Inhomogeneous ODE's

The essence of mathematics is not to make simple things complicated, but to make complicated things simple, S. Gudder

I’d far rather be happy than right any day, Douglas Adams, The Hitchhiker’s Guide to the Galaxy.

Recall

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

• Dependent variables: Variables that depend on one or more other variables (y).
• Independent variables: Variables upon which the dependent variables depend (x).
• Derivatives: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

• y = y(x) is the unknown function of the independent variable x.
• y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
• a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

• The function f(x, y) (the right-hand side of the ODE) in the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

Second-order Linear Homogeneous ODE’s with Constant Coefficients

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0, where A and B are constants.

To solve this ODE, we seek two linearly independent solutions y₁(t) and y₂(t). The general solution is then a linear combination of these solutions: $c_1y_1 + c_2y_2$ where c₁ and c₂ are two arbitrary constants determined by initial conditions. The key to solving the ODE is the characteristic equation, whose roots determine the behavior of the solutions.

Second-order Linear Homogeneous ODE’s

Second-order linear homogeneous ordinary differential equations (ODEs) are fundamental tools in mathematical modeling and appear frequently in physics, engineering, and other sciences. They describe a wide range of phenomena, such as mechanical vibrations, electrical circuits, and fluid dynamics where the rate of change of a quantity depends linearly on the quantity and its derivatives.

Definition. A second-order linear homogeneous ODE is a differential equation of the form: y'' + p(x)y' + q(x)y = 0 where:

• y = y(x) is the unknown function of x.
• y’ and y’’ are the first and second derivative of y with respect to x, respectively.
• p(x) and q(x) are given functions of x, which may be constants or variable functions, depending on the problem.

The term linear refers to the fact that the unknown y and its derivatives y’ and y′′ appear to the first power only and are not multiplied together. That is, there are no squares (y²), cubes (y’⁽³⁾) terms, or other nonlinear expression.

The term homogeneous means the right-hand side of the equation is zero. In other words, the equation is set equal to zero, indicating that there are no external forcing functions. This contrasts with nonhomogeneous (or inhomogeneous) equations, where the right-hand side is a non-zero function f(x), leading to an equation of the form: y’’ + p(x)y’ + q(x)y = f(x).

General Solution of a Second-Order Linear Homogeneous ODE

The goal when solving such ODEs is to find the general solution, which represents all possible solutions that satisfy the differential equation.

The general solution to a second-order linear homogeneous ODEs is: y = c₁y₁ + c₂y₂ where:

• y₁ and y₂ are two linearly independent solutions to the ODE.
• c₁ and c₂ are arbitrary constants determined by initial or boundary conditions specific to a given problem.

Inhomogeneous Second-Order Linear Differential Equations

An inhomogeneous second-order linear differential equation is a differential equation of the form: y'' + p(x)y' + q(x)y = f(x) where:

• y is the unknown function of x.
• y’ and y’’ are the first and second derivative of y with respect to x, respectively.
• f(x), p(x) and q(x) are given functions of x.
• f(x) is called the input, forcing term, driving term, or sometimes the signal.

The goal is to find y(x), known as the response or output, that satisfies this equation.

Solving the Inhomogeneous Equation

To solve the inhomogeneous equation, we proceed in three main steps:

1. Solve the Associated Homogeneous Equation. Before addressing the inhomogeneous equation, we first solve the corresponding homogeneous equation: y’’ + p(x)y’ + q(x)y = 0.

   This is called the associated or reduced homogeneous equation. Its solutions are denoted as y_c(x) = c₁y₁(x) + c₂y₂(x), where: y₁ and y₂ are linearly independent solutions of the homogeneous equation. These solutions are known as the complementary solutions. c₁ and c₂ are arbitrary constants determined by initial or boundary conditions.

   Linear independence means that one solution cannot be expressed as a scalar multiple of the other. The Wronskian W(y₁, y₂)(x) is non-zero if y₁ and y₂ are linearly independent on the interval of interest.

2. Find a Particular Solution. Next, we find a particular solution y_p​(x) to the inhomogeneous equation: y’’ + p(x)y’ + q(x)y = f(x). The particular solution y_p(x) satisfies the inhomogeneous equation and accounts for the presence of the forcing term f(x).

3. The general solution to the inhomogeneous equation is the sum of the complementary solution y_c and a particular solution y_p: y(x) = y_p(x) + y_c(x) = y_p(x) + c₁y₁(x) + c₂y₂(x).

• y_p(x) is any particular solution to the inhomogeneous equation. It represents a specific solution that responds to the forcing term f(x) (it models the system’s response directly due to the external input).
• y_c(x) = c₁y₁(x) + c₂y₂(x) is the general solution to the associated homogeneous equation. It accounts for the behavior of the system without the forcing term f(x). It represents the system’s natural behavior without external forcing or input.
• By combining these, we capture all possible solutions to the inhomogeneous equation.

Examples

In this section, we explore real-world systems that can be modeled using inhomogeneous second-order linear differential equations. Specifically, we will discuss:

• A spring-mass-dashpot system models the motion of a mass attached to a spring and a damper (dashpot). The equation governing this system is: mx’’+bx’ + kx = f(t) where:

m: Mass of the object (in kilograms).
b: Damping coefficient (in kilograms per second), related to the dashpot’s resistance.
k: Spring constant (in newtons per meter), representing the stiffness of the spring.
x(t): Displacement of the mass from its equilibrium position at time t.
f(t): External force applied to the system (in newtons). The system is said to be forced or influenced by an external force when f(t) ≠ 0, and passive or free when f(t) = 0, meaning that the system evolves solely based on its initial conditions and inherent properties.

mx’’ (inertia term) represents the mass’s resistance to acceleration (Newton’s second law); bx’ (damping term) accounts for the energy loss due to friction or resistance in the dashpot; kx (spring term) represents the restoring force exerted by the spring, proportional to the displacement x; f(t) (external force) is an input function that drives the system, such as an applied push or pull.

Applications: Mechanical engineering (design of suspension system in vehicles), civil engineering (analysis of building responses to earthquakes), and physics (study of oscillatory motion and vibrations).

• Simple Electrical Circuit (RLC Circuit). An RLC circuit consists of:

A Resistor (R) provides resistance to the flow of current.
An Inductor (L) opposes changes in current due to its magnetic field.
A Capacitor (C) stores electrical energy in an electric field.
A Voltage source ε(t) supplies electromotive force (EMF), which can be constant or time-varying.

According to Kirchhoff’s Voltage Law (KVL), the sum of voltage drops around a closed loop equals the applied voltage: $L\frac{di}{dt}+Ri+\frac{q}{C} = ε(t)$ where:

i(t) is the current flowing through the circuit (the rate of flow of electric charge)
q(t) is the amount of electric charge stored in the capacitor. Relationship: $\frac{dq}{dt} = i(t)$, meaning the rate of flow of electric charge (dq/dt) with respect to time is equal to the current (i).

Voltage Drops Across Circuit Components:

1. Inductor (L): The voltage drop across an inductor is proportional to the rate of change of current flowing through it: V_L = L * di/dt, where i is the current, di/dt represents the rate of change of current, and t is time.
2. Resistor (R): Following Ohm’s Law, the voltage drop across a resistor is proportional to the current flowing through it: V_R = R * i(t).
3. Capacitor (C): The voltage drop across a capacitor is proportional to the charge stored on its plates: V_C = q/C, where q is the charge of the capacitor and C is the capacitance, a measure of the capacitor’s ability to store charge.
4. Voltage source (ε(t)): The voltage source provides the driving electromotive force (EMF) for the circuit, which can be constant or time-varying, influencing the circuit’s response (Refer to Figure 5 for a visual representation and aid in understanding it).

Deriving the Differential Equation

Therefore, $\frac{dq}{dt} = i$ and applying KVL to the RLC circuit gives us the equation: $V_L + V_R + V_C = ε(t) ↭ L\frac{di}{dt}+Ri+\frac{q}{C} = ε(t)$. Differentiating this equation gives: $L\frac{d^2i}{dt^2}+R\frac{di}{dt} + \frac{i}{C} = ε’(t)$ ↭[Notation] Li’’ + Ri’ + ⁱ⁄_C = ε’(t). This equation governs the flow of current in the circuit in response to an applied voltage ε(t).

General Solution to the Inhomogeneous Equation

An inhomogeneous second-order linear differential equation is an equation of the form: y′′ +p(x)y′ +q(x)y = f(x).

We can express this equation more concisely using a linear differential operator L. Let L = D² +pD + q =[Notation] $\frac{d^2}{dx^2}+p(x)\frac{d}{dx} + q(x)$ be the differential operator, so our inhomogeneous second-order linear differential equation becomes: Ly = f(x) and the associated homogeneous equation is: Ly = 0.

The general solution to the inhomogeneous equation Ly = f(x) is: y(x) = y_p + y_c = y_p + c₁y₁ + c₂y₂ where:

• y_c(x) = c₁y₁ + c₂y₂ is the general solution to the homogeneous equation Ly = 0. It represents all possible solutions to the homogeneous equation and accounts for the natural behavior of the system without external forcing.
• y₁ and y₂ are linear independent solutions to the associated homogeneous equation Ly = 0.
• c₁ and c₂ are arbitrary constants determined by initial or boundary conditions.
• y_p is a particular solution to the inhomogeneous equation, i.e., it satisfies y_p’’ + p(x)y_p’ + q(x)y_p = f(x) or equivalently Ly = f(x) (particular solution meaning that any solution will do, i.e., it is not unique). It accounts for the effect of (or captures the system’s response to) the external forcing term f(x).

The reason for adding the complementary solution is that the inhomogeneous equation consists of two parts:

1. The forcing term f(x), which is responsible for the inhomogeneity; it represents external influences or inputs to the system.
2. The homogeneous part y′′ +p(x)y′ +q(x)y = 0, which describes the system’s inherent behaviour without external forces.

The complementary solution y_c accounts for the behavior of the system without the forcing term (system's natural behaviour), while the particular solution y_c responds directly to the input or external forcing term f(x).

Proof.

Let’s prove why the general solution is of the form y(x) = y_p + c₁y₁ + c₂y₂

Step 1: Proving the form is a solution

We need to prove that y = y_p + y_c = y_p + c₁y₁ + c₂y₂ is a solution to the original equation y’’ + p(x)y’ + q(x)y = f(x) ↭ Ly = f(x)

L(y) = L(y_p + c₁y₁ + c₂y₂) =[L is a linear operator] L(y_p) + L(c₁y₁ + c₂y₂) =[y_c = c₁y₁ + c₂y₂ is a solution to the associated homogeneous equation, i.e., L(y_c) = L(c₁y₁ + c₂y₂) = 0. We knot that L(y_p) = f(x) because y_p is a particular solution ] f(x) + 0 = f(x)

Step 2: Proving no other solutions exist

Second, we need to show that there are no other solutions. Now, assume that u(x) is another solution to the inhomogeneous equation ⇒[By definition, u is a solution and y_p is a particular solution] L(u) = f(x), L(y_p) = f(x).

Subtract these two equations: L(u)-L(y_p) = f(x) -f(x) = 0.

Since L is a linear operator, this simplifies to: L(u-y_p) = 0. This means that u-y_p is a solution to the associated homogeneous solution Ly = 0.

Therefore, u-y_p must be a linear combination of y₁ and y₂: $∃\tilde{c_1},\tilde{c_2}: u-y_p = \tilde{c_1}y_1 + \tilde{c_2}y_2 ⇒u = y_p + \tilde{c_1}y_1 + \tilde{c_2}y_2$, which is exactly of the form y_p + c₁y₁ + c₂y₂.

This proves that the general solution y = y_p + c₁y₁ + c₂y₂ includes all possible solutions to the inhomogeneous equation ∎

Solved examples

• y’’ -3y’ + 2y = e^2x

1. Solve the Homogeneous Equation. The associated homogeneous equation is obtained by setting f(x) = 0: y’’ -3y’ + 2y = 0. To solve this linear differential equation, we assume a solution of the form y = e^rx, leading to the characteristic equation r² -3r + 2 = 0⇒[Solving the quadratic equation.] Roots: r₁ = 1 and r₂ = 2. Since we have two distinct and real roots, the general solution to the homogeneous equation is (complementary Solution): $y_c(x) = c_1e^x + c_2e^{2x}$ where c₁ and c₂ are arbitrary constants.

2. Find a Particular Solution. Since the right-hand side is f(x) = e^2x, which is similar to one term in y_c(x) (y₂), we cannot use y_p(x) = Ae^2x as our particular solution because it would be linearly dependent on the complementary solution, we modify our trial solution to account for this.

   Assumed Form: y_p(x) = $xe^{2x}$. Compute Derivatives: y_p’(x) = $e^{2x} + 2xe^{2x}$, y_p’’(x) = $2e^{2x} + 2e^{2x} + 2xe^{2x}·2 = 4e^{2x}+4xe^{2x}$

   Substitute into the Inhomogeneous Equation: $[4e^{2x}+4xe^{2x}] -3[e^{2x} + 2xe^{2x}] +2xe^{2x} = e^{2x} ↭ (4-3)e^{2x} + (4-6+2)xe^{2x} = e^{2x}↭ e^{2x} = e^{2x}$, confirming that $y_p(x) = xe^{2x}$ is a valid particular solution.

3. General Solution: The general solution is the sum of the complementary and particular solutions: $y(x) = y_c(x) + y_p(x) = c_1e^x + c_2e^{2x} + xe^{2x}$

• y’’ -2y’ -3y = 3e^2t

1. Solve the Homogeneous Equation. The associated homogeneous equation is: y’’ -2y’ -3y = 0. Find the characteristic equation r² -2r -3 = 0⇒[Solve for r] Roots (two real distinct roots): r = 3 and r = -1. Complementary Solution: $y_c(t) = c_1e^{3t} + c_2e^{-t}$.

2. Find a Particular Solution. We observe that e^2t is not a solution to the homogeneous equation (since r = 2 is not a root of the characteristic equation). Therefore, we can assume: $y_p = Ae^{2t}$.

   Compute derivatives: $y_p’(t) = 2Ae^{2t}, y_p’’(t) = 4Ae^{2t}$.

   Substitute into the Inhomogeneous Equation: $4Ae^{2t} -4Ae^{2t} -3Ae^{2t} = 3e^{2t} ↭[\text{Simplify}] -Ae^{2t} = e^{2t}⇒[\text{Divide both sides by }e^{2t}] A = -1$. Therefore, the particular solution is $y_p = -e^{2t}$.

3. Write the General solution. The general solution is $y(x) = y_c(x) + y_p(x) = c_1e^{3t} + c_2e^{-t} -e^{2t}$

Note on Choosing the Particular Solution

When selecting the form of the particular solution y_p(x), consider the nature of the non-homogeneous term f(x).

• If the non-homogeneous term is of the form e^rt and r is not a root of the characteristic equation, we can assume y_p = Ae^rt.
• If r is a root of the characteristic equation, we must multiply the assumed particular solution by t until it becomes linearly independent from the complementary solution.

Table of Common Non-Homogeneous Terms and Corresponding Particular Solution Forms

Multiply the assumed form by x^s as needed to avoid duplication with terms in the complementary solution.

• y’’ + 5y’ + 6y = x²

1. Solve the Homogeneous Equation. The associated homogeneous equation is: y’’ + 5y’ + 6y = 0. Find the characteristic equation r² +5r + 6 = 0 ↭ (r + 2)(r + 3) = 0 ⇒ Roots (two real distinct roots) r₁ = -2, r₂ = -3. Complementary Solution: $y_c(x) = c_1e^{-2x} + c_2e^{-3x}$.

2. Find a particular Solution. Since the right hand is x², a polynomial of degree two, we assume the particular solution y_p(x) is a polynomial of the same degree. y_p(x) = Ax² + Bx + C.

   Compute derivatives: y_p’(x) = 2Ax + B, y_p’’(x) = 2A.

   Substitute into the Inhomogeneous Equation: 2A + 5(2Ax + B) + 6(Ax² + Bx + C) = x² ↭ 2A + 10Ax + 5B + 6Ax² + 6Bx + 6C = x² ↭[Group like terms:] (6A)x² + (10A +6B)x + (2A + 5B + 6C) = x²

   Set up a system of equations by equating coefficients:

   $\begin{cases} 6A = 1~ (i) \\ 10A +6B = 0 ~ (ii)\\ 2A + 5B + 6C~ (iii)\end{cases}$

   Solve for A, B, and C:

   (i) ⇒ A = ¹⁄₆. (ii) $10·\frac{1}{6} + 6B = 0↭[\text{Multiply by 6}] 10 + 36B = 0 ⇒ B = \frac{-10}{36} = \frac{-5}{18}$. (iii) $2·\frac{1}{6}+5\frac{-5}{18} +6C = 0 ↭[\text{Multiply by 18}] 6 -25 + 108C = 0↭ -19 +108C = 0 ⇒ C = \frac{19}{108}$. Therefore, the particular solution is $y_p = Ax^2+Bx+C = \frac{1}{6}x^2+\frac{-5}{18}x+\frac{19}{108}$.

3. Write the general solution: $y(x) = y_c(x) + y_p(x) = c_1e^{-2x} + c_2e^{-3x} + \frac{1}{6}x^2+\frac{-5}{18}x+\frac{19}{108}$.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.