Depending on $g(t)$ often it is difficult to come up with a suitable trial solution of a given non-homogeneous equation in the method of undetermined coefficient discussed here. In this note, we discuss an alternative method called *variation of parameters*. First assume that we know the general solution

$$x_{\mathrm{hom}}(t)=c_1x_1(t)+c_2x_2(t)$$

of the homogeneous equation $$\ddot{x}+p(t)\dot{x}+q(t)x=0$$ Let

\begin{equation}

\label{eq:vp}

x(t)=u_1(t)x_1(t)+u_2(t)x_2(t)

\end{equation}

be a solution of the non-homogeneous equation $$\ddot{x}+p(t)\dot{x}+q(t)x=g(t)$$ Differentiating \eqref{eq:vp}

$$\dot{x}(t)=\dot{u}_1x_1+u_1\dot{x}_1+\dot{u}_2x_2+u_2\dot{x}_2$$

Require that

\begin{equation}

\label{eq:vp2}

\dot{u}_1x_1+\dot{u}_2x_2=0

\end{equation}

so that we have

\begin{equation}

\label{eq:vp3}

\dot{x}(t)=u_1\dot{x}_1+u_2\dot{x}_2

\end{equation}

Differentiating \eqref{eq:vp3}

\begin{equation}

\label{eq:vp4}

\ddot{x}(t)=\dot{u}_1\dot{x}_1+u_1\ddot{x}_1+\dot{u}_2\dot{x}_2+u_2\ddot{x}_2

\end{equation}

Substituting $\ddot{x}$, $\dot{x}$, $x$ in the non-homogeneous equation with the corresponding expressions in \eqref{eq:vp}, \eqref{eq:vp3} and \eqref{eq:vp4}, respectively results in the equation

\begin{equation}

\label{eq:vp5}

\dot{u}_1\dot{x}_1+\dot{u}_2\dot{x}_2=g(t)

\end{equation}

Let

$$W(x_1,x_2)(t):=\begin{vmatrix}

x_1(t) & x_2(t)\\

\dot{x}_1(t) & \dot{x}_2(t)

\end{vmatrix}=x_1(t)\dot{x}_2(t)-x_2(t)\dot{x}_1(t)$$

$W(x_1,x_2)(t)$ is called the* Wronskian* of $x_1(t)$ and $x_2(t)$. If $x_1(t)$ and $x_2(t)$ are linearly dependent, so are the columns of $W(x_1,x_2)(t)$ hence $W(x_1,x_2)(t)=0$ for all $t$. This means that If $W(x_1,x_2)(t)\ne 0$ for some $t$, $x_1(t)$ and $x_2(t)$ are linearly independent. Also we have the following theorem holds.

*Theorem*. Let $x_1(t)$ and $x_2(t)$ be solutions of a homogeneous second-order linear differential equation. If $x_1(t)$ and $x_2(t)$ are linearly independent, then $W(x_1,x_2)(t)\ne 0$ for all $t$.

Since $x_1(t)$ and $x_2(t)$ are linearly independent, $W(x_1,x_2)\ne 0$ for all $t$ so by Cramer’s rule the solution of the system of linear equations \eqref{eq:vp2}, \eqref{eq:vp5} in $\dot{u}_1$ and $\dot{u}_2$ is given by

\begin{equation}

\begin{aligned}

\dot{u}_1(t)&=\frac{\begin{vmatrix}

0 & x_2\\

g(t) & \dot{x}_2

\end{vmatrix}

}{W(x_1,x_2)(t)}=-\frac{g(t)x_2(t)}{W(x_1,x_2)(t)}\\

\dot{u}_2(t)&=\frac{\begin{vmatrix}

x_1 & 0\\

\dot{x}_1 & g(t)

\end{vmatrix}

}{W(x_1,x_2)(t)}=\frac{g(t)x_1}{W(x_1,x_2)(t)}

\end{aligned}\label{eq:vp6}

\end{equation}

Integrating \eqref{eq:vp6}, $u_1(t)$ and $u_2(t)$ are determined to be

\begin{equation}

\label{eq:vp7}

\begin{aligned}

u_1(t)&=-\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt+c_1\\

u_2(t)&=\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt+c_2

\end{aligned}

\end{equation}

where $c_1$ and $c_2$ are constant.

Therefore

\begin{equation}

\label{eq:vp8}

\begin{aligned}

x(t)=c_1x_1(t)+&c_2x_2(t)+\left\{-x_1(t)\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt+\right.\\

&\left.x_2(t)\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt\right\}

\end{aligned}

\end{equation}

is indeed the general solution of the non-homegeneous equation for

\begin{equation}

\label{eq:vp9}

X(t)=-x_1(t)\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt+x_2(t)\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt

\end{equation}

being a particular solution of the non-homogeneous equation.

*Example*. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=2\sin t$.

*Solution*. Recall that $x_1(t)=e^{-t}$, $x_2(t)=e^{4t}$. The Wronskian is

$$W(x_1,x_2)(t)=\begin{vmatrix}

e^{-t} & e^{4t}\\

-e^{-t} & 4e^{4t}

\end{vmatrix}=5e^{3t}$$

\begin{align*}

\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{e^{4t}(2\sin t)}{5e^{3t}}dt\\

&=-\frac{1}{5}e^t\cos t+\frac{1}{5}e^t\sin t

\end{align*}

and

\begin{align*}

\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{e^{-t}(2\sin t)}{5e^{3t}}dt\\

&=-\frac{2}{85}e^{-4t}\cos t-\frac{8}{85}e^{-4t}\sin t

\end{align*}

The general solution is

\begin{align*}

x(t)&=c_1e^{-t}+c_2e^{4t}\\

&-e^{-t}\left(-\frac{1}{5}e^t\cos t+\frac{1}{5}e^t\sin t\right)+e^{4t}\left(-\frac{2}{85}e^{-4t}\cos t-\frac{8}{85}e^{-4t}\sin t\right)\\

&=c_1e^{-t}+c_2e^{4t}+\frac{3}{17}\cos t-\frac{5}{17}\sin t.

\end{align*}

*Example*. Solve the non-homogeneous $\ddot{x}+4x=8\tan t$, $-\frac{\pi}{2}<t<\frac{\pi}{2}$

*Solution*. $x_1(t)=\cos(2t)$, $x_2(t)=\sin(2t)$. The Wronskian is

$$W(x_1,x_2)(t)=\begin{vmatrix}

\cos(2t) & \sin(2t)\\

-2\sin(2t) & 2\cos(2t)

\end{vmatrix}=2$$

\begin{align*}

\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{\sin(2t)(8\tan t)}{2}dt\\

&=4t-2\sin(2t)

\end{align*}

and

\begin{align*}

\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{\cos(2t)(8\tan t)}{2}dt\\

&=-2\cos(2t)+4\ln(\cos t)

\end{align*}

Thus the general solution is

$$x(t)=c_1\cos(2t)+c_2\sin(2t)-4t\cos(2t)+4\sin(2t)\ln(\cos t)$$