📂Odinary Differential Equations

General Solution to Second-Order Linear Nonhomogeneous Differential Equations

Auxiliary Lemma¹

Consider the following nonhomogeneous/homogeneous second-order linear differential equation:

$$\begin{align} y^{\prime \prime}+p(t)y^\prime + q(t)y &=g(t) \label{eq1} \\ y^{\prime \prime}+p(t)y^\prime + q(t)y &=0 \label{eq2} \end{align}$$

Assume that $y_{1} (t)$ and $y_{2} (t)$ are solutions to the nonhomogeneous differential equation $\eqref{eq1}$, and that $y_{1}(t)$ and $y_{2}(t)$ are the fundamental set of solutions to the homogeneous differential equation $\eqref{eq2}$. Then, the following equation holds:

$$y_{1} (t) – y_{2} (t)= c_{1}y_{1} (t)+c_2y_{2}(t)$$

Where $c_{1}, c_{2}$ is a constant.

Proof

Let’s define the differential operator $L$ as follows:

$$L[y] := y^{\prime \prime} + p(t)y^\prime + q(t)y$$

Since $y_{1}$ and $y_{2}$ are solutions to the nonhomogeneous differential equation $\eqref{eq1}$,

$$L[y_{1}]=g(t) \quad \text{and} \quad L[y_{2}]=g(t)$$

Subtracting the two equations yields $L[y_{1}]-L[y_{2}]=g(t)-g(t)=0$. At this point, as $L$ is a linear operator, it satisfies the following:

$$L[y_{1}] - L[y_{2}] = L[ y_{1} – y_{2}]=0$$

Since $y_{1} – y_{2}$ satisfies $L[y_{1} – y_{2}]=0$, $y_{1} – y_{2}$ is a solution to the homogeneous differential equation $\eqref{eq2}$.

Moreover, any solution to the homogeneous differential equation can be represented as a linear combination of the fundamental set. Therefore, the following holds:

$$y_{1} – y_{2}=c_{1}y_{1} + c_2y_{2}$$

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The auxiliary lemma allows us to derive the general solution of the nonhomogeneous differential equation.

Theorem

Consider the following nonhomogeneous/homogeneous second-order linear differential equation:

$$\begin{align} y^{\prime \prime}+p(t)y^\prime + q(t)y &=g(t) \tag{1} \\ y^{\prime \prime}+p(t)y^\prime + q(t)y &=0 \tag{2} \end{align}$$

Assume that $\phi (t), Y(t)$ is a solution to the nonhomogeneous differential equation $\eqref{eq1}$, and that $y_{1}, y_{2}$ is the fundamental set of solutions to the homogeneous differential equation $\eqref{eq2}$. The general solution to the nonhomogeneous differential equation $\eqref{eq2}$ is as follows:

$$\phi (t) = c_{1}y_{1} (t) +c_2y_{2} (t) +Y(t)$$

Where $c_{1}, c_{2}$ is a constant.

Explanation

$c_{1}y_{1} + c_2y_{2}$ is referred to as the complementary solution. $Y$ is referred to as the particular solution. Summarizing from the two theorems, the process of finding the general solution of the nonhomogeneous differential equation is as follows:

1. Assume the nonhomogeneous differential equation as a homogeneous one and find its general solution. Call it the complementary solution.
2. Find any particular solution that satisfies the nonhomogeneous differential equation.
3. The sum of the complementary and particular solutions is the general solution of the nonhomogeneous differential equation.

Proof

Let’s assume $\phi$ and $Y$ as arbitrary solutions to the nonhomogeneous differential equation $\eqref{eq2}$. Instead of $y_{1}, y_{2}$, we insert $\phi, Y$ into the auxiliary lemma. Then, we obtain the following result:

$$\phi – Y = c_{1}y_{1} +c_2y_{2} \implies \phi (t) = c_{1}y_{1} (t) +c_2y_{2}(t) +Y (t)$$

Therefore, the general solution of the nonhomogeneous differential equation is expressed as the sum of the general solution to the homogeneous differential equation and any particular solution of the nonhomogeneous differential equation.

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