Let us consider the following second-order linear differential equation
\begin{equation}
\label{eq:nhde}
\ddot{x}+p(t)\dot{x}+q(t)x=g(t)
\end{equation}
Let $X_1(t)$ and $X_2(t)$ be two solution of \eqref{eq:nhde}. Then $X_1(t)-X_2(t)$ is a solution of the homogeneous equation
\begin{equation}
\label{eq:hde}
\ddot{x}+p(t)\dot{x}+q(t)x=0
\end{equation}
Thus $X_1(t)-X_2(t)=c_1x_!(t)+c_2x_2(t)$ where $x_1(t)$ and $x_2(t)$ are a fundamental set of solutions of \eqref{eq:hde}. This implies that the general solution of \eqref{eq:nhde} is given by
$$x(t)=c_1x_1(t)+c_2x_2(t)+X(t)$$
where $X(t)$ is a solution of \eqref{eq:nhde}. That is, solving the non-homogeneous equation \eqref{eq:nhde} boils down to finding a solution of \eqref{eq:nhde}. There are two methods of finding a solution of \eqref{eq:nhde}:

1.  Method of Undetermined Coefficients
2. Variation of Parameters

The Method of Undetermined Coefficients

This method finds a solution by guessing a particular solution with undetermined coefficients.

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

Solution. Let $X(t)=Ae^{2t}$ be a solution of the non-homogeneous equation. Then $\dot{X}=2Ae^{2t}$ and $\ddot{X}=4Ae^{2t}$. By substitution we obtain $\ddot{X}-3\dot{X}-4X=-6Ae^{2t}$ and so $A=-\frac{1}{2}$. Hence the general solution of the non-homogeneous equation is
$$x(t)=c_1e^{-t}+c_2e^{4t}-\frac{1}{2}e^{2t}$$

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

Solution. Let $X(t)=A\sin t$. Then $\dot{X}(t)=A\cos t$ and $\ddot{X}=-A\sin t$. By substitution we obtain $\ddot{X}-3\dot{X}-4X=-5A\sin t -3A\cos t$. This has to be the same as $2\sin t$ which leads to the equation $(5A+2)\sin t+3A\cos t$. Since $\sin t$ and $\cos t$ are linearly independent, $5A+2=0$ and $3A=0$, a contradiction! This time we assume that $X(t)=A\sin t+B\cos t$. Then $\dot{X}(t)=A\cos t-B\sin t$, $\ddot{X}(t)=-A\sin t-B\cos t$ and
$$\ddot{X}-3\dot{X}-4X=(-5A+3B)\sin t+(-3A-5B)\cos t$$
Comparing this with $2\sin t$ we get the system of linear equations
$$\left\{\begin{aligned}-5A+3B&=2\\
-3A-5B&=0
\end{aligned}\right.$$
of which solution is $A=-\frac{5}{17}$ and $B=\frac{3}{17}$. Therefore, the general solution is
$$x(t)=c_1e^{-t}+c_2e^{4t}-\frac{5}{17}\sin t+\frac{3}{17}\cos t$$

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=4t^2-1$.

Solution. Let $X(t)=At^2+Bt+C$ be a solution of the non-homogeneous equation. Then $A=-1$, $B=\frac{3}{2}$ and $C=-\frac{11}{8}$. Hence the general solution is
$$x(t)=c_1e^{-t}+c_2e^{4t}-t^2+\frac{3}{2}t-\frac{11}{8}$$

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=-8e^t\sin 2t$.

Solution.
\begin{align*}
\dot{X}(t)&=(-2A+B)e^t\sin 2t+(A+2B)e^t\cos 2t\\
\ddot{X}(t)&=(-3A+4B)e^t\cos 2t+(-4A-3B)e^t\sin 2t
\end{align*}
and
$$\ddot{X}-3\dot{X}-4X=(-10A-2B)e^t\cos 2t+(2A-10B)e^t\cos 2t$$
Comparing this with $-8e^t\cos 2t$ we get the system of linear equations
$$\left\{\begin{aligned}
-10A-2B&=-8\\
2A-10B&=0
\end{aligned}\right.$$
of which solution is $A=\frac{10}{13}$ and $B=\frac{1}{13}$.
Hence the general solution is
$$x(t)=c_1e^{-t}+c_2e^{4t}+\frac{10}{13}e^t\cos 2t+\frac{2}{13}e^t\sin t$$