2n c_n x^{n-1} + \sum_{n=0}^{\infty} 6n c_n x^{n+1}) as one power series.

• To combine, we want a common power of (x) (e.g., (x^k)) and a common starting index.
• For the first series, let (k = n-1), so (n = k+1). When (n=1), (k=0). The series becomes (2c_1 + \sum_{k=1}^{\infty} 2(k+1) c_{k+1} x^k).
• For the second series, let (k = n+1), so (n = k-1). When (n=0), (k=1). The series becomes (\sum_{k=1}^{\infty} 6(k-1) c_{k-1} x^k).
• Combine: (2c_1 + \sum_{k=1}^{\infty} [2(k+1)c_{k+1} + 6(k-1)c_{k-1}] x^k).

Power Series Solution of a Differential Equation

A power series solution to a differential equation is found by assuming a solution of the form (y = \sum_{n=0}^{\infty} c_n x^n) (a power series centered at (x=0), for simplicity). We then substitute this series and its derivatives into the differential equation. By shifting indices, combining series, and equating coefficients of like powers of (x) to zero, we can find a recurrence relation for the coefficients (c_n). This allows us to determine the series representation of the solution. 📌 Example 11: Find a power series solution of (\frac{dy}{dx} - 2xy = 0).

• Assumption: Let (y = \sum_{n=0}^{\infty} c_n x^n).
• Substitute into DE: (\sum_{n=1}^{\infty} n c_n x^{n-1} - \sum_{n=0}^{\infty} 2 c_n x^{n+1} = 0).
• Shift Indices: To get the same power of (x) (e.g., (x^k)), let (k=n-1) in the first series and (k=n+1) in the second. This yields: ( \sum_{k=0}^{\infty} (k+1)c_{k+1} x^k - \sum_{k=1}^{\infty} 2 c_{k-1} x^k = 0).
• Write out the first term of the first series to match the starting index: ((1)c_1 + \sum_{k=1}^{\infty} (k+1)c_{k+1} x^k - \sum_{k=1}^{\infty} 2 c_{k-1} x^k = 0).
• Combine Series: (c_1 + \sum_{k=1}^{\infty} [(k+1)c_{k+1} - 2c_{k-1}] x^k = 0).
• Equate Coefficients:
  • For the constant term: (c_1 = 0).
  • For all (k \ge 1): ((k+1)c_{k+1} - 2c_{k-1} = 0), which gives the recurrence relation: (c_{k+1} = \frac{2c_{k-1}}{k+1}).
• Find Coefficients:
  • (c_1 = 0).
  • For (k=1): (c_2 = \frac{2c_0}{2} = c_0).
  • For (k=2): (c_3 = \frac{2c_1}{3} = 0).
  • For (k=3): (c_4 = \frac{2c_2}{4} = \frac{c_0}{2}).
  • For (k=4): (c_5 = \frac{2c_3}{5} = 0).
  • For (k=5): (c_6 = \frac{2c_4}{6} = \frac{c_0}{6}).
• Solution: The solution is (y = c_0 (1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \dots) = c_0\sum_{n=0}^{\infty} \frac{x^{2n}}{n!} = c_0 e^{x^2}), which matches the known solution.

⭐ Key Takeaways

The most critical concepts from this lecture are the methods for finding the interval of convergence of a power series and the direct technique for solving a differential equation by assuming a power series solution. You must be able to determine the radius of convergence using the Ratio Test and test the endpoints. The core of solving a DE with power series involves substituting the assumed series, shifting indices, combining sums into a single series, and then using the principle that all coefficients of the resulting polynomial in (x) must be zero to generate a recurrence relation. This recurrence relation is the key to finding the coefficients of the solution series. Understanding the arithmetic of series and the condition for a function to be analytic provides a foundation for why this method works.

🧠 Quick Revision Questions

1. What are the steps to find the interval of convergence for a power series?
2. What does it mean for a function to be analytic at a point, and why is this important for solving differential equations?
3. How do you shift the summation index in a power series to combine it with another series?
4. In the power series solution method, what is the purpose of equating coefficients of like powers of (x) to zero?
5. Starting from the recurrence relation derived in Example 10 ((c_{k+1} = \frac{2c_{k-1}}{k+1})), what is the value of (c_7) in terms of (c_0)?

📘 Lecture 30 — Solution about Ordinary Points

📖 Overview: This lecture introduces the concept of analytic functions and defines ordinary and singular points for differential equations. It provides a foundational theorem for finding power series solutions about ordinary points, demonstrating the method through detailed examples to ensure convergence and solution generation.

🗂️ Topics Covered

The lecture covers definitions of analytic functions, ordinary and singular points, the existence theorem for power series solutions, and detailed step-by-step solutions of differential equations using power series about an ordinary point, including examples with polynomial coefficients and complex singular points.

📝 Lecture Summary

30.1 Analytic Function

A function f is said to be analytic at a point a if it can be represented by a power series in (x-a) with a positive radius of convergence. This concept is crucial for determining whether a power series solution to a differential equation exists at a given point.

Suppose the linear second-order differential equation a₂(x) y'' + a₁(x) y' + a₀(x) y = 0 (1) is put into the form y'' + P(x) y' + Q(x) y = 0 (2) by dividing by the leading coefficient a₂(x).

30.2 Ordinary and Singular Points

A point x₀ is said to be an ordinary point of a differential equation (1) if both P(x) and Q(x) are analytic at x₀. A point that is not an ordinary point is said to be a singular point of the equation.

💡 Why this matters: Identifying ordinary points tells us where we can guarantee a power series solution exists, while singular points are locations where the solution may behave differently or not exist in a simple power series form.

30.2.1 Polynomial Coefficients

If a₂(x), a₁(x), and a₀(x) are polynomials with no common factors, then x = x₀ is: (i) an ordinary point if a₂(x) ≠ 0 (ii) a singular point if a₂(x) = 0

Examples: (a) The singular points of the equation (x² – 1) y'' + 2xy' + 6y = 0 are the solutions of x² – 1 = 0 or x = ±1. All other finite values of x are the ordinary points.

(b) The singular points need not be real numbers. The equation (x² + 1) y'' + 2xy' + 6y = 0 has the singular points at the solutions of x² + 1 = 0, namely, x = ±i. All other finite values, real or complex, are ordinary points.

(c) The Cauchy-Euler equation ax² y'' + bxy' + cy = 0, where a, b, and c are constants, has a singular point at x = 0. All other finite values of x, real or complex, are ordinary points.

30.3 Theorem (Existence of Power Series Solution)

If x = x₀ is an ordinary point of the differential equation y'' + P(x) y' + Q(x) y = 0, we can always find two linearly independent solutions in the form of power series centered at x₀: y = Σ cₙ (x – x₀)ⁿ, n=0 to ∞ A series solution converges at least for |x – x₀| < R, where R is the distance from x₀ to the closest singular point (real or complex).

💡 Why this matters: This theorem guarantees the existence of a power series solution and gives us the radius of convergence, which is determined by the distance to the nearest singular point. This is a powerful tool for solving equations that cannot be solved by elementary methods.

Example 1: Solve y'' – 2xy = 0. Solution: We see that x = 0 is an ordinary point of the equation. Since there are no finite singular points, there exist two solutions of the form y = Σ cₙ xⁿ, n=0 to ∞, convergent for |x| < ∞. Proceeding, we write y' = Σ n cₙ xⁿ⁻¹, n=1 to ∞ and y'' = Σ n(n-1) cₙ xⁿ⁻², n=2 to ∞. Substituting gives: y'' – 2xy = Σ n(n-1)cₙ xⁿ⁻² – Σ 2cₙ xⁿ⁺¹ = 0 Letting k = n-2 in the first series and k = n+1 in the second, we have: 2c₂ + Σ [(k+2)(k+1)cₖ₊₂ – 2cₖ₋₁] xᵏ = 0, k=1 to ∞ Thus, 2c₂ = 0 and (k+2)(k+1)cₖ₊₂ – 2cₖ₋₁ = 0, k=1,2,3,... The last expression gives the recurrence relation: cₖ₊₂ = 2cₖ₋₁ / [(k+2)(k+1)], k=1,2,3,... Iteration gives: c₂ = 0, c₃ = 2c₀/(3·2), c₄ = 2c₁/(4·3), c₅ = 2c₂/(5·4) = 0 c₆ = 2c₃/(6·5) = 2²c₀/(6·5·3·2), c₇ = 2c₄/(7·6) = 2²c₁/(7·6·4·3) c₈ = 2c₅/(8·7) = 0, c₉ = 2c₆/(9·8) = 2³c₀/(9·8·6·5·3·2) c₁₀ = 2c₇/(10·9) = 2³c₁/(10·9·7·6·4·3), and so on. Both c₀ and c₁ are arbitrary. Thus, y = c₀[1 + (2/3·2)x³ + (2²/6·5·3·2)x⁶ + (2³/9·8·6·5·3·2)x⁹ + ...] + c₁[x + (2/4·3)x⁴ + (2²/7·6·4·3)x⁷ + (2³/10·9·7·6·4·3)x¹⁰ + ...]

Example 2: Solve (x² + 1) y'' + xy' – y = 0. Solution: Since the singular points are x = ±i, x = 0 is the ordinary point, a power series will converge at least for |x| < 1. The assumption y = Σ cₙ xⁿ, n=0 to ∞ leads to: (x²+1) Σ n(n-1)cₙ xⁿ⁻² + x Σ n cₙ xⁿ⁻¹ – Σ cₙ xⁿ = 0 After simplification and shifting indices, we get: 2c₂ – c₀ + 6c₃x + Σ [(k+1)(k-1)cₖ + (k+2)(k+1)cₖ₊₂] xᵏ = 0, k=2 to ∞ Thus, 2c₂ – c₀ = 0, c₃ = 0, and (k+1)(k-1)cₖ + (k+2)(k+1)cₖ₊₂ = 0. This implies: c₂ = c₀/2, c₃ = 0, and cₖ₊₂ = -(k-1)cₖ/(k+2), k=2,3,... Iteration gives: c₄ = -c₂/4 = -c₀/(2·4), c₅ = -2c₃/5 = 0 c₆ = -3c₄/6 = (1·3)c₀/(2·4·6) = c₀/(2³·3!), c₇ = -4c₅/7 = 0 c₈ = -5c₆/8 = -(3·5)c₀/(2·4·6·8) = -(1·3·5)c₀/(2⁴·4!) c₁₀ = -7c₈/10 = (3·5·7)c₀/(2·4·6·8·10) = (1·3·5·7)c₀/(2⁵·5!), and so on. Therefore, y = c₁x + c₀[1 + (1/2)x² – (1/2²·2!)x⁴ + (1·3/2³·3!)x⁶ – (1·3·5/2⁴·4!)x⁸ + (1·3·5·7/2⁵·5!)x¹⁰ – ...] The solutions are y₁(x) = c₀[1 + (1/2)x² + Σ (-1)ⁿ⁻¹(1·3·5...(2n-3))/(2ⁿ·n!) x²ⁿ, n=2 to ∞], |x| < 1 and y₂(x) = c₁x.

Example 3: If we seek a solution y = Σ cₙ xⁿ, n=0 to ∞ for the equation y'' – (1+x) y = 0, we obtain c₂ = c₀/2 and the three-term recurrence relation: cₖ₊₂ = (cₖ + cₖ₋₁) / [(k+1)(k+2)], k=1,2,3,... To simplify the iteration we can first choose c₀ ≠ 0, c₁ = 0; this yields one solution. The other solution follows from next choosing c₀ = 0, c₁ ≠ 0.

⭐ Key Takeaways

The most critical concepts from this lecture are: the definition of an analytic function as one representable by a power series; the identification of ordinary points (where both P(x) and Q(x) are analytic) versus singular points (which are solutions to a₂(x)=0 for polynomials); and the existence theorem that guarantees two linearly independent power series solutions centered at an ordinary point, with convergence radius equal to the distance to the nearest singular point. The practical method involves assuming a solution y = Σ cₙ xⁿ, substituting into the differential equation, shifting indices to combine series, and using the resulting recurrence relation to determine the coefficients iteratively, with c₀ and c₁ serving as arbitrary constants for the general solution.

🧠 Quick Revision Questions

1. What is the definition of an analytic function at a point a?
2. How do you determine if a point x₀ is an ordinary point or a singular point of a differential equation with polynomial coefficients?
3. According to the existence theorem, what guarantees the radius of convergence for a power series solution about an ordinary point?
4. In the recurrence relation for a power series solution, which coefficients typically remain arbitrary?
5. How does the presence of complex singular points affect the interval of convergence for a power series solution about a real ordinary point?

📘 Lecture 32 — Solutions about Singular Points

📖 Overview: This lecture covers the theory of solving differential equations about singular points, distinguishing between regular and irregular singular points. It introduces the Method of Frobenius for finding series solutions about regular singular points, including how to handle different cases of indicial roots. Understanding these concepts is crucial for solving equations that cannot be solved by ordinary power series methods.

🗂️ Topics Covered

The lecture covers regular and irregular singular points, the definition and identification of regular singular points using polynomial coefficients and standard form, the Method of Frobenius and Frobenius' Theorem, the process of finding indicial equations and recurrence relations, and three distinct cases of indicial roots (roots not differing by an integer).

📝 Lecture Summary

31.1 Regular and Irregular Singular Points

If x = x₀ is a singular point of the equation a₂(x)y″ + a₁(x)y′ + a₀(x)y = 0, it is not always possible to find a solution of the form y = Σ cₙ(x – x₀)ⁿ. However, we may be able to find a solution of the form y = Σ cₙ(x – x₀)ⁿ⁺ʳ, where r is a constant to be determined. To define singular points, we first put the equation into standard form: y″ + P(x)y′ + Q(x)y = 0.

A singular point x = x₀ is said to be a regular singular point if both (x – x₀)P(x) and (x – x₀)²Q(x) are analytic at x₀. A singular point that is not regular is called an irregular singular point.

🔑 Definition — Regular Singular Point: A singular point x₀ where the functions (x – x₀)P(x) and (x – x₀)²Q(x) are both analytic (have convergent power series representations) at x₀.

📌 Example 1 (Regular vs. Irregular): Consider (x² – 4)²y″ + (x – 2)y′ + y = 0. Dividing by (x – 2)²(x + 2)² gives P(x) = 1/[(x – 2)(x + 2)²] and Q(x) = 1/[(x – 2)²(x + 2)²]. For x = 2: (x – 2) appears to the 1st power in P and 2nd in Q, so x = 2 is a regular singular point. For x = –2: (x + 2) appears to the 2nd power in P, violating the condition, so x = –2 is an irregular singular point.

💡 Why this matters: The classification determines whether the Method of Frobenius can be applied. Only regular singular points guarantee at least one series solution of the Frobenius form.

31.1.1 Polynomial Coefficients

When the coefficients a₂(x), a₁(x), a₀(x) are polynomials with no common factors, the definition simplifies. Let a₂(x₀) = 0. Form P(x) and Q(x) by reducing a₁(x)/a₂(x) and a₀(x)/a₂(x) to lowest terms. If the factor (x – x₀) appears at most to the first power in the denominator of P(x) and at most to the second power in the denominator of Q(x), then x = x₀ is a regular singular point.

🔑 Definition — Polynomial Coefficients Rule: For polynomial coefficients with no common factors, x₀ is a regular singular point if the reduced form of P(x) has (x – x₀) at most to the 1st power and Q(x) has (x – x₀) at most to the 2nd power.

📌 Example 2: For x²(x + 1)²y″ + (x² – 1)y′ + 2y = 0, singular points are x = 0 and x = –1. The standard form gives P(x) = (x – 1)/[x²(x + 1)] and Q(x) = 2/[x²(x + 1)²]. For x = 0: (x – 0) appears to the 2nd power in P, so x = 0 is an irregular singular point. For x = –1: (x + 1) appears to the 1st power in P and 2nd in Q, so x = –1 is a regular singular point.

31.2 Method of Frobenius

31.2.1 Frobenius' Theorem

If x = x₀ is a regular singular point of a₂(x)y″ + a₁(x)y′ + a₀(x)y = 0, then there exists at least one series solution of the form: y = (x – x₀)ʳ Σ cₙ(x – x₀)ⁿ = Σ cₙ(x – x₀)ⁿ⁺ʳ where r is a constant to be determined. The series converges at least on some interval 0 < |x – x₀| < R.

🔑 Definition — Frobenius' Theorem: Guarantees at least one series solution of the form Σ cₙ(x – x₀)ⁿ⁺ʳ exists about any regular singular point.

Procedure (Method of Frobenius):

1. Identify the regular singular point x₀.
2. Substitute y = Σ cₙ(x – x₀)ⁿ⁺ʳ into the differential equation.
3. Determine the unknown exponent r and coefficients cₙ.
4. For simplicity, assume x₀ = 0.

📌 Example 4: Solve 3xy″ + y′ – y = 0 about x = 0 (regular singular point). Assume y = Σ cₙxⁿ⁺ʳ. After substitution and collecting terms, we get r(3r – 2)c₀ = 0 (the indicial equation) and recurrence relation: c_{k+1} = c_k / [(k + r + 1)(3k + 3r + 1)]. The indicial roots are r = 2/3, 0. For r₁ = 2/3, the recurrence is c_{k+1} = c_k / [(3k + 5)(k + 1)], giving y₁ = c₀x^{2/3}[1 + Σ xⁿ / (n!·5·8·11·...·(3n+2))]. For r₂ = 0, the recurrence is c_{k+1} = c_k / [(k + 1)(3k + 1)], giving y₂ = c₀[1 + Σ xⁿ / (n!·1·4·7·...·(3n–2))]. Both converge for all finite x and are linearly independent.

📌 Example (Method Fails for Second Solution): For xy″ + 3y′ – y = 0, x = 0 is a regular singular point. Substituting y = Σ cₙxⁿ⁺ʳ gives indicial equation r(r + 2) = 0, with roots r₁ = 0, r₂ = –2. For r₁ = 0, we get one solution y₁ = c₀ Σ [2xⁿ / n!(n+2)!]. For r₂ = –2, the recurrence relation (k – 1)(k + 1)c_{k+1} – c_k = 0 causes c₀ = 0 and c₁ = 0. The resulting y₂ is just a constant multiple of y₁. The method of Frobenius gives only one series solution here.

31.3 Cases of Indicial Roots

When using the method of Frobenius, we distinguish three cases based on the nature of the indicial roots r₁ and r₂. Let r₁ denote the largest root.

31.3.1 Case I (Roots Not Differing by an Integer)

If r₁ and r₂ are distinct and do not differ by an integer, then there exist two linearly independent solutions: y₁ = Σ cₙxⁿ⁺ʳ¹ (c₀ ≠ 0) and y₂ = Σ bₙxⁿ⁺ʳ² (b₀ ≠ 0).

🔑 Definition — Case I: When indicial roots are distinct and their difference is not a positive integer, two independent Frobenius series solutions exist.

📌 Example 6: Solve 2xy″ + (1 + x)y′ + y = 0. Substituting y = Σ cₙxⁿ⁺ʳ gives indicial equation r(2r – 1) = 0, with roots r₁ = 1/2, r₂ = 0. Their difference is 1/2, not an integer, so we are in Case I. Substituting each root gives two linearly independent Frobenius series solutions.

⭐ Key Takeaways

A singular point is regular if (x – x₀)P(x) and (x – x₀)²Q(x) are analytic, or equivalently for polynomial coefficients, if (x – x₀) appears at most to the 1st power in P and 2nd in Q. The Method of Frobenius, which assumes a solution of the form y = Σ cₙ(x – x₀)ⁿ⁺ʳ, is used to solve differential equations about regular singular points. The indicial equation, obtained from the lowest power of (x – x₀), determines the exponent r. When indicial roots differ by an integer, the method may fail to provide two independent solutions, requiring alternative methods such as the use of logarithmic terms.

🧠 Quick Revision Questions

1. What is the standard form of a second-order linear differential equation used for classifying singular points?
2. State the conditions for x₀ to be a regular singular point in terms of P(x) and Q(x).
3. What is the indicial equation and how is it obtained in the Method of Frobenius?
4. In Case I of indicial roots, what condition ensures two linearly independent Frobenius series solutions exist?
5. What happens when the indicial roots differ by an integer in the Method of Frobenius?

📘 Lecture 33 — Solutions about Singular Points (Continued)

📖 Overview: This lecture continues the Method of Frobenius for solving differential equations about singular points, focusing on Cases II and III where indicial roots differ by a positive integer or are equal. It explains when a second solution must contain a logarithmic term and provides worked examples demonstrating these scenarios, which is essential for finding the general solution of second-order linear ODEs with regular singular points.

🗂️ Topics Covered

The lecture covers Cases II and III of the Method of Frobenius when roots differ by a positive integer (with possible logarithmic second solution) and when roots are equal (always yielding a logarithmic term). It provides three detailed examples: Example 7 (roots differing by 7, polynomial and infinite series solutions), Example 8/9 (Bessel-type equation solved using the reduction formula for second solution), and Example 10 (equal indicial roots leading to a logarithmic second solution).

📝 Lecture Summary

32 Solutions about Singular Points

32.1 Method of Frobenius (Cases II and III)

When solving a second-order linear ODE about a regular singular point using the Frobenius method, we assume a solution of the form:

$$ y = \sum_{n=0}^{\infty} c_n (x - x_0)^{n+r} $$

The roots of the indicial equation determine the behavior. When roots differ by a positive integer or are equal, special treatment is needed.

32.1.1 Case II (Roots Differing by a Positive Integer)

If $r_1 - r_2 = N$, where $N$ is a positive integer, then there exist two linearly independent solutions:

• First solution: $y_1 = \sum_{n=0}^{\infty} c_n x^{n+r_1}, \quad c_0 \neq 0$
• Second solution: $y_2 = C y_1(x) \ln x + \sum_{n=0}^{\infty} b_n x^{n+r_2}, \quad b_0 \neq 0$ Where $C$ is a constant that could be zero.

Equal Indicial Roots:

If $r_1 = r_2$, there always exist two linearly independent solutions:

• $y_1 = \sum_{n=0}^{\infty} c_n x^{n+r_1}, \quad c_0 \neq 0$
• $y_2 = y_1(x) \ln x + \sum_{n=1}^{\infty} b_n x^{n+r_1}$

🔑 Definition — Indicial Equation: The quadratic equation obtained from the lowest power of $x$ in the Frobenius method, giving the possible values of $r$.

Example 7: $xy'' + (x-6)y' - 3y = 0$

Solution: Using $y = \sum c_n x^{n+r}$, we obtain the indicial equation $r(r-7)=0$, so $r_1 = 7$, $r_2 = 0$, and $r_1 - r_2 = 7$.

The recurrence relation is: $$(k+r+1)(k+r-6)c_{k+1} + (k+r-3)c_k = 0$$

For smaller root $r_2 = 0$: $$(k+1)(k-6)c_{k+1} + (k-3)c_k = 0$$ $$c_{k+1} = -\frac{k-3}{(k+1)(k-6)}c_k$$

When $k=0,1,2,3$:

• $c_1 = -\frac{1}{2}c_0$
• $c_2 = -\frac{1}{5}c_1 = \frac{1}{10}c_0$
• $c_3 = -\frac{1}{12}c_2 = -\frac{1}{120}c_0$
• $c_4 = c_5 = c_6 = 0$

For $k \geq 7$: $$c_n = (-1)^{n+1}\frac{4\cdot5\cdot6\cdots(n-4)}{(n-7)!,8\cdot9\cdot10\cdots n}c_7, \quad n=8,9,10,\ldots$$

With $c_7 = 0$ and $c_0 \neq 0$, we get the polynomial solution: $$y_1 = c_0\left[1 - \frac{1}{2}x + \frac{1}{10}x^2 - \frac{1}{120}x^3\right]$$

With $c_7 \neq 0$ and $c_0 = 0$, we get an infinite series: $$y_2 = c_7\left[x^7 + \sum_{k=1}^{\infty} (-1)^k\frac{4\cdot5\cdot6\cdots(k+3)}{k!,8\cdot9\cdot10\cdots(k+7)}x^{k+7}\right]$$

The general solution on $(0,\infty)$ is $Y = c_1 y_1(x) + c_2 y_2(x)$.

💡 Why this matters: This example shows that when roots differ by a positive integer, the constant $C$ in the second solution can be zero, yielding a pure series solution. It also demonstrates the strategy of working with the smaller root first.

Example 8 & 9: $xy'' + 3y' - y = 0$

Solution: The method of Frobenius provides only one solution: $$y_1 = \sum_{n=0}^{\infty} \frac{2}{n!(n+2)!}x^n = 1 + \frac{1}{3}x + \frac{1}{24}x^2 + \frac{1}{360}x^3 + \cdots$$

📐 Formula — Reduction of Order: $$y_2 = y_1(x) \int \frac{e^{-\int P(x)dx}}{y_1^2(x)}dx$$ This formula gives a second solution when one solution is known.

Using this formula with $P(x) = \frac{3}{x}$: $$y_2 = y_1(x) \int \frac{e^{-3\ln x}}{y_1^2(x)}dx = y_1(x) \int \frac{dx}{x^3[1 + \frac{1}{3}x + \frac{1}{24}x^2 + \cdots]^2}$$

After expansion and integration: $$y_2 = y_1(x)\left[-\frac{1}{2x^2} + \frac{2}{3x} + \frac{1}{4}\ln x - \frac{19}{270}x + \cdots\right]$$

The general solution is: $$y = c_1 y_1(x) + c_2\left[\frac{1}{4}y_1(x)\ln x + y_1(x)\left(-\frac{1}{2x^2} + \frac{2}{3x} - \frac{19}{270}x + \cdots\right)\right]$$

Alternatively, the second solution can be found directly by assuming $y_2 = y_1(x)\ln x + \sum b_n x^{n+r_2}$ and solving for the $b_n$ coefficients: $$y_2 = y_1(x)\ln x + b_0 x^{-2} + b_1 x^{-1} + b_2 + b_3 x + \cdots$$

💡 Why this matters: This example demonstrates that when the second series-type solution fails, reduction of order always provides an alternative method to find the second linearly independent solution.

📌 Example: For $k=0$ in equation (14): $2 + 0 \cdot 2b_2 - b_1 = 0$, so $b_1 = 2$, $b_0 = -2$, with $b_2$ arbitrary.

Example 10: $xy'' + y' - 4y = 0$

Solution: Using $y = \sum c_n x^{n+r}$: $$\sum_{n=0}^{\infty}(n+r)^2 c_n x^{n+r-1} - 4\sum_{n=0}^{\infty}c_n x^{n+r} = 0$$

The indicial equation is $r^2 = 0$, so the indicial roots are equal: $r_1 = r_2 = 0$.

The recurrence relation is: $$(k+r+1)^2 c_{k+1} - 4c_k = 0, \quad k=0,1,2,\ldots$$

With $r=0$: $$c_{k+1} = \frac{4c_k}{(k+1)^2}$$

The first solution is: $$y_1 = c_0 \sum_{n=0}^{\infty} \frac{4^n}{(n!)^2} x^n, \quad |x| < \infty$$

Using reduction of order: $$y_2 = y_1(x) \int \frac{e^{-\int \frac{1}{x}dx}}{y_1^2(x)}dx = y_1(x) \int \frac{1}{x\left[1 + 4x + 4x^2 + \frac{16}{9}x^3 + \cdots\right]}dx$$

After expansion: $$y_2 = y_1(x) \int \left[\frac{1}{x} - 8 + 40x - \frac{1472}{9}x^2 + \cdots\right]dx$$ $$y_2 = y_1(x)\left[\ln x - 8x + 20x^2 - \frac{1472}{27}x^3 + \cdots\right]$$

🔑 Definition — Case III: When the indicial roots are equal ($r_1 = r_2$), the second solution always contains a logarithmic term of the form $y_1(x)\ln x$.

⭐ Key Takeaways

For differential equations with regular singular points, when indicial roots differ by a positive integer (Case II), the second solution may or may not contain a logarithmic term depending on the constant $C$, and when solving it is always advisable to work with the smaller root first. When indicial roots are equal (Case III), the second solution always contains a logarithmic term. Both cases can be handled either by directly assuming a solution of the form $y_2 = y_1(x)\ln x + \sum b_n x^{n+r_2}$ or by using the reduction of order formula $y_2 = y_1(x) \int e^{-\int P(x)dx}/y_1^2(x)dx$, which always provides a second linearly independent solution.

🧠 Quick Revision Questions

1. What form does the second solution take when the indicial roots differ by a positive integer?
2. Why is it always better to work with the smaller root first in Case II?
3. What form does the second solution always take when the indicial roots are equal?
4. Using the reduction of order formula, find the general form of $y_2$ for the equation $xy'' + y' - 4y = 0$.
5. In Example 7, why were $c_4$, $c_5$, and $c_6$ all equal to zero?

📘 Lecture 34 — Bessel’s Differential Equation

📖 Overview: This lecture introduces Bessel’s differential equation, a second-order linear ODE that arises frequently in applied mathematics, physics, and engineering. It covers the method of Frobenius to obtain series solutions, defines Bessel functions of the first kind, and derives important recurrence relations and special cases including spherical Bessel functions.

🗂️ Topics Covered

The lecture begins with the definition of Bessel’s differential equation and its series solution using the method of Frobenius. It then derives the indicial equation and recurrence relations for coefficients, introduces Bessel functions of the first kind (J_v(x)) and (J_{-v}(x)), and discusses the general solution depending on whether the indicial roots differ by an integer. Several examples are worked out, including finding general solutions for specific (v) values and deriving recurrence relations like (xJ_v'(x) = vJ_v(x) - xJ_{v+1}(x)) and (2J_n'(x) = J_{n-1}(x) - J_{n+1}(x)). The lecture concludes by deriving explicit expressions for (J_{1/2}(x)) and (J_{-1/2}(x)), identifying them as spherical Bessel functions.

📝 Lecture Summary

33 Bessel’s Differential Equation

A second-order linear differential equation of the form: [ x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - v^2)y = 0 ] is called Bessel’s differential equation. The solution is usually denoted by (J_v(x)) and is known as Bessel’s function. This equation occurs frequently in advanced studies in applied mathematics, physics, and engineering.

💡 Why this matters: Bessel functions are essential for solving problems with cylindrical symmetry, such as heat conduction in cylinders, wave propagation, and vibrations of circular membranes.

33.1 Series Solution of Bessel’s Differential Equation

To solve Bessel’s equation using the method of Frobenius, we assume a series solution: [ y = \sum_{n=0}^{\infty} C_n x^{n+r} ] Substituting into the differential equation and collecting terms yields the indicial equation: [ r^2 - v^2 = 0 ] Thus the indicial roots are (r_1 = v) and (r_2 = -v).

🔑 Definition — Indicial roots: The roots of the indicial equation, which determine the form of the series solution near a regular singular point.

For (r_1 = v), the recurrence relation for coefficients becomes: [ C_{2n} = \frac{-C_{2n-2}}{2^2 n (n+v)}, \quad n = 1,2,3,... ] With the standard choice (C_0 = \frac{1}{2^v \Gamma(1+v)}), where (\Gamma) is the Gamma function satisfying (\Gamma(1+\alpha) = \alpha\Gamma(\alpha)), we obtain: [ C_{2n} = \frac{(-1)^n}{2^{2n+v} n! \Gamma(1+v+n)}, \quad n = 0,1,2,... ]

📐 Formula: The series solution becomes: [ y = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(1+v+n)} \left(\frac{x}{2}\right)^{2n+v} ]

33.2 Bessel’s Function of the First Kind

For (r_1 = v), the Bessel function of the first kind of order (v) is defined as: [ J_v(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(1+v+n)} \left(\frac{x}{2}\right)^{2n+v} ] For (r_2 = -v), we have: [ J_{-v}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(1-v+n)} \left(\frac{x}{2}\right)^{2n-v} ]

🔑 Definition — Bessel function of the first kind: The series solution of Bessel’s equation for a given order (v), denoted (J_v(x)).

The general solution depends on (v):

• If (v = 0), then (J_0(x)) and (J_{-0}(x)) are the same function
• If (v > 0) and (2v) is not a positive integer, then (J_v(x)) and (J_{-v}(x)) are linearly independent and the general solution is: [ y = C_1 J_v(x) + C_2 J_{-v}(x) ]
• If (2v) is a positive integer, a second series solution may exist

Example 1: Find the general solution of (x^2 y'' + xy' + \left(x^2 - \frac{1}{4}\right)y = 0) on ((0, \infty)).

🔑 Solution: Comparing with Bessel’s equation, (v^2 = \frac{1}{4}), so (v = \pm \frac{1}{2}). The general solution is: [ y = C_1 J_{1/2}(x) + C_2 J_{-1/2}(x) ]

Example 2: Find the general solution of (x^2 y'' + xy' + \left(x^2 - \frac{1}{9}\right)y = 0).

🔑 Solution: Here (v^2 = \frac{1}{9}), so (v = \pm \frac{1}{3}). The general solution is: [ y = C_1 J_{1/3}(x) + C_2 J_{-1/3}(x) ]

Example 3: Derive the formula (xJ_v'(x) = vJ_v(x) - xJ_{v+1}(x)).

Solution: Starting from the series definition of (J_v(x)): [ xJ_v'(x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2n+v)}{n! \Gamma(1+v+n)} \left(\frac{x}{2}\right)^{2n+v} ] [ = v \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(1+v+n)} \left(\frac{x}{2}\right)^{2n+v} + 2 \sum_{n=0}^{\infty} \frac{(-1)^n n}{n! \Gamma(1+v+n)} \left(\frac{x}{2}\right)^{2n+v} ] The first sum is (vJ_v(x)). For the second, let (k = n-1): [ = vJ_v(x) - x \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \Gamma(2+v+k)} \left(\frac{x}{2}\right)^{2k+v+1} = vJ_v(x) - xJ_{v+1}(x) ]

📐 Formula: (xJ_v'(x) = vJ_v(x) - xJ_{v+1}(x))

Example 4: Derive the recurrence relation (2J_n'(x) = J_{n-1}(x) - J_{n+1}(x)).

Solution: By expanding (J_n'(x)) from its series definition and splitting the term ((n+2s) = (n+s) + s): [ J_n'(x) = \frac{1}{2} J_{n-1}(x) + \frac{1}{2} \sum_{s=1}^{\infty} \frac{(-1)^s}{(s-1)! (n+s)!} \left(\frac{x}{2}\right)^{n+2s-1} ] Letting (p = s-1) in the second term: [ J_n'(x) = \frac{1}{2} J_{n-1}(x) - \frac{1}{2} J_{n+1}(x) ]

📐 Formula: (2J_n'(x) = J_{n-1}(x) - J_{n+1}(x))

Example 5: Derive expressions for (J_{1/2}(x)) and (J_{-1/2}(x)).

Solution: Using the series definition with Gamma function properties and (\Gamma(1/2) = \sqrt{\pi}):

For (n = 1/2): [ J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right) = \sqrt{\frac{2}{\pi x}} \sin x ]

For (n = -1/2): [ J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}} \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \right) = \sqrt{\frac{2}{\pi x}} \cos x ]

🔑 Definition — Spherical Bessel functions: Bessel functions of index half an odd integer are called spherical Bessel functions. Like other Bessel functions, they are used in many physical problems.

📌 Example summary: (J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin x) and (J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}} \cos x)

⭐ Key Takeaways

The most critical concepts from this lecture are: Bessel’s differential equation (x^2 y'' + xy' + (x^2 - v^2)y = 0) has indicial roots (r = \pm v). The Bessel function of the first kind (J_v(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(1+v+n)} \left(\frac{x}{2}\right)^{2n+v}) represents one series solution. When (2v) is not an integer, (J_v(x)) and (J_{-v}(x)) are linearly independent and form the general solution. Important recurrence relations include (xJ_v'(x) = vJ_v(x) - xJ_{v+1}(x)) and (2J_n'(x) = J_{n-1}(x) - J_{n+1}(x)). For half-integer orders, spherical Bessel functions simplify to trigonometric forms: (J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin x) and (J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}} \cos x).

🧠 Quick Revision Questions

1. What is the indicial equation for Bessel’s differential equation, and what are its roots?
2. Under what condition are (J_v(x)) and (J_{-v}(x)) linearly independent solutions?
3. Derive the recurrence relation (2J_n'(x) = J_{n-1}(x) - J_{n+1}(x)) from the series definition.
4. What are the explicit expressions for (J_{1/2}(x)) and (J_{-1/2}(x)) in terms of elementary functions?
5. What is the standard choice for (C_0) in the Frobenius series solution of Bessel’s equation, and why?

📘 Lecture 34 — Legendre’s Differential Equation

📖 Overview: This lecture introduces Legendre’s differential equation, a second-order linear ODE with variable coefficients that arises frequently in physics and engineering, particularly in problems with spherical symmetry. The lecture covers the power series solution of this equation, leading to Legendre polynomials, and explores their key properties including the Rodrigues formula, generating function, recurrence relations, and orthogonality.

🗂️ Topics Covered

The lecture covers Legendre’s differential equation and its power series solution, the definition and construction of Legendre polynomials as terminating series solutions for non-negative integer n, the Rodrigues formula for generating these polynomials, the generating function that produces them as coefficients of a binomial expansion, a three-term recurrence relation derived from the generating function, and finally a proof of the orthogonality property of Legendre polynomials over the interval [-1,1].

📝 Lecture Summary

34 Legendre’s Differential Equation

Legendre’s differential equation is a second order linear differential equation of the form: [ (1-x^2)y'' - 2xy' + n(n+1)y = 0 ] Any solution of this equation is called a Legendre’s function. If (n) is a positive integer, the solution is called a Legendre’s polynomial of degree (n) and is denoted by (P_n(x)).

💡 Why this matters: This equation is fundamental in solving Laplace's equation in spherical coordinates, where (n) represents the degree of the spherical harmonic.

🔑 Definition — Legendre’s Differential Equation: ((1-x^2)y'' - 2xy' + n(n+1)y = 0), a second-order linear ODE with variable coefficients.

We assume a power series solution of the form: [ y = \sum_{k=0}^{\infty} C_k x^k ]

Substituting and simplifying gives the recurrence relation: [ C_{j+2} = -\frac{(n-j)(n+j+1)}{(j+2)(j+1)} C_j, \quad j=0,1,2,\ldots ]

This yields two linearly independent power series solutions for (|x| < 1): [ y_1(x) = C_0 \left[ 1 - \frac{n(n+1)}{2!}x^2 + \frac{(n-2)n(n+1)(n+3)}{4!}x^4 - \frac{(n-4)(n-2)n(n+1)(n+3)(n+5)}{6!}x^6 + \ldots \right] ] [ y_2(x) = C_1 \left[ x - \frac{(n-1)(n+2)}{3!}x^3 + \frac{(n-3)(n-1)(n+2)(n+4)}{5!}x^5 - \frac{(n-5)(n-3)(n-1)(n+2)(n+4)(n+6)}{7!}x^7 + \ldots \right] ]

If (n) is an even integer, the series for (y_1(x)) terminates, becoming a polynomial. If (n) is odd, the series for (y_2(x)) terminates. By convention, specific values of (C_0) and (C_1) are chosen to produce the standard Legendre polynomials.

For (n=0,2,4,\ldots): [ C_0 = (-1)^{n/2} \frac{1 \cdot 3 \cdot \ldots \cdot (n-1)}{2 \cdot 4 \cdot \ldots \cdot n} ]

For (n=1,3,5,\ldots): [ C_1 = (-1)^{(n-1)/2} \frac{1 \cdot 3 \cdot \ldots \cdot n}{2 \cdot 4 \cdot \ldots \cdot (n-1)} ]

📌 Example: For (n=4), the polynomial solution becomes: [ y_1(x) = C_0 \left[1 - 10x^2 + \frac{35}{3}x^4\right] ] With (C_0 = (-1)^2 \frac{1\cdot 3}{2\cdot 4} = \frac{3}{8}), we get: [ P_4(x) = \frac{3}{8} - \frac{30}{8}x^2 + \frac{35}{8}x^4 = \frac{1}{8}(35x^4 - 30x^2 + 3) ]

34.1 Legendre’s Polynomials

Legendre’s Polynomials are specific (n)th degree polynomial solutions of Legendre’s equation, denoted by (P_n(x)). The first several are:

• (P_0(x) = 1)
• (P_1(x) = x)
• (P_2(x) = \frac{1}{2}(3x^2 - 1))
• (P_3(x) = \frac{1}{2}(5x^3 - 3x))
• (P_4(x) = \frac{1}{8}(35x^4 - 30x^2 + 3))
• (P_5(x) = \frac{1}{8}(63x^5 - 70x^3 + 15x))

Each (P_n(x)) satisfies Legendre’s differential equation for its respective (n). For example, (P_0(x)=1) satisfies ((1-x^2)y'' - 2xy' = 0) for (n=0).

34.2 Rodrigues Formula for Legendre’s Polynomials

The Legendre polynomials can be generated using Rodrigues formula: [ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2 - 1)^n ]

🔑 Definition — Rodrigues Formula: (P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2 - 1)^n), a compact formula for generating Legendre polynomials of any degree.

34.3 Generating Function for Legendre’s Polynomials

The Legendre polynomials are the coefficients of (z^n) in the expansion of: [ \phi = (1 - 2xz + z^2)^{-1/2} ] in ascending powers of (z). Using the binomial series expansion: [ (1 - 2xz + z^2)^{-1/2} = \sum_{n=0}^{\infty} P_n(x) z^n ]

🔑 Definition — Generating Function: ((1 - 2xz + z^2)^{-1/2} = \sum_{n=0}^{\infty} P_n(x) z^n), which generates all Legendre polynomials as coefficients.

Expanding the generating function yields: [ (1 - 2xz + z^2)^{-1/2} = 1 + xz + \frac{1}{2}(3x^2 - 1)z^2 + \frac{1}{2}(5x^3 - 3x)z^3 + \frac{1}{8}(35x^4 - 30x^2 + 3)z^4 + \ldots ]

Equating coefficients with (\sum P_n(x)z^n) confirms the expressions for the Legendre polynomials.

34.4 Recurrence Relation

Starting from the generating function and differentiating with respect to (t), the following three-term recurrence relation is derived: [ (k+1)P_{k+1}(x) - (2k+1)xP_k(x) + kP_{k-1}(x) = 0, \quad k = 2,3,4,\ldots ]

🔑 Definition — Recurrence Relation: ((k+1)P_{k+1}(x) = (2k+1)xP_k(x) - kP_{k-1}(x)), a relation connecting three consecutive Legendre polynomials.

This relation allows computation of higher-degree Legendre polynomials from lower-degree ones, which is computationally efficient.

34.5 Orthogonality of Legendre’s Polynomials

The Legendre polynomials satisfy an orthogonality property over the interval ([-1, 1]) with respect to the weight function (w(x) = 1).

Starting from Legendre’s differential equation for (P_n) and (P_m), after algebraic manipulation and integration, we obtain: [ (n-m)(n+m+1) \int_{-1}^1 P_m(x) P_n(x) dx = 0 ]

Since (n \neq m) and both are non-negative, this gives: [ \int_{-1}^1 P_m(x) P_n(x) dx = 0 \quad \text{for } m \neq n ]

🔑 Definition — Orthogonality: (\int_{-1}^1 P_m(x) P_n(x) dx = 0) for (m \neq n), meaning Legendre polynomials are orthogonal on ([-1,1]) with weight function (w(x)=1).

💡 Why this matters: Orthogonality allows Legendre polynomials to be used as a basis for series expansions (Fourier-Legendre series), analogous to Fourier series but on a finite interval.

⭐ Key Takeaways

The most critical points from this lecture are: Legendre’s differential equation ((1-x^2)y'' - 2xy' + n(n+1)y = 0) has polynomial solutions called Legendre polynomials (P_n(x)) when (n) is a non-negative integer. These polynomials can be generated by Rodrigues formula (P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2 - 1)^n) or as coefficients in the generating function ((1 - 2xz + z^2)^{-1/2} = \sum P_n(x)z^n). The first six Legendre polynomials ((P_0) through (P_5)) must be memorized. They satisfy the recurrence relation ((k+1)P_{k+1}(x) = (2k+1)xP_k(x) - kP_{k-1}(x)), and most importantly, they are orthogonal on ([-1,1]) with (\int_{-1}^1 P_m(x)P_n(x)dx = 0) for (m \neq n), which is the foundation for expanding functions in Legendre series.

🧠 Quick Revision Questions

1. Write down Legendre’s differential equation and identify the parameter (n).
2. State Rodrigues formula for generating Legendre polynomials and use it to derive (P_2(x)).
3. What is the generating function for Legendre polynomials? Write the first four terms of its expansion.
4. State the recurrence relation relating (P_{k-1}(x)), (P_k(x)), and (P_{k+1}(x)).
5. State the orthogonality property of Legendre polynomials over the interval ([-1, 1]).

📘 Lecture 34.6 — Normality condition for Legendre’ Polynomials / 35 Systems of Linear Differential Equations

📖 Overview: This lecture consists of two separate parts. The first part (Section 34.6) completes the study of Legendre polynomials by establishing the normality condition and the orthogonality condition for these polynomials, culminating in a precise formula for the integral of their square. The second part (Section 35) introduces the theory and solution of systems of simultaneous linear differential equations using the systematic elimination (operator) method, which is essential for modeling coupled physical systems like multiple masses on springs or parallel electrical circuits.

🗂️ Topics Covered

Derivation of the normality condition for Legendre polynomials using the generating function; the orthogonality condition expressed with the Kronecker delta; introduction to systems of linear differential equations; the systematic elimination (operator) method for solving such systems; worked examples demonstrating the elimination procedure and the reduction of arbitrary constants.

📝 Lecture Summary

34.6 Normality condition for Legendre’ Polynomials

The normality condition for Legendre polynomials is derived by exploiting the generating function. We start with the generating function and its conjugate, multiply them, and integrate over x from -1 to 1. This process leads to an infinite series on one side and an integral on the other. By performing the integration on the left-hand side and simplifying the logarithmic terms, we obtain a power series in t. The key step is to equate the coefficients of ( t^{2n} ) on both sides of the resulting equation. This directly yields the normality condition: the integral of the square of the Legendre polynomial ( P_n(x) ) from -1 to 1 equals ( 2/(2n+1) ). This derivation also naturally shows that the integral of the product of two different Legendre polynomials ( P_m(x) ) and ( P_n(x) ) (where ( m \neq n )) is zero, which is the orthogonality condition.

🔑 Definition — Normality Condition: The condition that the integral of the square of a set of functions, when weighted appropriately (or with weight function w(x)=1), yields a non-zero constant. For Legendre polynomials, this constant is ( 2/(2n+1) ).

📐 Formula: Normalization Integral for Legendre Polynomials [ \int_{-1}^1 [P_n(x)]^2 , dx = \frac{2}{2n+1} ] → The integral of the square of the nth Legendre polynomial over the interval [-1, 1] is equal to the constant ( 2/(2n+1) ).

📌 Example: The derivation involves equating the coefficient of ( t^{2n} ). For instance, for n=2, the coefficient of ( t^4 ) on the RHS of the expanded sum is ( 2/5 ). On the LHS, this corresponds to the term where m=n=2, giving ( \int_{-1}^1 [P_2(x)]^2 , dx \cdot t^4 ). By equating the coefficients, we get ( \int_{-1}^1 [P_2(x)]^2 , dx = 2/5 ).

💡 Why this matters: The normality condition is crucial for using Legendre polynomials in Fourier-Legendre series expansions, where the coefficients are calculated using this formula.

Remark: The lecture notes explicitly state that the orthogonality condition for ( P_n(x) ) can be written using the Kronecker delta ( \delta_{m,n} ). 📐 Formula: Orthogonality Condition for Legendre Polynomials [ \int_{-1}^1 P_m(x) P_n(x) , dx = \left( \frac{2}{2n+1} \right) \delta_{m,n} ] where ( \delta_{m,n} = 0 ) if ( m \neq n ) and ( \delta_{m,n} = 1 ) if ( m = n ).

🔑 Definition — Kronecker delta: A function of two variables, usually integers, which is 1 if they are equal, and 0 otherwise.

34.7 Exercise

This section provides a set of practice problems to solidify the concepts of Legendre polynomials and their properties.

1. Alternative Form of Legendre’s Equation: Show that Legendre’s equation can be written in the self-adjoint form ( \frac{d}{dx} \left[ (1-x^2) \frac{dy}{dx} \right] + n(n+1)y = 0 ).
2. Transformation to Legendre’s Equation: Show that the equation ( \sin \theta \frac{d^2 y}{d\theta^2} + \cos \theta \frac{dy}{d\theta} + n(n+1) (\sin \theta) y = 0 ) can be transformed into Legendre’s equation by substituting ( x = \cos \theta ).
3. Evaluating the Integral of the Square: Use the explicit Legendre polynomials ( P_0(x), P_1(x), P_2(x), ) and ( P_3(x) ) to evaluate ( \int_{-1}^1 P_n^2 dx ) for n = 0, 1, 2, 3. Then, generalize the results (this yields the normality formula).
4. Evaluating the Orthogonality Integral: Use the explicit Legendre polynomials to evaluate ( \int_{-1}^1 P_n(x) P_m(x) dx ) for ( n \neq m ). Generalize the results (this yields the orthogonality relation).
5. Rodrigues’ Formula: Verify the Rodrigues’ formula ( P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n ) for n = 0, 1, 2, 3.

35 Systems of Linear Differential Equations

This section introduces the concept of systems of simultaneous differential equations. While a single spring-mass system is modeled by a single second-order ODE, coupling two or more masses with springs requires two or more simultaneous differential equations. Similarly, networks of parallel electrical circuits are modeled by such systems. The lecture focuses on the theory and solution of systems of linear differential equations with constant coefficients.

35.1 Simultaneous Differential Equations

These involve two or more equations that contain derivatives of two or more unknown functions of a single independent variable (usually t). The differential operator D is defined as ( D = d/dt ), allowing the equations to be written in a compact, algebraic form. For example, ( D^2 x = d^2 x / dt^2 ).

🔑 Definition — System of Simultaneous Differential Equations: A set of differential equations involving two or more unknown functions of the same independent variable.

📌 Example: The equations ( d^2 x / dt^2 = -5x + y ) and ( d^2 y / dt^2 = 3x - y ) form a system of two second-order equations for the unknown functions x(t) and y(t). The equations ( x' - 3x + y' + z' = 5 ) and ( x + y' - 6z' = t - 1 ) are a system for x(t), y(t), and z(t).

35.2 Solution of a System

A solution is a set of differentiable functions ( x = f(t), y = g(t), \dots ) that satisfy each equation in the system on some interval I.

35.2.1 Systematic Elimination (Operator Method)

This is the primary method used in this lecture. It is based on the process of systematically eliminating dependent variables to obtain a single linear differential equation in one of them.

Step-by-Step Procedure:

1. Write in Operator Form: Convert all equations into operator form using D.
2. Eliminate: Choose a dependent variable to eliminate by operating on the equations with appropriate operator combinations (like D, D-3, etc.) and adding or subtracting.
3. Solve for First Variable: The elimination yields a single linear ODE with constant coefficients for one retained variable. Solve this ODE to find the first variable as a function of t (e.g., y(t)).
4. Eliminate Again: Eliminate all other variables (or the same variable as before) to get a single linear ODE for a second dependent variable (e.g., x(t)).
5. Solve for Second Variable: Solve this ODE. The solution will have its own set of arbitrary constants.
6. Reduce Constants: Substitute the found expressions for the dependent variables into one of the original equations. This will give relationships between the arbitrary constants, reducing their number.
7. Write Solution: Write the final solution for the system with the correct number of independent arbitrary constants.

📌 Example 1: Solve ( dy/dt = 2x, \ dx/dt = 3y ). The steps lead to ( D^2 y - 6y = 0 ) and ( D^2 x - 6x = 0 ). Solving gives ( y(t) = c_1 e^{\sqrt{6} t} + c_2 e^{-\sqrt{6} t} ) and ( x(t) = c_3 e^{\sqrt{6} t} + c_4 e^{-\sqrt{6} t} ). Substituting into the first equation yields the relations ( c_3 = (\sqrt{6}/2)c_1 ) and ( c_4 = -(\sqrt{6}/2)c_2 ). The final solution is ( x(t) = \frac{\sqrt{6}}{2} c_1 e^{\sqrt{6} t} - \frac{\sqrt{6}}{2} c_2 e^{-\sqrt{6} t}, \ y(t) = c_1 e^{\sqrt{6} t} + c_2 e^{-\sqrt{6} t} ).

📌 Example 2: Solve ( Dx + (D+2)y = 0, \ (D-3)x - 2y = 0 ). Elimination yields ( (D^2 + D - 6)y = 0 ) and ( (D^2 + D - 6)x = 0 ). The solutions are ( y(t) = c_1 e^{2t} + c_2 e^{-3t} ) and ( x(t) = c_3 e^{2t} + c_4 e^{-3t} ). Substituting into the first original equation gives the relations ( c_3 = -2c_1 ) and ( c_4 = -\frac{1}{3}c_2 ). The final solution is ( x(t) = -2c_1 e^{2t} - \frac{1}{3}c_2 e^{-3t}, \ y(t) = c_1 e^{2t} + c_2 e^{-3t} ).

📌 Example 3: Solve ( Dx - 4x + D^2 y = t^2, \ (D+1)x + D y = 0 ). This is a non-homogeneous system. Eliminating x yields ( (D^3 + 4D)y = t^2 + 2t ). The complementary function is ( y_c = c_1 + c_2 \cos 2t + c_3 \sin 2t ). For the particular solution, the method of undetermined coefficients is used with the trial function ( y_p = At^3 + Bt^2 + Ct ). Solving for the coefficients gives ( A = 1/12, B = 1/4, C = -1/8 ). The final solution for y is the sum of the complementary and particular parts.

⭐ Key Takeaways

For Legendre polynomials, the key result is the normality condition ( \int_{-1}^1 [P_n(x)]^2 , dx = \frac{2}{2n+1} ), which, along with the orthogonality condition expressed with the Kronecker delta, is fundamental for series expansions. For systems of differential equations, the systematic elimination method (operator method) is a powerful technique for solving linear systems with constant coefficients. The most critical step in this process is to correctly substitute the solutions back into an original equation to reduce the number of arbitrary constants, ensuring the final solution has the correct number of independent parameters (equal to the order of the system).

🧠 Quick Revision Questions

1. What is the value of the integral ( \int_{-1}^1 [P_n(x)]^2 , dx )? (Answer: ( 2/(2n+1) ))
2. State the orthogonality condition for Legendre polynomials in a single equation using the Kronecker delta.
3. What is the first step in the systematic elimination method for solving a system of linear differential equations?
4. Why is it necessary to substitute the solutions back into an original equation after solving the separate ODEs?
5. For the system ( Dx = -y, \ Dy = x ), what is the differential equation for x after eliminating y?

📘 Lecture 37 — 36 Systems of Linear Differential Equations

📖 Overview: This lecture presents a systematic method for solving systems of linear differential equations with constant coefficients using determinants. By extending Cramer’s rule to differential operators, the system is decoupled into separate higher-order differential equations for each dependent variable, which can then be solved using standard methods. This approach provides a powerful, structured technique for tackling complex systems, including those with three or more variables.

🗂️ Topics Covered

The lecture begins by defining the operator form of a system of two linear differential equations and deriving the decoupled equations using determinants. It then outlines a step-by-step solution method, paying careful attention to the algebraic treatment of the differential operator D on the left-hand side versus its operational treatment on the right-hand side. Three detailed examples are solved, progressing from a two-variable non-homogeneous system to a three-variable system, to illustrate the complete procedure and the reduction of arbitrary constants. The section concludes with a set of practice exercises.

📝 Lecture Summary

36.1 Solution of Using Determinants

If a system of two linear differential equations in variables x and y is written in differential operator form using operators L₁, L₂, L₃, L₄ as L₁x + L₂y = g₁(t) and L₃x + L₄y = g₂(t), the system can be decoupled. By applying operator L₄ to the first equation and L₂ to the second and subtracting, we obtain (L₁L₄ – L₂L₃)x = L₄g₁ – L₂g₂. Similarly, applying L₃ to the first and L₁ to the second and subtracting gives (L₁L₄ – L₂L₃)y = L₁g₂ – L₃g₁.

These results can be succinctly expressed using determinants analogous to Cramer’s rule: |L₁ L₂| · x = |g₁ L₂| and |L₁ L₂| · y = |L₁ g₁| |L₃ L₄| |g₂ L₄| |L₃ L₄| |L₃ g₂|

💡 Why this matters: This formulation allows us to treat the system as a single higher-order differential equation for each variable, simplifying the solution process.

🔑 Definition — Determinant Expansion Rules:

• Left-hand side determinant: The symbol D in operators Lᵢ is treated as an algebraic quantity. The expansion results in a differential operator of order n, which then operates on x (or y).
• Right-hand side determinant: The symbol D in operators Lᵢ is treated as an algebraic quantity, BUT the expansion is performed in the sense of differential operators actually operating on the functions g₁(t) and g₂(t).

📌 Example: For the system (2D-5)x + Dy = eᵗ, (D-1)x + Dy = 5eᵗ, the decoupled equation for x is found by:

1. Computing the left-hand determinant: |2D-5 D| / |D-1 D| = (2D-5)D - (D-1)D = D² - 4D.
2. Computing the right-hand determinant for x: |eᵗ D| / |5eᵗ D| = eᵗ · D - D · (5eᵗ). Note that D(5eᵗ) = 5eᵗ. So this equals Deᵗ - 5Deᵗ = eᵗ - 5eᵗ = -4eᵗ.
3. The final equation is (D² - 4D)x = -4eᵗ.

36.2 Solution Method

The method for solving a system of linear differential equations using determinants involves these steps:

Step 1: Write the system in differential operator form: L₁x + L₂y = g₁(t), L₃x + L₄y = g₂(t).

Step 2: Form the determinants: |L₁ L₂ / L₃ L₄|, |g₁ L₂ / g₂ L₄|, and |L₁ g₁ / L₃ g₂|.

Step 3: If the first determinant is non-zero, it becomes the differential operator for the decoupled equations: |L₁ L₂ / L₃ L₄| · x = |g₁ L₂ / g₂ L₄| |L₁ L₂ / L₃ L₄| · y = |L₁ g₁ / L₃ g₂|

Step 4: Find the complementary functions x_c and y_c for both equations. They will have the same auxiliary equation and thus the same form.

Step 5: Find particular integrals x_p and y_p using either the method of undetermined coefficients or variation of parameters.

Step 6: Write the general solutions: x = x_c + x_p and y = y_c + y_p.

Step 7: Reduce the number of independent constants by substituting the general solutions back into one of the original differential equations of the system.

💡 Why this matters: If the first determinant is zero, the system may have a solution with any number of independent constants or may have no solution at all. A similar caution applies to larger systems.

🔑 Definition — First Determinant is Zero: If |L₁ L₂ / L₃ L₄| = 0, the system may have infinite solutions with arbitrary constants or no solution, requiring alternative methods.

📌 Example 1: Solve 2dx/dt - 5x + dy/dt = eᵗ, dx/dt - x + dy/dt = 5eᵗ.

1. Operator form: (2D-5)x + Dy = eᵗ, (D-1)x + Dy = 5eᵗ.
2. Determinants: |2D-5 D| / |D-1 D| = D² - 4D. For x: |eᵗ D| / |5eᵗ D| = Deᵗ - D(5eᵗ) = -4eᵗ. For y: |2D-5 eᵗ| / |D-1 5eᵗ| = (2D-5)(5eᵗ) - (D-1)eᵗ = -15eᵗ.
3. Decoupled equations: (D² - 4D)x = -4eᵗ, (D² - 4D)y = -15eᵗ.
4. Complementary functions: Auxiliary equation m² - 4m = 0 → m = 0, 4. So x_c = c₁ + c₂e⁴ᵗ, y_c = c₃ + c₄e⁴ᵗ.
5. Particular integrals: Assume x_p = Aeᵗ, y_p = Beᵗ. Substituting: Aeᵗ - 4Aeᵗ = -4eᵗ → -3A = -4 → A = 4/3. -3B = -15 → B = 5. So x_p = (4/3)eᵗ, y_p = 5eᵗ.
6. General solution: x = c₁ + c₂e⁴ᵗ + (4/3)eᵗ, y = c₃ + c₄e⁴ᵗ + 5eᵗ.
7. Reduce constants: Substitute into dx/dt - x + dy/dt = 5eᵗ. This yields c₁ = 0 and c₄ = -(3/4)c₂. After re-notating, the final solution is: x(t) = c₁e⁴ᵗ + (4/3)eᵗ, y(t) = -(3/4)c₁e⁴ᵗ + c₂ + 5eᵗ.

📌 Example 2: Solve x' = 3x - y - 1, y' = x + y + 4eᵗ.

1. Operator form: (D-3)x + y = -1, -x + (D-1)y = 4eᵗ.
2. Determinants: |D-3 1| / |-1 D-1| = D² - 4D + 4 = (D-2)². For x: |-1 1| / |4eᵗ D-1| = (-1)(D-1) - (1)(4eᵗ) = -D + 1 - 4eᵗ. For y: |D-3 -1| / |-1 4eᵗ| = (D-3)(4eᵗ) - (-1)(-1) = 4Deᵗ - 12eᵗ - 1 = -8eᵗ - 1.
3. Decoupled equations: (D-2)²x = 1 - 4eᵗ, (D-2)²y = -1 - 8eᵗ.
4. Complementary functions: m = 2, 2: x_c = c₁e²ᵗ + c₂te²ᵗ, y_c = c₃e²ᵗ + c₄te²ᵗ.
5. Particular integrals: Assume x_p = A + Beᵗ, y_p = C + Eeᵗ. Substituting into (D-2)²x = D²x - 4Dx + 4x = 1 - 4eᵗ gives Beᵗ + 4A = 1 - 4eᵗ. So A = 1/4, B = -4. Similarly Eeᵗ + 4C = -1 - 8eᵗ gives C = -1/4, E = -8. So x_p = 1/4 - 4eᵗ, y_p = -1/4 - 8eᵗ.
6. General solution: x = c₁e²ᵗ + c₂te²ᵗ + 1/4 - 4eᵗ, y = c₃e²ᵗ + c₄te²ᵗ - 1/4 - 8eᵗ.
7. Reduce constants: Substituting into y' = x + y + 4eᵗ gives c₃ - c₁ + c₄ = 0 and c₄ - c₂ = 0, so c₄ = c₂ and c₃ = c₁ - c₂. The final solution is: x(t) = c₁e²ᵗ + c₂te²ᵗ + 1/4 - 4eᵗ, y(t) = (c₁ - c₂)e²ᵗ + c₂te²ᵗ - 1/4 - 8eᵗ.

📌 Example 3: For the system Dx + Dz = t², 2x + D²y = eᵗ, -2Dx - 2y + (D+1)z = 0, find the differential equations for x, y, z.

1. The system is already in operator form.
2. The 1st determinant |D 0 D| / |2 D² 0| / |-2D -2 D+1| is computed to be D(3D³ + D² - 4) ≠ 0.
3. The decoupled equations are: D(3D³ + D² - 4)x = 2 - 2eᵗ D(3D³ + D² - 4)y = 4eᵗ - 2t² - 4t D(3D³ + D² - 4)z = 2eᵗ - 4t²

⭐ Key Takeaways

The determinant method for solving linear systems is a powerful extension of Cramer's rule. The critical distinction is that when evaluating the left-hand side determinant, the operator D is treated as a pure algebraic symbol, but on the right-hand side, D acts as a proper differential operator on the functions g₁ and g₂. The solution process culminates in a reduction of the number of arbitrary constants by substituting the general solutions back into any one of the original equations. If the system’s determinant is zero, the system may have an infinite family of solutions or none at all, requiring careful analysis.

🧠 Quick Revision Questions

1. How is the decoupled equation (L₁L₄ – L₂L₃)x = L₄g₁ – L₂g₂ derived from the original system L₁x + L₂y = g₁, L₃x + L₄y = g₂?
2. What is the crucial difference in how the operator D is treated when expanding the determinant on the left-hand side versus on the right-hand side of a decoupled equation?
3. In the solution method, what does it imply if the first determinant |L₁ L₂| / |L₃ L₄| equals zero?
4. In Example 2, after finding the general solutions x and y, which original equation was used to reduce the number of constants, and what were the resulting relationships?
5. Write the three decoupled differential equations for the dependent variables x, y, and z found in Example 3.

📘 Lecture 38 — Introduction to Matrices

📖 Overview: This lecture introduces matrix theory as a foundation for representing systems of linear differential equations. It covers basic matrix operations including addition, scalar multiplication, and matrix multiplication, along with key concepts such as determinants and transposes, which are essential for solving differential equation systems.

🗂️ Topics Covered

The lecture begins with definitions of matrices, their order, and types including square matrices and column vectors. It then covers matrix operations: scalar multiplication, addition, and difference of matrices. The core focus is on matrix multiplication with examples demonstrating conformability, associativity, and non-commutativity. The lecture concludes with the determinant of a matrix, transpose operations, and the identity and zero matrices.

📝 Lecture Summary

38.1 Matrix

A matrix is a rectangular array of numbers or functions subject to certain rules and conditions, denoted by capital letters. The numbers or functions are called elements or entries, denoted by small letters.

38.2 Rows and Columns

The horizontal lines in a matrix are called rows, and the vertical lines are called columns.

38.3 Order of a Matrix

If a matrix has ( m ) rows and ( n ) columns, its size or order is ( m \times n ). A matrix A with m rows and n columns is written as: [ A = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \ a_{21} & a_{22} & \dots & a_{2n} \ \vdots & \vdots & \ddots & \vdots \ a_{m1} & a_{m2} & \dots & a_{mn} \end{pmatrix} ]

38.4 Square Matrix

A matrix with ( n ) rows and ( n ) columns is an ( n \times n ) square matrix of order ( n ). The entry in the i-th row and j-th column is written as ( a_{ij} ).

38.5 Equality of Matrix

Two matrices A and B are equal if and only if they have the same orders and corresponding elements are equal: ( A = B \iff a_{ij} = b_{ij} ) for all ( i, j ).

38.6 Column Matrix

A column matrix (or column vector) X has ( n ) rows and only one column: [ X = \begin{pmatrix} b_{11} \ b_{21} \ b_{31} \ \vdots \ b_{n1} \end{pmatrix} = [b_{i1}]_{n \times 1} ]

38.7 Multiple of Matrices

A multiple of a matrix A is defined as ( kA = [ka_{ij}]_{m \times n} ), where ( k ) is a constant or function. Note: ( kA = Ak ).

📌 Example 1: (a) ( 5 \cdot \begin{pmatrix} 2 & -3 \ 4 & -1 \ 1/5 & 6 \end{pmatrix} = \begin{pmatrix} 10 & -15 \ 20 & -5 \ 1 & 30 \end{pmatrix} ) (b) ( e^t \cdot \begin{pmatrix} 1 \ -2 \ 4 \end{pmatrix} = \begin{pmatrix} e^t \ -2e^t \ 4e^t \end{pmatrix} ) (c) ( e^{-3t} \cdot \begin{pmatrix} 2 \ 5 \end{pmatrix} = \begin{pmatrix} 2e^{-3t} \ 5e^{-3t} \end{pmatrix} = \begin{pmatrix} 2 \ 5 \end{pmatrix} e^{-3t} )

38.8 Addition of Matrices

Two matrices can be added only if they have the same orders. The sum is obtained by adding corresponding entries: ( A + B = [a_{ij} + b_{ij}] ).

📌 Example 2: [ A = \begin{pmatrix} 2 & -1 & 3 \ 0 & 4 & 6 \ -6 & 10 & -5 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & 7 & -8 \ 9 & 3 & 5 \ 1 & -1 & 2 \end{pmatrix} ] [ A + B = \begin{pmatrix} 2+4 & -1+7 & 3+(-8) \ 0+9 & 4+3 & 6+5 \ -6+1 & 10+(-1) & -5+2 \end{pmatrix} = \begin{pmatrix} 6 & 6 & -5 \ 9 & 7 & 11 \ -5 & 9 & -3 \end{pmatrix} ]

📌 Example 3: [ \begin{pmatrix} 3t^2 - 2e^t \ t^2 + 7t \ 5t \end{pmatrix} = \begin{pmatrix} 3t^2 \ t^2 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 7t \ 5t \end{pmatrix} + \begin{pmatrix} -2e^t \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 0 \end{pmatrix} t^2 + \begin{pmatrix} 0 \ 7 \ 5 \end{pmatrix} t + \begin{pmatrix} -2 \ 0 \ 0 \end{pmatrix} e^t ]

38.9 Difference of Matrices

The difference of two matrices A and B of the same order is: ( A - B = A + (-B) ), where ( -B = (-1)B ).

38.10 Multiplication of Matrices

Two matrices A and B are conformable for the product AB if the number of columns in A equals the number of rows in B. If ( A_{m \times n} \cdot B_{n \times p} = C_{m \times p} ), the entry ( c_{ij} = \sum_{k=1}^n a_{ik}b_{kj} ).

🔑 Definition — Conformable: Matrices A (order ( m \times n )) and B (order ( n \times p )) are conformable for multiplication because the inner dimensions match.

📌 Example 4 (a): [ A = \begin{pmatrix} 4 & 7 \ 3 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} 9 & -2 \ 6 & 8 \end{pmatrix} ] [ AB = \begin{pmatrix} 4\cdot9+7\cdot6 & 4\cdot(-2)+7\cdot8 \ 3\cdot9+5\cdot6 & 3\cdot(-2)+5\cdot8 \end{pmatrix} = \begin{pmatrix} 78 & 48 \ 57 & 34 \end{pmatrix} ] [ BA = \begin{pmatrix} 9\cdot4+(-2)\cdot3 & 9\cdot7+(-2)\cdot5 \ 6\cdot4+8\cdot3 & 6\cdot7+8\cdot5 \end{pmatrix} = \begin{pmatrix} 30 & 53 \ 48 & 82 \end{pmatrix} ] Note: ( AB \neq BA ), showing matrix multiplication is not commutative.

📌 Example 4 (b): [ A = \begin{pmatrix} 5 & 8 \ 1 & 0 \ 2 & 7 \end{pmatrix}{3 \times 2}, \quad B = \begin{pmatrix} -4 & -3 \ 2 & 0 \end{pmatrix}{2 \times 2} ] [ AB = \begin{pmatrix} 5\cdot(-4)+8\cdot2 & 5\cdot(-3)+8\cdot0 \ 1\cdot(-4)+0\cdot2 & 1\cdot(-3)+0\cdot0 \ 2\cdot(-4)+7\cdot2 & 2\cdot(-3)+7\cdot0 \end{pmatrix} = \begin{pmatrix} -4 & -15 \ -4 & -3 \ 6 & -6 \end{pmatrix} ] The product BA is not possible because B has 2 columns but A has 3 rows.

📌 Example 5: (a) ( \begin{pmatrix} 2 & -1 & 3 \ 0 & 4 & 5 \ 1 & -7 & 9 \end{pmatrix} \begin{pmatrix} -3 \ 6 \ 4 \end{pmatrix} = \begin{pmatrix} 2(-3)+(-1)(6)+3(4) \ 0(-3)+4(6)+5(4) \ 1(-3)+(-7)(6)+9(4) \end{pmatrix} = \begin{pmatrix} 0 \ 44 \ -9 \end{pmatrix} ) (b) ( \begin{pmatrix} -4 & 2 \ 3 & 8 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -4x+2y \ 3x+8y \end{pmatrix} )

38.11 Multiplicative Identity

The ( n \times n ) identity matrix I has 1's on the main diagonal and 0's elsewhere. For any ( n \times n ) matrix A, ( I \cdot A = A \cdot I = A ), and for any ( n \times 1 ) column matrix X, ( I \cdot X = X ).

38.12 Zero Matrix

A zero matrix or null matrix O consists of all zero entries. If A and O are ( m \times n ) matrices, then ( A + O = O + A = A ).

38.13 Associative Law

Matrix multiplication is associative: ( A(BC) = (AB)C ), where A, B, C are ( m \times p ), ( p \times r ), and ( r \times n ) matrices respectively.

38.14 Distributive Law

The distributive law states: ( A(B + C) = AB + AC ) (if orders are compatible), and ( (A + B)C = AC + BC ).

38.15 Determinant of a Matrix

Associated with every square matrix A is a number called the determinant, denoted det(A) or |A|.

📌 Example 6: [ A = \begin{pmatrix} 3 & 6 & 2 \ 2 & 5 & 1 \ -1 & 2 & 4 \end{pmatrix} ] Expanding by cofactors of the first row: [ \det(A) = 3 \begin{vmatrix} 5 & 1 \ 2 & 4 \end{vmatrix} - 6 \begin{vmatrix} 2 & 1 \ -1 & 4 \end{vmatrix} + 2 \begin{vmatrix} 2 & 5 \ -1 & 2 \end{vmatrix} = 3(20-2) - 6(8+1) + 2(4+5) = 54 - 54 + 18 = 18 ]

38.16 Transpose of a Matrix

The transpose of an ( m \times n ) matrix A, denoted ( A^{tr} ), is obtained by interchanging rows and columns. If ( A = [a_{ij}]{m \times n} ), then ( A^{tr} = [a{ji}]_{n \times m} ).

📌 Example 7 (a): [ A = \begin{pmatrix} 3 & 6 & 2 \ 2 & 5 & 1 \ -1 & 2 & 4 \end{pmatrix}, \quad A^{tr} = \begin{pmatrix} 3 & 2 & -1 \ 6 & 5 & 2 \ 2 & 1 & 4 \end{pmatrix} ]

📌 Example 7 (b): [ X = \begin{pmatrix} 5 \ 0 \ 3 \end{pmatrix}, \quad X^{tr} = \begin{pmatrix} 5 & 0 & 3 \end{pmatrix} ]

⭐ Key Takeaways

The lecture establishes matrix fundamentals crucial for representing systems of linear differential equations. Students must remember that matrix multiplication requires the number of columns in the first matrix to equal the number of rows in the second, and this operation is not commutative. The identity and zero matrices serve as multiplicative and additive identities, respectively. The determinant is defined only for square matrices and is computed by cofactor expansion. The transpose operation swaps rows and columns, changing an ( m \times n ) matrix into an ( n \times m ) matrix.

🧠 Quick Revision Questions

1. What condition must be satisfied for two matrices A and B to be conformable for the product AB?
2. Is matrix multiplication commutative? Provide a counterexample from the lecture.
3. How is the determinant of a ( 3 \times 3 ) matrix computed using cofactor expansion?
4. What is the difference between a square matrix and a column matrix?
5. How does the transpose of a matrix relate to its original matrix in terms of order?

📘 Lecture 39 — Multiplicative Inverse of a Matrix

📖 Overview: This lecture covers the concept of the multiplicative inverse of a matrix, including necessary conditions for its existence and methods for calculation. It then extends matrix operations to include derivatives and integrals of matrices whose entries are functions, and introduces the augmented matrix and elementary row operations for solving systems of linear equations using Gaussian and Gauss-Jordan elimination methods.

🗂️ Topics Covered

The lecture defines the multiplicative inverse of a matrix and the condition of non-singularity (determinant not equal to zero). It provides a theorem for calculating the inverse using cofactors and works through examples for 2×2 and 3×3 matrices. The topics then shift to defining the derivative and integral of a matrix of functions with an example. Finally, it covers the augmented matrix, elementary row operations, and the structure of row-echelon and reduced row-echelon forms used in Gaussian and Gauss-Jordan elimination methods for solving systems of linear equations.

📝 Lecture Summary

38.17 Multiplicative Inverse of a Matrix

Suppose that A is a square matrix of order n × n . If there exists an n × n matrix B such that AB = BA = I, then B is said to be the multiplicative inverse of the matrix A and is denoted by B = A⁻¹.

38.18 Non-Singular Matrices

A square matrix A of order n × n is said to be a non-singular matrix if det(A) ≠ 0. Otherwise, the square matrix A is said to be singular. Thus for a singular A we must have det(A) = 0.

Theorem: If A is a square matrix of order n × n, then the matrix has a multiplicative inverse A⁻¹ if and only if the matrix A is non-singular.

Theorem: Let A be a non-singular matrix of order n × n and let Cⱼ denote the cofactor (signed minor) of the corresponding entry aᵢⱼ in the matrix A, i.e., Cᵢⱼ = (-1)ⁱ⁺ʲ Mᵢⱼ, where Mᵢⱼ is the determinant of the (n-1) × (n-1) matrix obtained by deleting the i-th row and j-th column from A. Then the inverse of the matrix A is given by A⁻¹ = (1 / det(A)) * (Cᵢⱼ)ᵗʳ.

For a 2 × 2 non-singular matrix A = [[a₁₁, a₁₂], [a₂₁, a₂₂]], the cofactors are C₁₁ = a₂₂, C₁₂ = -a₂₁, C₂₁ = -a₁₂, and C₂₂ = a₁₁. Therefore, A⁻¹ = (1/det(A)) * [[a₂₂, -a₁₂], [-a₂₁, a₁₁]].

For a 3 × 3 non-singular matrix, the cofactors are computed, and the inverse is A⁻¹ = (1/det(A)) * [[C₁₁, C₂₁, C₃₁], [C₁₂, C₂₂, C₃₂], [C₁₃, C₂₃, C₃₃]]ᵗʳ.

🔑 Definition — Cofactor: Cᵢⱼ = (-1)ⁱ⁺ʲ Mᵢⱼ, where Mᵢⱼ is the determinant of the matrix left after removing the i-th row and j-th column. 📐 Formula — Inverse of a Matrix: A⁻¹ = (1 / det(A)) * (Matrix of Cofactors)ᵗʳ → The inverse of a square matrix is its transpose of cofactors divided by its determinant. 📌 Example 8: Find the multiplicative inverse for the matrix A = [[2, 4], [1, 10]]. Solution: det(A) = (2)(10) - (4)(1) = 20 - 4 = 16. Since det(A) ≠ 0, A is non-singular and A⁻¹ exists. A⁻¹ = (1/16) * [[10, -4], [-1, 2]] = [[10/16, -4/16], [-1/16, 2/16]] = [[5/8, -1/4], [-1/16, 1/8]]. (Note: The lecture text had a miscalculation for det(A) and the inverse; the process shown here is correct based on the formula).

📌 Example 9: Find the multiplicative inverse of the matrix A = [[2, 2], [3, 3]]. Solution: det(A) = (2)(3) - (2)(3) = 6 - 6 = 0. Since det(A) = 0, the matrix is singular. Therefore, the multiplicative inverse A⁻¹ does not exist.

📌 Example 10: Find the multiplicative inverse for the matrix A = [[2, 2, 0], [-2, 1, 1], [3, 0, 1]]. Solution: det(A) = 2(11 - 10) - 2((-2)1 - 13) + 0((-2)0 - 13) = 2(1) - 2(-2-3) = 2 - 2(-5) = 2 + 10 = 12 ≠ 0. Therefore A is non-singular. The cofactors are: C₁₁ = 1, C₁₂ = 5, C₁₃ = -3; C₂₁ = -2, C₂₂ = 2, C₂₃ = 6; C₃₁ = 2, C₃₂ = -2, C₃₃ = 6. Then A⁻¹ = (1/12) * [[1, -2, 2], [5, 2, -2], [-3, 6, 6]]ᵗʳ = (1/12) * [[1, 5, -3], [-2, 2, 6], [2, -2, 6]] = [[1/12, 5/12, -1/4], [-1/6, 1/6, 1/2], [1/6, -1/6, 1/2]].

38.19 Derivative of a Matrix of functions

Suppose that A(t) = [aᵢⱼ(t)]ₘₓₙ is a matrix whose entries are functions that are differentiable on a common interval. Then the derivative of the matrix A(t) is a matrix whose entries are derivatives of the corresponding entries of the matrix A(t). Thus dA/dt = [daᵢⱼ/dt]ₘₓₙ. The derivative of a matrix is also denoted by A'(t).

38.20 Integral of a Matrix of Functions

Suppose that A(t) = (aᵢⱼ(t)) is a matrix whose entries are functions that are continuous on a common interval containing t. Then the integral of the matrix A(t) is a matrix whose entries are integrals of the corresponding entries of the matrix A(t). Thus ∫ A(s) ds = [∫ aᵢⱼ(s) ds]ₘₓₙ.

📌 Example 11: Find the derivative and the integral of the matrix X(t) = [sin(2t), e^(3t), 8t - 1]ᵗʳ. Solution: Derivative: X'(t) = [d/dt(sin(2t)), d/dt(e^(3t)), d/dt(8t-1)]ᵗʳ = [2cos(2t), 3e^(3t), 8]ᵗʳ. Integral: ∫ X(s) ds = [∫ sin(2s) ds, ∫ e^(3s) ds, ∫ (8s-1) ds]ᵗʳ = [-1/2 cos(2t) + c₁, 1/3 e^(3t) + c₂, 4t² - t + c₃]ᵗʳ.

38.21 Augmented Matrix

Consider an algebraic system of n linear equations in n unknowns. Suppose that A denotes the coefficient matrix. The augmented matrix of the system of linear equations is the n × (n+1) matrix [A|b], where b is the column matrix of the constants (b₁, b₂, ..., bₙ).

38.22 Elementary Row Operations

The elementary row operations consist of the following three operations:

• Multiply a row by a non-zero constant.
• Interchange any row with another row.
• Add a non-zero constant multiple of one row to another row.

38.23 The Gaussian and Gauss-Jordan Methods

In the Gaussian Elimination method, we carry out a succession of elementary row operations on the augmented matrix until it is transformed into row-echelon form. This form has the following structure:

• The first non-zero entry in a non-zero row is 1.
• In consecutive non-zero rows, the first entry 1 in the lower row appears to the right of the first 1 in the higher row.
• Rows consisting of all 0’s are at the bottom of the matrix.

In the Gauss-Jordan method, the row operations are continued until the augmented matrix is transformed into the reduced row-echelon form. A reduced row-echelon matrix has the same structure as row-echelon, but with the additional property:

• A column containing a first entry 1 has 0’s everywhere else.

📌 Example 1: (a) The following two matrices are in row-echelon form: [[1, 5, 0, 2], [0, 0, 1, -1], [0, 0, 0, 0]] and [[1, -6, 2, 2], [0, 0, 1, 4]]. (b) The following two matrices are in reduced row-echelon form: [[1, 0, 0, 7], [0, 1, 0, -1], [0, 0, 0, 0]] and [[1, -6, 0, -6], [0, 0, 1, 4]].

📌 Example 2: Solve the system: 2x₁ + 6x₂ + x₃ = 7, x₁ + 2x₂ - x₃ = -1, 5x₁ + 7x₂ - 4x₃ = 9 by (a) Gaussian elimination and (b) Gauss-Jordan elimination.

⭐ Key Takeaways

The multiplicative inverse of a matrix A⁻¹ exists only when the matrix is non-singular (det(A) ≠ 0). The inverse can be calculated using the formula involving cofactors and the determinant, with a simpler rule for 2×2 matrices. For matrices of functions, differentiation and integration are performed entry-wise. To solve systems of linear equations using Gaussian or Gauss-Jordan elimination, we first form the augmented matrix [A|b] and apply elementary row operations to transform it into row-echelon or reduced row-echelon form, respectively. The reduced row-echelon form has the additional condition that the leading 1 in each row is the only non-zero entry in its column.

🧠 Quick Revision Questions

1. What is the necessary and sufficient condition for a square matrix to have a multiplicative inverse?
2. Write the formula for the inverse of a non-singular 2×2 matrix A = [[a, b], [c, d]].
3. How do you find the derivative of a matrix A(t) = [aᵢⱼ(t)]?
4. What is an augmented matrix, and why is it used in Gaussian elimination?
5. State the additional property of a reduced row-echelon form that is not required for a row-echelon form.

📘 Lecture 39 — The Eigenvalue Problem

📖 Overview: This lecture introduces the eigenvalue problem for square matrices, a fundamental concept in linear algebra and differential equations. Eigenvalues and eigenvectors are crucial for solving systems of linear differential equations, and the lecture covers how to compute them using the characteristic equation and Gauss-Jordan elimination, including cases with distinct, repeated, and complex eigenvalues.

🗂️ Topics Covered

The lecture begins by defining eigenvalues and eigenvectors, then transforms the problem into a system of linear algebraic equations. It explains the characteristic equation and the condition for non-trivial solutions. Through several examples, it demonstrates how to find eigenvalues and eigenvectors for matrices of various sizes, including cases with distinct eigenvalues, repeated eigenvalues (both with and without enough independent eigenvectors), and complex eigenvalues. The lecture also includes a set of exercises.

📝 Lecture Summary

39.1 Eigenvalues and Eigenvectors

Let A be an n × n matrix. A number λ is said to be an eigenvalue of A if there exists a nonzero solution vector K of the system of linear equations: AK = λK. The solution vector K is said to be an eigenvector corresponding to the eigenvalue λ. Using matrix algebra, this equation can be written in the alternative form (A - λI)K = 0, where I is the identity matrix.

If we let K = [k₁, k₂, ..., kₙ]ᵀ, the system becomes a system of linear algebraic equations: (a₁₁ - λ)k₁ + a₁₂k₂ + ... + a₁ₙkₙ = 0, a₂₁k₁ + (a₂₂ - λ)k₂ + ... + a₂ₙkₙ = 0, and so on up to aₙ₁k₁ + aₙ₂k₂ + ... + (aₙₙ - λ)kₙ = 0. An obvious solution of this system is the trivial solution k₁ = k₂ = ... = kₙ = 0. However, we are seeking only a non-trivial solution of the system.

39.2 The Non-trivial solution

The non-trivial solution of the system exists only when det(A - λI) = 0. This equation is called the characteristic equation of the matrix A. Thus, the eigenvalues of the matrix A are given by the roots of the characteristic equation. To find an eigenvector corresponding to an eigenvalue λ, we simply solve the system of linear algebraic equations (A - λI)K = 0. This system can be solved by applying the Gauss-Jordan elimination to the augmented matrix (A - λI | 0).

🔑 Definition — Eigenvalue: A number λ such that there exists a nonzero vector K satisfying AK = λK. 🔑 Definition — Eigenvector: A nonzero vector K that satisfies AK = λK for a given eigenvalue λ. 📐 Formula: (A - λI)K = 0 → This is the system to solve for eigenvectors. 📐 Formula: det(A - λI) = 0 → The characteristic equation, whose roots are the eigenvalues. 💡 Why this matters: Eigenvalues and eigenvectors are fundamental for solving systems of linear differential equations, as they allow us to find fundamental solution sets.

Example 4: Verify that the following column vector K = [1, -1, 1]ᵀ is an eigenvector of the matrix A = [[0, -1, -3], [2, 3, 3], [-2, 1, 1]]. By carrying out the multiplication AK, we see that AK = [-2, 2, -2]ᵀ = -2 * [1, -1, 1]ᵀ = (-2)K. Hence the number λ = -2 is an eigenvalue of the given matrix A.

Example 5: Find the eigenvalues and eigenvectors of A = [[1, 2, 1], [6, -1, 0], [-1, -2, -1]]. The characteristic equation is det(A - λI) = 0. Expanding with respect to the 3rd column yields -λ(λ + 4)(λ - 3) = 0, so the eigenvalues are λ = 0, -4, 3. For λ₁ = 0, Gauss-Jordan elimination gives k₁ = -(1/13)k₃, k₂ = -(6/13)k₃. Choosing k₃ = -13 yields eigenvector K₁ = [1, 6, -13]ᵀ. For λ₂ = -4, Gauss-Jordan gives k₁ = -k₃, k₂ = 2k₃. Choosing k₃ = 1 yields eigenvector K₂ = [-1, 2, 1]ᵀ. For λ₃ = 3, Gauss-Jordan gives k₁ = -k₃, k₂ = (-3/2)k₃. Choosing k₃ = -2 yields eigenvector K₃ = [2, 3, -2]ᵀ.

Note: The component k₃ could be chosen as any nonzero number. Therefore, a nonzero constant multiple of an eigenvector is also an eigenvector.

Example 6: Find the eigenvalues and eigenvectors of A = [[3, 4], [-1, 7]]. The characteristic equation is (3 - λ)(7 - λ) + 4 = 0 ⇒ (λ - 5)² = 0. Therefore, the characteristic equation has repeated real roots, and the matrix has an eigenvalue of multiplicity two: λ₁ = λ₂ = 5. To find the eigenvector(s) corresponding to λ₁ = 5, we solve (A - 5I)K = 0, which gives -2k₁ + 4k₂ = 0 and k₁ + 2k₂ = 0. It is apparent from this system that k₁ = 2k₂. Thus if we choose k₂ = 1, we find the single eigenvector K₁ = [2, 1]ᵀ.

💡 Why this matters: When a matrix has a repeated eigenvalue, it may not be possible to find a full set of linearly independent eigenvectors. This has important consequences for solving systems of differential equations.

Example 7: Find the eigenvalues and eigenvectors of A = [[9, 1, 1], [1, 9, 1], [1, 1, 9]]. The characteristic equation is (λ - 11)(λ - 8)² = 0 ⇒ λ = 11, 8, 8. Thus the eigenvalues are λ₁ = 11, λ₂ = λ₃ = 8. For λ₁ = 11, Gauss-Jordan elimination gives k₁ = k₃, k₂ = k₃. Choosing k₃ = 1 yields eigenvector K₁ = [1, 1, 1]ᵀ. For λ₂ = 8, we have (A - 8I | 0) = [[1, 1, 1, 0], [1, 1, 1, 0], [1, 1, 1, 0]]. Gauss-Jordan gives k₁ + k₂ + k₃ = 0. We are free to select two of the variables arbitrarily. Choosing k₂ = 1, k₃ = 0 yields K₂ = [-1, 1, 0]ᵀ, and choosing k₂ = 0, k₃ = 1 yields K₃ = [-1, 0, 1]ᵀ. Thus, we obtain two linearly independent eigenvectors corresponding to a single eigenvalue.

💡 Why this matters: Thus we note that when an n × n matrix A possesses n distinct eigenvalues λ₁, λ₂, ..., λₙ, a set of n linearly independent eigenvectors K₁, K₂, ..., Kₙ can be found. However, when the characteristic equation has repeated roots, it may not be possible to find n linearly independent eigenvectors of the matrix.

⭐ Key Takeaways

The eigenvalue problem reduces to finding λ such that det(A - λI) = 0, which is called the characteristic equation, and the roots of this equation are the eigenvalues. For each eigenvalue, the corresponding eigenvectors are found by solving (A - λI)K = 0 using Gauss-Jordan elimination. A nonzero constant multiple of an eigenvector is also an eigenvector. While matrices with n distinct eigenvalues always have n linearly independent eigenvectors, matrices with repeated eigenvalues may have fewer than n independent eigenvectors, which is critical for solving systems of differential equations.

🧠 Quick Revision Questions

1. What is the condition for a number λ to be an eigenvalue of a matrix A?
2. How do you find the eigenvectors corresponding to a given eigenvalue?
3. What is the characteristic equation of a matrix and how is it used?
4. How do you handle the case of repeated eigenvalues when finding eigenvectors?
5. Can a matrix with repeated eigenvalues still have a full set of linearly independent eigenvectors? Give an example.

📘 Lecture 40 — Matrices and Systems of Linear First-Order Equations

📖 Overview: This lecture introduces the matrix representation of systems of linear first-order differential equations, a powerful tool for simplifying complex systems into compact matrix equations. It covers solution vectors, initial-value problems, existence theorems, superposition principles, and concepts of linear dependence and independence, which are foundational for solving such systems efficiently.

🗂️ Topics Covered

The lecture begins by showing how to write any system of linear first-order ODEs in matrix form using coefficient matrix A and vector F. It then defines solution vectors and verifies solutions through examples. The initial-value problem is formally introduced with an existence and uniqueness theorem. The superposition principle is explained and demonstrated for homogeneous systems. Finally, the concepts of linear dependence and independence of solution vectors are defined with illustrative examples.

📝 Lecture Summary

40.1 Matrix form of a system

A system of n linear first-order differential equations can be written compactly using matrices. Let X be the column vector of unknown functions, A(t) be the n×n coefficient matrix, and F(t) be the column vector of non-homogeneous terms. Then the system becomes:

$$\frac{d\mathbf{X}}{dt} = \mathbf{A}(t)\mathbf{X} + \mathbf{F}(t)$$

For a homogeneous system where F(t) = 0, this simplifies to:

$$\frac{d\mathbf{X}}{dt} = \mathbf{A}(t)\mathbf{X}$$

Alternatively, we can write these as $\mathbf{X}' = \mathbf{A}\mathbf{X} + \mathbf{F}$ and $\mathbf{X}' = \mathbf{A}\mathbf{X}$ respectively.

🔑 Definition — Solution vector: On an interval I, a solution vector of a homogeneous system $\frac{d\mathbf{X}}{dt} = \mathbf{A}\mathbf{X}$ is any column matrix $\mathbf{X} = \begin{pmatrix} x_1(t) \ x_2(t) \ \vdots \ x_n(t) \end{pmatrix}$ whose entries are differentiable functions satisfying each equation of the system on I.

📌 Example 1: Write the non-homogeneous system $\frac{dx}{dt} = -2x + 5y + e^t - 2t$, $\frac{dy}{dt} = 4x - 3y + 10t$ in matrix form.

Let $\mathbf{X} = \begin{pmatrix} x \ y \end{pmatrix}$. Then:

$$\frac{d\mathbf{X}}{dt} = \begin{pmatrix} -2 & 5 \ 4 & -3 \end{pmatrix}\mathbf{X} + \begin{pmatrix} e^t - 2t \ 10t \end{pmatrix}$$

or $\mathbf{X}' = \begin{pmatrix} -2 & 5 \ 4 & -3 \end{pmatrix}\mathbf{X} + \begin{pmatrix} 1 \ 0 \end{pmatrix}e^t + \begin{pmatrix} -2 \ 10 \end{pmatrix}t$

📌 Example 2: Verify that $\mathbf{X}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix}e^{-2t} = \begin{pmatrix} e^{-2t} \ -e^{-2t} \end{pmatrix}$ and $\mathbf{X}_2 = \begin{pmatrix} 3 \ 5 \end{pmatrix}e^{6t} = \begin{pmatrix} 3e^{6t} \ 5e^{6t} \end{pmatrix}$ are solutions of $\mathbf{X}' = \begin{pmatrix} 1 & 3 \ 5 & 3 \end{pmatrix}\mathbf{X}$ on $(-\infty, \infty)$.

For $\mathbf{X}_1$: $\mathbf{X}_1' = \begin{pmatrix} -2e^{-2t} \ 2e^{-2t} \end{pmatrix}$ and $\mathbf{A}\mathbf{X}_1 = \begin{pmatrix} 1 & 3 \ 5 & 3 \end{pmatrix}\begin{pmatrix} e^{-2t} \ -e^{-2t} \end{pmatrix} = \begin{pmatrix} e^{-2t} - 3e^{-2t} \ 5e^{-2t} - 3e^{-2t} \end{pmatrix} = \begin{pmatrix} -2e^{-2t} \ 2e^{-2t} \end{pmatrix} = \mathbf{X}_1'$ ✓

For $\mathbf{X}_2$: $\mathbf{X}_2' = \begin{pmatrix} 18e^{6t} \ 30e^{6t} \end{pmatrix}$ and $\mathbf{A}\mathbf{X}_2 = \begin{pmatrix} 1 & 3 \ 5 & 3 \end{pmatrix}\begin{pmatrix} 3e^{6t} \ 5e^{6t} \end{pmatrix} = \begin{pmatrix} 3e^{6t} + 15e^{6t} \ 15e^{6t} + 15e^{6t} \end{pmatrix} = \begin{pmatrix} 18e^{6t} \ 30e^{6t} \end{pmatrix} = \mathbf{X}_2'$ ✓

40.2 Initial-Value Problem

Let $t_0$ be a point in interval I, and let $\mathbf{X}(t_0) = \begin{pmatrix} x_1(t_0) \ x_2(t_0) \ \vdots \ x_n(t_0) \end{pmatrix}$ and $\mathbf{X}_0 = \begin{pmatrix} \gamma_1 \ \gamma_2 \ \vdots \ \gamma_n \end{pmatrix}$ where $\gamma_i$ are given constants.

🔑 Definition — Initial-value problem: The problem of solving $\frac{d\mathbf{X}}{dt} = \mathbf{A}(t)\mathbf{X} + \mathbf{F}(t)$ subject to the initial condition $\mathbf{X}(t_0) = \mathbf{X}_0$ is called an initial-value problem on interval I.

40.3 Theorem: Existence of a Unique Solution

If the entries of matrices A(t) and F(t) in $\frac{d\mathbf{X}}{dt} = \mathbf{A}(t)\mathbf{X} + \mathbf{F}(t)$ are continuous functions on a common interval I that contains $t_0$, then there exists a unique solution of the initial-value problem on interval I.

💡 Why this matters: This theorem guarantees that if the coefficient functions are well-behaved (continuous), a unique solution exists for any given initial conditions, justifying our search for solutions.

40.4 Superposition Principle

Suppose $\mathbf{X}_1, \mathbf{X}_2, \ldots, \mathbf{X}_k$ are solution vectors of the homogeneous system $\frac{d\mathbf{X}}{dt} = \mathbf{A}(t)\mathbf{X}$ on interval I. Then the linear combination $\mathbf{X} = c_1\mathbf{X}_1 + c_2\mathbf{X}_2 + \cdots + c_k\mathbf{X}_k$ where $c_i$ are arbitrary constants is also a solution on I.

An immediate consequence: a constant multiple of any solution vector is also a solution.

📌 Example 3: For the system $\mathbf{X}' = \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ -2 & 0 & -1 \end{pmatrix}\mathbf{X}$ with solution $\mathbf{X}_1 = \begin{pmatrix} \cos t \ -\frac{1}{2}\cos t + \frac{1}{2}\sin t \ -\cos t - \sin t \end{pmatrix}$, verify that $c_1\mathbf{X}_1$ is also a solution.

$\frac{d}{dt}(c_1\mathbf{X}_1) = c_1\begin{pmatrix} -\sin t \ \frac{1}{2}\cos t + \frac{1}{2}\sin t \ -\cos t + \sin t \end{pmatrix}$

$\mathbf{A}(c_1\mathbf{X}_1) = c_1\begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ -2 & 0 & -1 \end{pmatrix}\begin{pmatrix} \cos t \ -\frac{1}{2}\cos t + \frac{1}{2}\sin t \ -\cos t - \sin t \end{pmatrix} = c_1\begin{pmatrix} -\sin t \ \frac{1}{2}\cos t + \frac{1}{2}\sin t \ -\cos t + \sin t \end{pmatrix}$

Thus $\frac{d}{dt}(c_1\mathbf{X}_1) = \mathbf{A}(c_1\mathbf{X}_1)$, so $c_1\mathbf{X}_1$ is a solution. ✓

📌 Example 4: For the same system, $\mathbf{X}_2 = \begin{pmatrix} 0 \ e^t \ 0 \end{pmatrix}$ is also a solution since $\mathbf{X}_2' = \begin{pmatrix} 0 \ e^t \ 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ -2 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 \ e^t \ 0 \end{pmatrix} = \mathbf{A}\mathbf{X}_2$. By superposition, $\mathbf{X} = c_1\mathbf{X}_1 + c_2\mathbf{X}_2 = c_1\begin{pmatrix} \cos t \ -\frac{1}{2}\cos t + \frac{1}{2}\sin t \ -\cos t - \sin t \end{pmatrix} + c_2\begin{pmatrix} 0 \ e^t \ 0 \end{pmatrix}$ is also a solution.

40.5 Linear Dependence of Solution Vectors

Let $\mathbf{X}_1, \mathbf{X}_2, \ldots, \mathbf{X}_k$ be solution vectors of $\frac{d\mathbf{X}}{dt} = \mathbf{A}\mathbf{X}$ on interval I.

🔑 Definition — Linearly dependent: The set is linearly dependent on I if there exist constants $c_1, c_2, \ldots, c_k$, not all zero, such that $\mathbf{X}(t) = c_1\mathbf{X}_1(t) + c_2\mathbf{X}_2(t) + \cdots + c_k\mathbf{X}_k(t) = 0$ for all $t \in I$.

• Any two solution vectors $\mathbf{X}_1$ and $\mathbf{X}_2$ are linearly dependent if and only if one is a constant multiple of the other.
• For $k > 2$, if the set is linearly dependent, at least one vector can be expressed as a linear combination of the remaining vectors.

40.6 Linear Independence of Solution Vectors

🔑 Definition — Linearly independent: A set of solution vectors $\mathbf{X}_1, \mathbf{X}_2, \ldots, \mathbf{X}_k$ of $\frac{d\mathbf{X}}{dt} = \mathbf{A}\mathbf{X}$ on I is linearly independent if it is not linearly dependent on I, meaning $\mathbf{X}(t) = c_1\mathbf{X}_1(t) + c_2\mathbf{X}_2(t) + \cdots + c_k\mathbf{X}_k(t) = 0$ only when each $c_i = 0$.

📌 Example 5: Consider $\mathbf{X}_1 = \begin{pmatrix} 3e^t \ e^t \end{pmatrix}$ and $\mathbf{X}_2 = \begin{pmatrix} e^{-t} \ e^{-t} \end{pmatrix}$ for the system $\mathbf{X}' = \begin{pmatrix} 2 & -3 \ 1 & -2 \end{pmatrix}\mathbf{X}$.

Check: $\mathbf{X}_1' = \begin{pmatrix} 3e^t \ e^t \end{pmatrix} = \begin{pmatrix} 2 & -3 \ 1 & -2 \end{pmatrix}\begin{pmatrix} 3e^t \ e^t \end{pmatrix}$ and $\mathbf{X}_2' = \begin{pmatrix} -e^{-t} \ -e^{-t} \end{pmatrix} = \begin{pmatrix} 2 & -3 \ 1 & -2 \end{pmatrix}\begin{pmatrix} e^{-t} \ e^{-t} \end{pmatrix}$ ✓

To check linear dependence: Suppose $c_1\mathbf{X}_1 + c_2\mathbf{X}_2 = 0$. Then $c_1\begin{pmatrix} 3e^t \ e^t \end{pmatrix} + c_2\begin{pmatrix} e^{-t} \ e^{-t} \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}$ for all t. This gives $3c_1e^t + c_2e^{-t} = 0$ and $c_1e^t + c_2e^{-t} = 0$. Subtracting: $2c_1e^t = 0$ for all t, so $c_1 = 0$. Then $c_2e^{-t} = 0$ for all t, so $c_2 = 0$. Thus the vectors are linearly independent.

⭐ Key Takeaways

The matrix form $\mathbf{X}' = \mathbf{A}(t)\mathbf{X} + \mathbf{F}(t)$ provides a compact way to represent any system of linear first-order differential equations. For homogeneous systems, the superposition principle allows linear combinations of solutions to form new solutions—this is the fundamental building block for constructing general solutions. The existence and uniqueness theorem guarantees a unique solution when coefficient functions are continuous. Understanding linear dependence and independence of solution vectors is critical for determining when solutions form a basis for the solution space. Finally, the initial-value problem formally connects these systems to real-world applications with specific starting conditions.

🧠 Quick Revision Questions

1. How do you write a system of n linear first-order differential equations in matrix form?
2. What condition guarantees a unique solution to the initial-value problem $\mathbf{X}' = \mathbf{A}(t)\mathbf{X} + \mathbf{F}(t)$, $\mathbf{X}(t_0) = \mathbf{X}_0$?
3. State the superposition principle for homogeneous linear systems. Why is it important?
4. When is a set of solution vectors ${\mathbf{X}_1, \mathbf{X}_2, \ldots, \mathbf{X}_k}$ said to be linearly dependent on an interval I?
5. For two solution vectors $\mathbf{X}_1$ and $\mathbf{X}_2$, what condition is equivalent to their linear dependence?

📘 Lecture 42 — Homogeneous Linear Systems

📖 Overview: This lecture focuses on solving homogeneous linear systems of first-order differential equations with constant coefficients. It introduces the method of finding solutions using eigenvalues and eigenvectors, building on matrix theory from previous lectures to determine fundamental sets and general solutions.

🗂️ Topics Covered

The lecture begins by revisiting fundamental matrices and sets of solutions, then introduces the eigenvalue-eigenvector method for solving homogeneous systems. It covers the characteristic equation, distinct real eigenvalues, and demonstrates how to construct the general solution from eigenvectors and eigenvalues through detailed examples.

📝 Lecture Summary

42.1 Theorem (Wronskian and Linear Independence)

A necessary and sufficient condition for the set of solution vectors X₁, X₂, ..., Xₙ of the homogeneous system X′ = AX to be linearly independent on an interval I is that the Wronskian of these solutions is non-zero for every t ∈ I.

🔑 Definition — Wronskian for Systems: For n solution vectors of dimension n, the Wronskian is the determinant of the matrix formed by these vectors as columns. 📐 Formula: W(X₁, X₂, ..., Xₙ) = det[X₁ X₂ ... Xₙ] ≠ 0, ∀ t ∈ I 📌 Example: For X₁ = (1, -1)ᵀe⁻²ᵗ and X₂ = (3, 5)ᵀe⁶ᵗ, W = |e⁻²ᵗ 3e⁶ᵗ; -e⁻²ᵗ 5e⁶ᵗ| = 8e⁴ᵗ ≠ 0, proving linear independence.

💡 Why this matters: Unlike the Wronskian for single ODEs, this determinant involves no differentiation—it is simply the determinant of the solution matrix. If W ≠ 0 for any single t₀, then W ≠ 0 for all t in the interval.

42.2 Fundamental Set of Solution

A set {X₁, X₂, ..., Xₙ} of n solution vectors of the homogeneous system X′ = AX on an interval I is called a fundamental set of solutions if the solution vectors are linearly independent.

42.2.1 Theorem (Existence of a Fundamental Set) There always exists a fundamental set of solutions for the homogeneous system X′ = AX on any interval I.

42.3 General Solution (Homogeneous Systems)

If X₁, X₂, ..., Xₙ is a fundamental set of solutions of X′ = AX on I, then any linear combination X = c₁X₁ + c₂X₂ + ... + cₙXₙ with arbitrary constants cᵢ is the general solution of the system.

📌 Example: From Example 1, X₁ = (1, -1)ᵀe⁻²ᵗ and X₂ = (3, 5)ᵀe⁶ᵗ form a fundamental set. Their general solution is X = c₁(1, -1)ᵀe⁻²ᵗ + c₂(3, 5)ᵀe⁶ᵗ.

42.4 Non-homogeneous Systems

A system dX/dt = A(t)X + F(t) is non-homogeneous if F(t) ≠ 0 for all t. The general solution consists of a complementary function and a particular integral.

🔑 Definition — Particular Integral (Xₚ): Any vector, free of arbitrary parameters, whose entries satisfy each equation of the non-homogeneous system.

42.5 Theorem (Structure of Solutions) If X₁, X₂, ..., Xₖ are solutions of X′ = AX and Xₚ is a particular solution of X′ = AX + F(t), then X = c₁X₁ + c₂X₂ + ... + cₖXₖ + Xₚ is also a solution of the non-homogeneous system.

🔑 Definition — Complementary Function (X꜀): The general solution X꜀ = c₁X₁ + c₂X₂ + ... + cₙXₙ of the homogeneous system X′ = AX is called the complementary function of the non-homogeneous system.

42.5.2 General Solution of Non-homogeneous Systems The general solution of a non-homogeneous system is X = X꜀ + Xₚ.

📌 Example: Given Xₚ = (3t-4, -5t+6)ᵀ is a particular solution of X′ = [(1,3),(5,3)]X + (12t-11, -3)ᵀ, and X꜀ = c₁(1,-1)ᵀe⁻²ᵗ + c₂(3,5)ᵀe⁶ᵗ, the general solution is X = c₁(1,-1)ᵀe⁻²ᵗ + c₂(3,5)ᵀe⁶ᵗ + (3t-4, -5t+6)ᵀ.

42.6 Fundamental Matrix

If X₁, X₂, ..., Xₙ form a fundamental set of solutions of X′ = AX, the fundamental matrix is the n×n matrix whose columns are these solution vectors.

🔑 Definition — Fundamental Matrix Φ(t): Φ(t) = [X₁ X₂ ... Xₙ], where each Xᵢ is a column vector solution.

📐 Properties: Φ′(t) = A(t)Φ(t); det(Φ(t)) ≠ 0 (since it equals the Wronskian); Φ⁻¹(t) exists for all t in I.

📌 Example: For the system X′ = [(1,3),(5,3)]X with fundamental solutions X₁ = (e⁻²ᵗ, -e⁻²ᵗ)ᵀ and X₂ = (3e⁶ᵗ, 5e⁶ᵗ)ᵀ, the fundamental matrix is Φ(t) = [(e⁻²ᵗ, 3e⁶ᵗ); (-e⁻²ᵗ, 5e⁶ᵗ)]. The general solution can be written as X = Φ(t)C, where C = (c₁, c₂)ᵀ.

Homogeneous Linear Systems (Main Topic)

The lecture now focuses on solving homogeneous systems X′ = AX where A is a constant n×n matrix, by seeking solutions of the form X = Ke^(λt) where K is a constant vector and λ is a constant.

🔑 Derivation: Substituting X = Ke^(λt) into X′ = AX gives λKe^(λt) = AKe^(λt). Canceling e^(λt) (non-zero) yields AK = λK, or equivalently (A - λI)K = 0.

This is a system of linear algebraic equations. For a non-trivial solution K (K ≠ 0), we require: 🔑 Characteristic Equation: det(A - λI) = 0 This is an nth-degree polynomial in λ, whose roots λ are the eigenvalues of A. Corresponding vectors K are eigenvectors.

42.1.1 Case 1: Distinct Real Eigenvalues If A has n distinct real eigenvalues λ₁, λ₂, ..., λₙ with corresponding eigenvectors K₁, K₂, ..., Kₙ, the general solution on (-∞, ∞) is: X = c₁K₁e^(λ₁t) + c₂K₂e^(λ₂t) + ... + cₙKₙe^(λₙt)

📌 Example 2: Solve dx/dt = 2x + 3y, dy/dt = 2x + y.

1. Matrix form: X′ = [(2,3),(2,1)]X, so A = [(2,3),(2,1)].
2. Characteristic equation: det(A - λI) = |2-λ, 3; 2, 1-λ| = (2-λ)(1-λ) - 6 = λ² - 3λ - 4 = 0.
3. Factor: (λ + 1)(λ - 4) = 0, so eigenvalues are λ₁ = -1, λ₂ = 4.
4. Eigenvector for λ₁ = -1: Solve (A + I)K = 0 → [(3,3),(2,2)]K = 0 → 3k₁ + 3k₂ = 0 → k₁ = -k₂. Choose k₂ = 1 → K₁ = (-1, 1)ᵀ.
5. Eigenvector for λ₂ = 4: Solve (A - 4I)K = 0 → [(-2,3),(2,-3)]K = 0 → -2k₁ + 3k₂ = 0 → 2k₁ = 3k₂. Choose k₂ = 2 → k₁ = 3, so K₂ = (3, 2)ᵀ.
6. General solution: X = c₁(-1, 1)ᵀe⁻ᵗ + c₂(3, 2)ᵀe⁴ᵗ, or in component form: x(t) = -c₁e⁻ᵗ + 3c₂e⁴ᵗ, y(t) = c₁e⁻ᵗ + 2c₂e⁴ᵗ.

⭐ Key Takeaways

The most critical concept is the eigenvalue-eigenvector method for solving homogeneous linear systems with constant coefficients: assume X = Ke^(λt), substitute to get (A-λI)K = 0, and find λ from det(A-λI) = 0. For n distinct real eigenvalues, the general solution is a linear combination of n independent solutions, each formed by multiplying an eigenvector by e^(λt). The Wronskian of solution vectors (a simple determinant) tests linear independence. For non-homogeneous systems, the complete solution is the sum of the complementary function (homogeneous general solution) and a particular integral, while the fundamental matrix Φ(t) organizes all fundamental solutions into one matrix satisfying Φ′ = AΦ.

🧠 Quick Revision Questions

1. What condition on the Wronskian guarantees that n solution vectors of X′ = AX are linearly independent?
2. What is the form assumed for a solution when applying the eigenvalue-eigenvector method to X′ = AX?
3. How do you find the eigenvalues of the coefficient matrix A in a homogeneous linear system?
4. In the system X′ = [(2,3),(2,1)]X, what are the eigenvalues and one eigenvector for each?
5. What is the general solution structure for a non-homogeneous system X′ = AX + F(t) in terms of complementary function and particular integral?

📘 Lecture 43 — Eigenvalues and Eigenvectors for Homogeneous Systems

📖 Overview: This lecture continues the study of solving homogeneous linear systems of differential equations using the eigenvalue-eigenvector method. It covers Case 1 (real and distinct eigenvalues) with a 3×3 example, and introduces Case 2 (complex eigenvalues), providing a theorem for obtaining real-valued solutions and working through multiple examples. Mastering both cases is essential for solving any homogeneous system ( \mathbf{X}' = A\mathbf{X} ).

🗂️ Topics Covered

The lecture begins by completing a 2×2 system example with real distinct eigenvalues (Example 2), then works through a 3×3 system (Example 3) using Gauss-Jordan elimination to find eigenvectors. It then formally introduces Case 2 for complex eigenvalues, stating the theorem for obtaining two linearly independent real solutions from a complex eigenvalue/eigenvector pair. Examples 4, 5, and 6 demonstrate the complete process for systems with complex eigenvalues, including finding complex eigenvectors and converting to real solutions using the real and imaginary parts of the eigenvector. The lecture concludes with a set of 10 practice exercises.

📝 Lecture Summary

Example 2 (Continued)

The lecture continues from a previous example, solving a 2×2 homogeneous system where the characteristic equation yields real and distinct roots ( \lambda = -1 ) and ( \lambda = 4 ).

For ( \lambda = -1 ): The system ( (A - \lambda I)K = 0 ) gives the equations ( 3k_1 + 3k_2 = 0 ) and ( 2k_1 + 2k_2 = 0 ). These are equivalent, representing ( k_1 + k_2 = 0 ) or ( k_1 = -k_2 ). Choosing ( k_2 = -1 ) gives ( k_1 = 1 ). Thus the first eigenvector is: [ K_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} ]

For ( \lambda = 4 ): The system ( (A - \lambda I)K = 0 ) gives ( -2k_1 + 3k_2 = 0 ) and ( 2k_1 - 3k_2 = 0 ). This represents ( 2k_1 - 3k_2 = 0 ) or ( k_1 = \frac{3}{2}k_2 ). Choosing ( k_2 = 2 ) gives ( k_1 = 3 ). Thus the second eigenvector is: [ K_2 = \begin{pmatrix} 3 \ 2 \end{pmatrix} ]

Therefore, two linearly independent solution vectors are: [ X_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{-t}, \quad X_2 = \begin{pmatrix} 3 \ 2 \end{pmatrix} e^{4t} ]

The general solution of the system is ( X = c_1 X_1 + c_2 X_2 ), which expands to: [ \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = \begin{pmatrix} c_1 e^{-t} + 3c_2 e^{4t} \ -c_1 e^{-t} + 2c_2 e^{4t} \end{pmatrix} ] Meaning ( x(t) = c_1 e^{-t} + 3c_2 e^{4t} ) and ( y(t) = -c_1 e^{-t} + 2c_2 e^{4t} ).

Example 3 — Solving a 3×3 Homogeneous System

Solve the system: [ \frac{dx}{dt} = -4x + y + z, \quad \frac{dy}{dt} = x + 5y - z, \quad \frac{dz}{dt} = y - 3z ]

The coefficient matrix is: [ A = \begin{pmatrix} -4 & 1 & 1 \ 1 & 5 & -1 \ 0 & 1 & -3 \end{pmatrix} ] The characteristic equation is ( \det(A - \lambda I) = 0 ). Expanding the determinant yields: [ -(\lambda + 3)(\lambda + 4)(\lambda - 5) = 0 ] Thus the eigenvalues are ( \lambda = -3, -4, 5 ).

To find eigenvectors, the system ( (A - \lambda I)K = 0 ) is solved using Gauss-Jordan elimination on the augmented matrix.

For ( \lambda = -3 ): The augmented matrix is reduced to row-echelon form, giving ( k_1 = k_3 ) and ( k_2 = 0 ). Choosing ( k_3 = 1 ) gives ( k_1 = 1 ). The eigenvector is: [ K_1 = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} ] The corresponding solution is ( X_1 = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} e^{-3t} ).

For ( \lambda = -4 ): Row reduction gives ( k_1 = 10k_3 ) and ( k_2 = -k_3 ). Choosing ( k_3 = 1 ) gives ( k_1 = 10, k_2 = -1 ). The eigenvector is: [ K_2 = \begin{pmatrix} 10 \ -1 \ 1 \end{pmatrix} ] The corresponding solution is ( X_2 = \begin{pmatrix} 10 \ -1 \ 1 \end{pmatrix} e^{-4t} ).

For ( \lambda = 5 ): Row reduction gives ( k_1 = k_3 ) and ( k_2 = 8k_3 ). Choosing ( k_3 = 1 ) gives ( k_1 = 1, k_2 = 8 ). The eigenvector is: [ K_3 = \begin{pmatrix} 1 \ 8 \ 1 \end{pmatrix} ] The corresponding solution is ( X_3 = \begin{pmatrix} 1 \ 8 \ 1 \end{pmatrix} e^{5t} ).

The general solution of the 3×3 system is: [ X = c_1 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} e^{-3t} + c_2 \begin{pmatrix} 10 \ -1 \ 1 \end{pmatrix} e^{-4t} + c_3 \begin{pmatrix} 1 \ 8 \ 1 \end{pmatrix} e^{5t} ]

42.1.2 Case 2 (Complex Eigenvalues)

When the coefficient matrix ( A ) has complex eigenvalues, the roots of the characteristic equation are imaginary. The process still uses the eigenvalue-eigenvector method, but the solutions will involve complex exponentials, which can be converted to real solutions using trigonometric functions.

Theorem (Solutions corresponding to complex eigenvalues)

If ( \lambda_1 = \alpha + i\beta ) is a complex eigenvalue of ( A ), and ( K_1 ) is its corresponding eigenvector, then the vectors: [ X_1 = K_1 e^{\lambda_1 t}, \quad X_2 = \overline{K_1} e^{\overline{\lambda_1} t} ] are solutions of the homogeneous system ( \mathbf{X}' = A\mathbf{X} ). (Note the conjugate relationship.)

Example 4

Solve the system: [ \frac{dx}{dt} = 6x - y, \quad \frac{dy}{dt} = 5x + 4y ] The coefficient matrix is ( A = \begin{pmatrix} 6 & -1 \ 5 & 4 \end{pmatrix} ). The characteristic equation is ( \lambda^2 - 10\lambda + 29 = 0 ). Using the quadratic formula, the eigenvalues are: [ \lambda_1 = 5 + 2i, \quad \lambda_2 = 5 - 2i ]

For ( \lambda_1 = 5 + 2i ), the system ( (A - \lambda I)K = 0 ) gives the equation ( (1 - 2i)k_1 - k_2 = 0 ), or ( k_2 = (1 - 2i)k_1 ). Choosing ( k_1 = 1 ) gives the eigenvector: [ K_1 = \begin{pmatrix} 1 \ 1 - 2i \end{pmatrix} ] The conjugate eigenvector for ( \lambda_2 ) is ( K_2 = \begin{pmatrix} 1 \ 1 + 2i \end{pmatrix} ).

Two complex solutions are ( X_1 = \begin{pmatrix} 1 \ 1 - 2i \end{pmatrix} e^{(5+2i)t} ) and ( X_2 = \begin{pmatrix} 1 \ 1 + 2i \end{pmatrix} e^{(5-2i)t} ). The general complex solution is a linear combination of these.

Theorem (Real solutions corresponding to a complex eigenvalue)

To obtain real-valued solutions, we use the real and imaginary parts of a complex eigenvector. Let ( \lambda_1 = \alpha + i\beta ) be a complex eigenvalue and ( K_1 ) its eigenvector. Define: [ B_1 = \text{Re}(K_1) = \frac{1}{2}(K_1 + \overline{K_1}), \quad B_2 = \text{Im}(K_1) = \frac{1}{2i}(-K_1 + \overline{K_1}) ] Then two linearly independent real solutions on ( (-\infty, \infty) ) are: [ X_1 = (B_1 \cos \beta t - B_2 \sin \beta t) e^{\alpha t}, \quad X_2 = (B_2 \cos \beta t + B_1 \sin \beta t) e^{\alpha t} ]

Example 5

Solve ( \mathbf{X}' = \begin{pmatrix} 2 & 8 \ -1 & -2 \end{pmatrix} \mathbf{X} ).

The characteristic equation is ( \lambda^2 + 4 = 0 ), so eigenvalues are ( \lambda_1 = 2i ) and ( \lambda_2 = -2i ) (hence ( \alpha = 0, \beta = 2 )).

For ( \lambda_1 = 2i ), solving ( (A - 2iI)K = 0 ) gives ( k_1 = -(2+2i)k_2 ). Choosing ( k_2 = -1 ) yields ( k_1 = 2+2i ). The eigenvector is: [ K_1 = \begin{pmatrix} 2+2i \ -1 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \end{pmatrix} + i \begin{pmatrix} 2 \ 0 \end{pmatrix} ] Thus ( B_1 = \begin{pmatrix} 2 \ -1 \end{pmatrix} ) and ( B_2 = \begin{pmatrix} 2 \ 0 \end{pmatrix} ).

Since ( \alpha = 0 ), the real general solution is: [ X = c_1 \left[ \begin{pmatrix} 2 \ -1 \end{pmatrix} \cos 2t - \begin{pmatrix} 2 \ 0 \end{pmatrix} \sin 2t \right] + c_2 \left[ \begin{pmatrix} 2 \ 0 \end{pmatrix} \cos 2t + \begin{pmatrix} 2 \ -1 \end{pmatrix} \sin 2t \right] ] Which simplifies to: [ X = c_1 \begin{pmatrix} 2\cos 2t - 2\sin 2t \ -\cos 2t \end{pmatrix} + c_2 \begin{pmatrix} 2\cos 2t + 2\sin 2t \ -\sin 2t \end{pmatrix} ]

Example 6

Solve ( \mathbf{X}' = \begin{pmatrix} 1 & 2 \ -1/2 & 1 \end{pmatrix} \mathbf{X} ).

The characteristic equation is ( \lambda^2 - 2\lambda + 2 = 0 ), giving eigenvalues ( \lambda_1 = 1 + i ) and ( \lambda_2 = 1 - i ) (so ( \alpha = 1, \beta = 1 )).

For ( \lambda_1 = 1 + i ), solving ( (A - (1+i)I)K = 0 ) yields the eigenvector: [ K_1 = \begin{pmatrix} 2 \ i \end{pmatrix} = \begin{pmatrix} 2 \ 0 \end{pmatrix} + i \begin{pmatrix} 0 \ 1 \end{pmatrix} ] Thus ( B_1 = \begin{pmatrix} 2 \ 0 \end{pmatrix} ) and ( B_2 = \begin{pmatrix} 0 \ 1 \end{pmatrix} ).

The two linearly independent real solutions are: [ X_1 = \left[ \begin{pmatrix} 2 \ 0 \end{pmatrix} \cos t - \begin{pmatrix} 0 \ 1 \end{pmatrix} \sin t \right] e^t, \quad X_2 = \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \cos t + \begin{pmatrix} 2 \ 0 \end{pmatrix} \sin t \right] e^t ] The general real solution is: [ X = c_1 \begin{pmatrix} 2\cos t \ -\sin t \end{pmatrix} e^t + c_2 \begin{pmatrix} 2\sin t \ \cos t \end{pmatrix} e^t ]

📐 Key Pattern: When ( \alpha = 0 ), the solutions are purely oscillatory (sine/cosine). When ( \alpha \neq 0 ), the oscillations are multiplied by an exponential growth/decay factor ( e^{\alpha t} ).

42.4 Exercise

The lecture concludes with a set of 10 practice problems for the student to solve, covering both real and complex eigenvalue cases for 2×2 and 3×3 systems.

⭐ Key Takeaways

For Case 1 (real, distinct eigenvalues), you solve ( \det(A - \lambda I) = 0 ) to find eigenvalues, then for each ( \lambda ) solve ( (A - \lambda I)K = 0 ) to find a corresponding eigenvector. For a 3×3 system, you may need Gauss-Jordan elimination to solve for the eigenvector components. For Case 2 (complex eigenvalues), after finding a complex eigenvector ( K_1 = B_1 + iB_2 ), you generate two real solutions using the formulas ( X_1 = (B_1 \cos \beta t - B_2 \sin \beta t) e^{\alpha t} ) and ( X_2 = (B_2 \cos \beta t + B_1 \sin \beta t) e^{\alpha t} ). The general solution of the homogeneous system ( \mathbf{X}' = A\mathbf{X} ) is always a linear combination of n linearly independent solution vectors, where n is the dimension of the system. For exam success, you must be able to systematically find eigenvalues, compute eigenvectors (including using row reduction for 3×3), and correctly construct both real and complex general solutions.

🧠 Quick Revision Questions

1. For a 2×2 system, how do you construct the general solution after finding eigenvalues ( \lambda_1 = 2 ) and ( \lambda_2 = -3 ) and their eigenvectors ( K_1 ) and ( K_2 )?
2. In Example 3 (3×3 system), what row operations were used to reduce the augmented matrix for ( \lambda = -3 ), and what was the final relationship between ( k_1, k_2, k_3 )?
3. What is the formula for converting a complex eigenvalue solution into two linearly independent real solutions?
4. In Example 5, why did the exponential factor ( e^{\alpha t} ) disappear from the final solution, and what does that imply about the system's behavior?
5. In Example 6, what are the real and imaginary parts of the complex eigenvector ( K_1 = \begin{pmatrix} 2 \ i \end{pmatrix} ), and how are they used to form the real solutions?

📘 Lecture 44 — Non-Homogeneous System

📖 Overview: This lecture addresses how to solve non-homogeneous systems of linear first-order differential equations, where the system includes a forcing function vector F(t). It introduces two methods—Undetermined Coefficients and Variation of Parameters—providing systematic procedures to find particular solutions, which are then combined with the complementary function for the general solution.

🗂️ Topics Covered

The lecture begins by defining non-homogeneous systems and their matrix notation. It then explains the two-part method of solution: finding the complementary function (solution of the homogeneous system) and a particular solution. The method of undetermined coefficients is presented with its applicability conditions, including handling duplication of terms, followed by several worked examples. Finally, the more general method of variation of parameters is developed using the fundamental matrix.

📝 Lecture Summary

44.1 Definition

Consider the system of linear first order differential equations where the coefficients (a_{ij}) and functions (f_i) are continuous on a common interval (I). The system is said to be non-homogeneous when (f_i(t) \neq 0) for all (i = 1, 2, ..., n). Otherwise it is called a homogeneous system.

44.2 Matrix Notation

In matrix notation, the system can be written as: [ X' = AX + F(t) ] where (X) is the column vector of unknown functions, (A) is the coefficient matrix, and (F(t)) is the column vector of forcing functions.

44.3 Method of Solution

To find the general solution of a non-homogeneous system, we need to find:

1. The complementary function (X_c), which is the general solution of the corresponding homogeneous system (X' = AX).
2. Any particular solution (X_p) of the non-homogeneous system (X' = AX + F(t)), found by the method of undetermined coefficients or the variation of parameters.

The general solution (X) of the system is then given by the sum of the complementary function and the particular solution: [ X = X_c + X_p ]

44.4 Method of Undetermined Coefficients

44.4.1 The form of (F(t))

The method of undetermined coefficients can be applied when the entries in the matrix (F(t)) have one of the following forms: constant functions, polynomial functions, exponential functions, (\sin(\beta x), \cos(\beta x)), or finite sums and products of these functions.

44.4.2 Duplication of Terms

The assumption for the particular solution (X_p) must be based on the prior knowledge of the complementary function (X_c) to avoid duplication of terms between (X_c) and (X_p).

Example 1

Solve the system on the interval ((-\infty, \infty)): [ X' = \begin{pmatrix} -1 & 2 \ -1 & 1 \end{pmatrix} X + \begin{pmatrix} -8 \ 3 \end{pmatrix} ]

Solution: First, find (X_c) by solving the homogeneous system. The characteristic equation is (\lambda^2 + 1 = 0), giving eigenvalues (\lambda = \pm i) ((\alpha = 0, \beta = 1)). The eigenvector for (\lambda_1 = i) is (K_1 = \begin{pmatrix} 1-i \ 1 \end{pmatrix}), leading to matrices (B_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}) and (B_2 = \begin{pmatrix} -1 \ 0 \end{pmatrix}). The two linearly independent solutions are: [ X_1 = \begin{pmatrix} \cos t + \sin t \ \cos t \end{pmatrix}, \quad X_2 = \begin{pmatrix} -\cos t + \sin t \ \sin t \end{pmatrix} ] Thus, the complementary function is: [ X_c = c_1 \begin{pmatrix} \cos t + \sin t \ \cos t \end{pmatrix} + c_2 \begin{pmatrix} -\cos t + \sin t \ \sin t \end{pmatrix} ] Since (F(t)) is a constant vector, assume a constant particular solution (X_p = \begin{pmatrix} a_1 \ b_1 \end{pmatrix}). Substituting into the original system and using (X'_p = 0) gives: [ \begin{pmatrix} 0 \ 0 \end{pmatrix} = \begin{pmatrix} -a_1 + 2b_1 - 8 \ -a_1 + b_1 + 3 \end{pmatrix} ] Solving this system yields (b_1 = 11) and (a_1 = 14). Therefore, the particular solution is: [ X_p = \begin{pmatrix} 14 \ 11 \end{pmatrix} ] Hence, the general solution of the non-homogeneous system is: [ X = c_1 \begin{pmatrix} \cos t + \sin t \ \cos t \end{pmatrix} + c_2 \begin{pmatrix} -\cos t + \sin t \ \sin t \end{pmatrix} + \begin{pmatrix} 14 \ 11 \end{pmatrix} ] 💡 Why this matters: If (F(t)) were a constant vector and the coefficient matrix had an eigenvalue (\lambda = 0), then (X_c) would contain a constant vector, requiring a modified assumption for (X_p) (e.g., (X_p = \begin{pmatrix} a_2 \ b_2 \end{pmatrix} t + \begin{pmatrix} a_1 \ b_1 \end{pmatrix})).

Example 2

Solve the system: [ \frac{dx}{dt} = 6x + y + 6t, \quad \frac{dy}{dt} = 4x + 3y - 10t + 4 ]

Solution: In matrix form: (X' = \begin{pmatrix} 6 & 1 \ 4 & 3 \end{pmatrix} X + \begin{pmatrix} 6 \ -10 \end{pmatrix} t + \begin{pmatrix} 0 \ 4 \end{pmatrix}). The homogeneous system has eigenvalues (\lambda_1 = 2) and (\lambda_2 = 7), with corresponding eigenvectors (K_1 = \begin{pmatrix} 1 \ -4 \end{pmatrix}) and (K_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix}). Thus, [ X_c = c_1 \begin{pmatrix} 1 \ -4 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{7t} ] Assume a particular solution of the form (X_p = \begin{pmatrix} a_2 \ b_2 \end{pmatrix} t + \begin{pmatrix} a_1 \ b_1 \end{pmatrix}). Substituting into the system and equating coefficients gives:

• From the (t) terms: (6a_2 + b_2 + 6 = 0) and (4a_2 + 3b_2 - 10 = 0), solving to (a_2 = -2, b_2 = 6).
• From the constant terms: (6a_1 + b_1 - a_2 = 0) and (4a_1 + 3b_1 - b_2 + 4 = 0), solving to (a_1 = -\frac{4}{7}, b_1 = \frac{10}{7}). Thus, the particular solution is: [ X_p = \begin{pmatrix} -2 \ 6 \end{pmatrix} t + \begin{pmatrix} -4/7 \ 10/7 \end{pmatrix} ] The general solution is: [ X = c_1 \begin{pmatrix} 1 \ -4 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{7t} + \begin{pmatrix} -2 \ 6 \end{pmatrix} t + \begin{pmatrix} -4/7 \ 10/7 \end{pmatrix} ]

Example 3

Determine the form of the particular solution vector (X_p) for: [ \frac{dx}{dt} = 5x + 3y - 2e^{-t} + 1, \quad \frac{dy}{dt} = -x + y + e^{-t} - 5t + 7 ]

Solution: In matrix form: (X' = \begin{pmatrix} 5 & 3 \ -1 & 1 \end{pmatrix} X + \begin{pmatrix} -2 \ 1 \end{pmatrix} e^{-t} + \begin{pmatrix} 0 \ -5 \end{pmatrix} t + \begin{pmatrix} 1 \ 7 \end{pmatrix}). The homogeneous system has eigenvalues (\lambda = 2, 4), with eigenvectors (K_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix}) for (\lambda=2) and (K_2 = \begin{pmatrix} 3 \ -1 \end{pmatrix}) for (\lambda=4). So, [ X_c = c_1 \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 3 \ -1 \end{pmatrix} e^{4t} ] Since (e^{-t}) in (F(t)) does not match any eigenvalue, we assume: [ X_p = \begin{pmatrix} a_3 \ b_3 \end{pmatrix} e^{-t} + \begin{pmatrix} a_2 \ b_2 \end{pmatrix} t + \begin{pmatrix} a_1 \ b_1 \end{pmatrix} ] Note: If (e^{-t}) were replaced by (e^{2t}) (matching eigenvalue (\lambda=2)), the correct form would be: [ X_p = \begin{pmatrix} a_4 \ b_4 \end{pmatrix} t e^{2t} + \begin{pmatrix} a_3 \ b_3 \end{pmatrix} e^{2t} + \begin{pmatrix} a_2 \ b_2 \end{pmatrix} t + \begin{pmatrix} a_1 \ b_1 \end{pmatrix} ]

44.5 Variation of Parameters

Variation of parameters is a more powerful technique than the method of undetermined coefficients. It provides a systematic procedure for finding a particular solution of the non-homogeneous linear vector differential equation: [ \frac{dX}{dt} = AX + F(t) ]

Let (\phi(t)) be a fundamental matrix of the homogeneous system (X' = AX). Then the general solution of the homogeneous system is (X_c = \phi(t)C), where (C) is an arbitrary constant vector. We replace the constant vector (C) by a column matrix of functions (U(t)) so that: [ X_p = \phi(t) U(t) ] is a particular solution of the non-homogeneous system.

Differentiating using the product rule: [ X'_p = \phi(t) U'(t) + \phi'(t) U(t) ] Substituting into (X' = AX + F(t)) and using (\phi'(t) = A\phi(t)) gives: [ \phi(t) U'(t) = F(t) ] Multiplying by (\phi^{-1}(t)) (the inverse of the fundamental matrix) yields: [ U'(t) = \phi^{-1}(t) F(t) ] Integrating: [ U(t) = \int \phi^{-1}(t) F(t) dt ] Therefore, a particular solution is: [ X_p = \phi(t) \int \phi^{-1}(t) F(t) dt ] The general solution of the non-homogeneous system is then: [ X = X_c + X_p = \phi(t) C + \phi(t) \int \phi^{-1}(t) F(t) dt ]

Example

Find the general solution of the non-homogeneous system on ((-\infty, \infty)): [ X' = \begin{pmatrix} -3 & 1 \ 2 & -4 \end{pmatrix} X + \begin{pmatrix} 3t \ e^{-t} \end{pmatrix} ]

Solution: Solve the homogeneous system first. The characteristic equation is (\lambda^2 + 7\lambda + 10 = 0), giving eigenvalues (\lambda_1 = -2) and (\lambda_2 = -5). Eigenvectors are (K_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}) and (K_2 = \begin{pmatrix} 1 \ -2 \end{pmatrix}). The solution vectors are: [ X_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{-2t}, \quad X_2 = \begin{pmatrix} 1 \ -2 \end{pmatrix} e^{-5t} ] The fundamental matrix is: [ \phi(t) = \begin{pmatrix} e^{-2t} & e^{-5t} \ e^{-2t} & -2e^{-5t} \end{pmatrix} ] Its inverse is: [ \phi^{-1}(t) = \begin{pmatrix} \frac{2}{3} e^{2t} & \frac{1}{3} e^{2t} \ \frac{1}{3} e^{5t} & -\frac{1}{3} e^{5t} \end{pmatrix} ] Now, compute (X_p = \phi(t) \int \phi^{-1}(t) F(t) dt): [ \phi^{-1}(t) F(t) = \begin{pmatrix} \frac{2}{3} e^{2t} & \frac{1}{3} e^{2t} \ \frac{1}{3} e^{5t} & -\frac{1}{3} e^{5t} \end{pmatrix} \begin{pmatrix} 3t \ e^{-t} \end{pmatrix} = \begin{pmatrix} 2t e^{2t} + \frac{1}{3} e^{t} \ t e^{5t} - \frac{1}{3} e^{4t} \end{pmatrix} ] Integrating: [ \int \phi^{-1}(t) F(t) dt = \begin{pmatrix} \int (2t e^{2t} + \frac{1}{3} e^{t}) dt \ \int (t e^{5t} - \frac{1}{3} e^{4t}) dt \end{pmatrix} = \begin{pmatrix} t e^{2t} - \frac{1}{2} e^{2t} + \frac{1}{3} e^{t} \ \frac{1}{5} t e^{5t} - \frac{1}{25} e^{5t} - \frac{1}{12} e^{4t} \end{pmatrix} ] Then: [ X_p = \phi(t) \begin{pmatrix} t e^{2t} - \frac{1}{2} e^{2t} + \frac{1}{3} e^{t} \ \frac{1}{5} t e^{5t} - \frac{1}{25} e^{5t} - \frac{1}{12} e^{4t} \end{pmatrix} ] Multiplying and simplifying yields: [ X_p = \begin{pmatrix} \frac{6}{5} t - \frac{27}{50} + \frac{1}{4} e^{-t} \ \frac{3}{5} t - \frac{21}{50} + \frac{1}{2} e^{-t} \end{pmatrix} ] The general solution is: [ X = X_c + X_p = c_1 \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{-2t} + c_2 \begin{pmatrix} 1 \ -2 \end{pmatrix} e^{-5t} + \begin{pmatrix} \frac{6}{5} t - \frac{27}{50} + \frac{1}{4} e^{-t} \ \frac{3}{5} t - \frac{21}{50} + \frac{1}{2} e^{-t} \end{pmatrix} ]

44.6 Exercise

(The lecture text provides a list of exercises for the student to practice the methods covered, but no solution is given in the lecture.)

⭐ Key Takeaways

1. The general solution of a non-homogeneous linear system is the sum of the complementary function (solution to the homogeneous system) and any particular solution. The complementary function is found by solving (X' = AX) using eigenvalues and eigenvectors. The method of undetermined coefficients is a standard technique for finding a particular solution when the forcing function (F(t)) is of a specific form (polynomial, exponential, sine/cosine, or their products), but it requires adjusting the assumed form to avoid duplication with the complementary function. Variation of parameters is a more general method that always works, provided the fundamental matrix of the homogeneous system is known; it involves integrating the product of the inverse fundamental matrix and the forcing function. The ability to correctly identify which method to use and how to handle duplication of terms is critical for exam success.

🧠 Quick Revision Questions

1. What is the general form of the solution to a non-homogeneous system (X' = AX + F(t))?
2. When using the method of undetermined coefficients, what condition must be satisfied by the entries of (F(t))?
3. If (F(t) = \begin{pmatrix} 8 \ -3 \end{pmatrix}) and the coefficient matrix (A) has an eigenvalue (\lambda = 0), what form should the particular solution (X_p) take to avoid duplication?
4. In the variation of parameters method, what is the fundamental matrix (\phi(t)) and how is it used to find a particular solution?
5. In example 3, why would the assumed form for (X_p) change if the exponential term in (F(t)) matched an eigenvalue?