Solving First-Order Linear Inhomogeneous ODEs and Stability Analysis

The only way to learn mathematics is to do mathematics, Paul Halmos.

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ODE involving an unknown function $y(x)$, its first derivative $y'$, and functions of the independent variable $x$. It 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, continuous functions of $x$.

These equations are termed linear because the unknown function $y$ and its derivative $y'$ appear only to the first power. They are never multiplied together (e.g., $yy'$), squared (e.g., $(y')^2$), or composed inside other functions in any nonlinear way (e.g., $\sin(y)$ or $e^y$).

By dividing the entire equation by $a(x)$ (assuming $a(x) \neq 0$), we get the Standard Form: $y' + p(x)y = q(x)$

If the function $q(x) = 0$ for all $x$ in the interval of interest, the equation simplifies to $y' + p(x)y = 0$, which is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem

If the functions $p(x)$ and $q(x)$ are continuous on an open interval $I$ containing the point $x_0$, then the initial value problem $y' + p(x)y = q(x)$ with $y(x_0) = y_0$ has a unique solution that exists on the entire interval $I$.

This is a powerful guarantee: the solution will not blow up or behave unpredictably as long as the coefficients $p(x)$ and $q(x)$ remain continuous.

Solving the First-Order Linear Inhomogeneous ODE

Let’s look at a linear inhomogeneous differential equation with a constant coefficient $k$ and an independent variable $t$ (often representing time): $y' + ky = q(t)$ where:

• y is the unknown function of the variable t.
• y′ is the derivative of y with respect to t.
• k is a constant coefficient (a specific case of $p(t)$).
• q(t) is a given function of t, known as the inhomogeneous term, forcing term, or input.

Our goal is to find the general solution, which consists of two parts: the complementary solution $y_c$ and the particular solution $y_p$.

Step 1: Solving the Homogeneous Equation

The homogeneous equation corresponding to the given differential equation is obtained by setting the inhomogeneous term q(t) = 0: $y' + ky = 0$. This equation describes the system’s natural, unforced behavior. We solve it using separation of variables:

Step 1. Solving the Homogeneous Equation.$ y' + ky = 0 \implies \frac{dy}{dt} = -ky$.

To solve this, we separate variables: $\frac{dy}{y}=-kdt$.

Integrate both sides: $\int \frac{dy}{y} = \int -kdt$. $\ln|y| = -kt + C_1$ where $C_1$ is the constant of integration.

Exponentiate both sides to solve for $y$: $|y| = e^{-kt + C_1} = e^{C_1}e^{-kt}$.

Let $C = \pm e^{C_1}$. Thus, the complementary solution is: $y_c(t) = Ce^{-kt}$ where C is an arbitrary constant determined by the initial conditions of the system.

Step 2: Solving the Inhomogeneous Equation

To find a particular solution $y_p(t)$ to the inhomogeneous equation $y' + ky = q(t)$, we use the Method of Integrating Factors.

The integrating factor, $\mu(t)$, is defined as $e^{\int p(t)dt}$. Since our $p(t)$ is just the constant $k$, the integrating factor is: $\mu(t) = e^{\int k dt} = e^{kt}$

Multiply every term in the original standard-form equation by $e^{kt}$: $e^{kt}y' + ke^{kt}y = e^{kt}q(t)$

This simplifies the left-hand side to: $\frac{d}{dt}(e^{kt}·y)=e^{kt}q(t)$.

Now, integrate both sides with respect to $t$: $e^{kt}y = \int e^{kt}q(t)dt$

Divide by $e^{kt}$ to isolate $y$, and it yields the particular solution: $y_p(t) = e^{-kt}\int e^{kt}q(t)dt$

Step 3: Writing the General Solution

The general solution to the linear inhomogeneous differential equation of the form is the sum of the complementary and particular solution:$y(t) = y_c(t) + y_p(t) = Ce^{-kt} + e^{-kt}\int e^{kt}q(t)dt$.

This structure holds deep physical significance:

• The Complementary Solution ($y_c$). It solves the corresponding homogeneous equation y’ + ky = 0. It represents the system’s inherent behavior based on its initial state, independent of external forces (how the system would behave in the absence of the external forcin). It contains the arbitrary constant $C$.
• The Particular Solution ($y_p$): Represents the system’s specific response to the external forcing term $q(t)$. It does not contain any arbitrary constants (the constant of integration is already accounted for in $y_c$).

Long-Term Behavior as $t \to \infty$

Understanding the limit of $y(t)$ as time progresses helps us interpret the physical implications of the constant $k$.

Case 1: $k > 0$ (Stable System)

If $k$ is positive, the exponential term $e^{-kt}$ decays to zero as $t \to \infty$.

Then, $\lim_{t \to \infty} y_c(t) = \lim_{t \to \infty} Ce^{-kt} = 0$

In this scenario, the complementary solution acts as a transient state that eventually fades away. Over time, the system “forgets” its initial conditions, and the solution $y(t)$ is entirely dominated by the particular solution $y_p(t)$ (the steady-state response driven by $q(t)$).

Case 2: $k < 0$ (Unstable System)

If $k$ is negative, the exponent $-kt$ becomes positive. As $t \to \infty$, the term $e^{-kt}$ grows exponentially.

Then, $\lim_{t \to \infty} y_c(t) = \lim_{t \to \infty} Ce^{-kt} = \infty \quad \text{(assuming) } C > 0)$

Here, the transient part of the solution grows without bound. The system exhibits explosive instability, and the initial conditions dominate the long-term behavior. The external forcing term $q(t)$ becomes negligible (it has little effect on the long-term behaviour) in comparison to the rapidly growing homogeneous response.

Solved Examples: First-Order Linear ODEs

• Example 1: Constant Coefficient with Exponential Forcing Term.

Solve: $y' + 2y = e^{3t}$

Step 1: Solve the Homogeneous Equation. The homogeneous equation is obtained by setting the right-hand side (the forcing term) to zero: $y' + 2y = 0$.

1. Separate Variables: $\frac{dy}{dt} = -2y \implies \frac{dy}{y} = -2dt$

2. Integrate both sides: $\int \frac{dy}{y} = \int-2dt ⇒ ln|y| = -2t + C'$

3. Solve for y. Exponentiate both sides to isolate y: $|y| = e^{-2t+C'} = C''e^{-2t}$ where $C'' = e^{C'}$. $y = \pm C''e^{-2t} = Ce^{-2t}$ where C = ±C’’(an arbitrary constant).

4. Complementary Solution: $y_c(t) = Ce^{-2t}$

Step 2: Find a Particular Solution

We will use the method of integrating factors to find a particular solution $y_p(t)$.

1. Compute the Integrating Factor: $μ(t) = e^{\int 2dt} = e^{2t}$

2. Multiply Both Sides. Multiply the original non-homogeneous equation by $e^{2t}$: $e^{2t}y' + 2e^{2t}y = e^{2t}e^{3t}$

3. Simplify Left-Hand Side. Recognize that the left side is the expanded product rule of $(e^{2t}y)$. $\frac{d}{dt}(e^{2t}y) = e^{5t}$

4. Integrate both sides: $e^{2t}y = \int e^{5t}dt + C ↭ e^{2t}y = \frac{1}{5}e^{5t} + C$. We ignore the constant of integration here because it is already accounted for in the complementary solution.

5. Solve for $y_p(t)$: Divide by $e^{2t}$: $y_p(t) = e^{-2t} \left( \frac{1}{5}e^{5t} \right) = \frac{1}{5}e^{3t}$

Step 3: Write the General Solution

The general solution is the sum of the complementary and particular solutions: $\boxed{y(t) = y_c(t) + y_p(t) = Ce^{-2t} + \frac{1}{5}e^{3t}}$.

Step 4: Analysis of the solution

1. Complementary solution. Since $e^{-2t}$ decays exponentially to zero as $t \to \infty$, the complementary solution approaches zero. This term represents the transient behavior of the system based on its initial state.
2. Particular Solution ($\frac{1}{5}e^{3t}$): Since $e^{3t}$ grows exponentially as $t \to \infty$, the particular solution increases without bound. This term represents the steady-state response of the system to the external forcing term $e^{3t}$.
3. Overall Behavior: For large $t$, the transient term vanishes and the particular solution dominates. Therefore, $y(t) \approx \frac{1}{5}e^{3t}$ as $t \to \infty$

• Example 2: Variable Coefficient with Linear Forcing Term.

Solve: $y' + 2ty = 4t$

Step 1: Solve the Homogeneous Equation. Write the Homogeneous Equation by dropping the forcing term: y’ + 2ty = 0.

1. Separate Variables: $\frac{dy}{dt} = -2ty ⇒ \frac{dy}{y} = -2tdt$

2. Integrate both sides: $\int \frac{dy}{y} = \int-2tdt \implies ln|y| = -t^2 + C'$

3. Solve for y: $|y| = e^{-t^2+C'} = C''e^{-t^2}$ where C’’ = e^C’. $y = ± C''e^{-t^2} = Ce^{-t^2}$ where C = ±C’'.

4. Complementary Solution: $y_c(t) = Ce^{-t^2}$

Step 2: Find a Particular Solution. We will use the method of integrating factors.

1. Compute the Integrating Factor: $μ(t) = e^{\int 2tdt} = e^{t^2}$

2. Multiply Both Sides. Multiply the original equation by $e^{t^2}$: $e^{t^2}y' + 2te^{t^2}y = 4te^{t^2}$

3. Simplify Left-Hand Side: $\frac{d}{dt}(e^{t^2}y) = 4te^{t^2}$

4. Integrate both sides: $e^{t^2}y = \int 4te^{t^2} dt$
   Use $u$-substitution: Let $u = t^2$, then $du = 2t dt$. The integral becomes: $\int 2(2t e^{t^2}) dt = \int 2e^u du = 2e^u = 2e^{t^2}$.
   So, $e^{t^2}y = 2e^{t^2}$

5. Solve for $y_p(t)$: Divide both sides by $e^{t^2}$. $y_p(t) = 2$ Ignoring the constant C (it will be accounted for in the complementary solution).

Step 3: Write the General Solution.

Combine the complementary and particular solutions: $\boxed{y(t) = y_c(t) + y_p(t) = Ce^{-t^2} + 2}$ where C is a constant determined by initial or boundary conditions.

Step 4. Analysis of the Solution

1. Complementary Solution ($Ce^{-t^2}$): This decays to zero as $t \to \infty$ because $e^{-t^2}$ decreases exponentially (and does so much faster than a standard exponential function).

2. Particular Solution ($2$): The particular solution is a constant. It represents the steady-state response to the forcing term $4t$ acting against the dampening term $2ty$.

3. Overall Behavior: As time progresses ($t \to \infty$), the initial conditions are “forgotten” as the transient term vanishes. The system approaches a stable equilibrium. Therefore, $y(t) \approx 2$ as $t \to \infty$.

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