Second Order Nonhomogenous Differential Equations

Thus far we have only really looked at methods to solving second order linear homogenousdifferential equations. We will now look at methods for solving second order linear nonhomogenous differential equations. Recall that for $g(t) \neq 0$, such a differential equation comes in the following form:

(1)

\begin{align} \quad \frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t) \end{align}

When solving a second order linear nonhomogenous differential equation, we will make use of the underlying second order homogenous differential equation which we define below.

Definition: If $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$ is a second order linear nonhomogenous differential equation then the Corresponding Homogenous Differential Equation is $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0$.

It should be noted that the general solution to the corresponding homogenous differential equation is sometimes called the Complementary Solution.

For example, consider the following second order linear nonhomogenous differential equation:

(2)

\begin{align} \quad \frac{d^2 y}{dt^2} + (t^2 - \sin t) \frac{dy}{dt} + (e^t - 3t)y = te^t \end{align}

Then the corresponding homogenous differential equation is:

(3)

\begin{align} \quad \frac{d^2 y}{dt^2} + (t^2 - \sin t) \frac{dy}{dt} + (e^t - 3t)y = 0 \end{align}

Theorem 1: Let $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$ be a second order linear nonhomogenous differential equation and suppose that $y = Y_1(t)$ and $y = Y_2(t)$ are solutions to this differential equation. Then the difference, $Y_1 - Y_2$ is a solution to the corresponding homogenous differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$. Furthermore, if $y = y_1(t)$ and $y = y_2(t)$ form a fundamental set of solutions for the corresponding homogenous differential equation, then for some constants $C$ and $D$ we have that $Y_1(t) - Y_2(t) = Cy_1(t) + Dy_2(t)$.

• Proof: Let $y = Y_1(t)$ and $y = Y_2(t)$ be solutions to the second order linear nonhomogenous differential equation $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$. Then plugging in the difference $Y_1 - Y_2$ into our second order linear nonhomogenous differential equation and we have that:

(4)

\begin{align} \quad \frac{d^2Y_1}{dt^2} + p(t) \frac{dY_1}{dt} + q(t) Y_1 = g(t) \quad \mathrm{and} \quad \frac{d^2Y_2}{dt^2} + p(t) \frac{dY_2}{dt} + q(t) Y_2 = g(t) \end{align}

• Subtracting the second equation from the first and we have that:

(5)

\begin{align} \quad \frac{d^2}{dt^2} (Y_1 - Y_2) + p(t) \frac{d}{dt} (Y_1 - Y_2) + q(t) (Y_1 - Y_2) = g(t) - g(t) = 0 \end{align}

• Thus we see that $y = Y_1(t) - Y_2(t)$ is a solution to the corresponding second order linear homogenous differential equation.

• Now suppose that $y = y_1(t)$ and $y = y_2(t)$ form a fundamental set of solutions for the corresponding second order linear homogenous differential equation. Then all solutions of this differential equation can be expressed as a linear combination of $y_1$ and $y_2$. But $Y_1(t) - Y_2(t)$ is a solution, and so there exists constants $C$ and $D$ such that:

(6)

\begin{align} \quad Y_1(t) - Y_2(t) = Cy_1(t) + Dy_2(t) \quad \blacksquare \end{align}

Theorem 2: If $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$ is a second order linear nonhomogenous differential equation, then the general solution for this differential equation is $\phi(t) = Cy_1(t) + Dy_2(t) + Y(t)$ where $y_1$ and $y_2$ form a fundamental set of solutions for the corresponding second order linear homogenous differential equation $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ and where $Y$ is a specific solution to the second order linear nonhomogenous differential equation.

• Proof: Let $y_1$ and $y_2$ form a fundamental set of solutions for the corresponding second order linear homogenous differential equation $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$. Consider an arbitrary solution $\phi$ and $Y$ is a specific solution to the second order linear nonhomogenous differential equation $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$. Then from Theorem 1 above, their difference is a linear combination of $y_1$ and $y_2$ for some constants $C$ and $D$:

(7)

\begin{align} \quad \phi(t) - Y(t) = Cy_1(t) + Dy_2(t) \end{align}

• Therefore an arbitrary solution $\phi$ can be obtained as $\phi(t) = Cy_1(t) + Dy_2(t) + Y(t)$. $\blacksquare$