The general solution for inhomogeneous differential equation I am working with the following inhomogeneous differential equation,
$$x''+x=3\cos (\omega t)$$
The general solution for this is $x(t)=x_h(t)+x_p(t)$
First step is to find $x_h(t):$
So the characteristic equation is,
$$\lambda^2+0 \lambda+1=0$$ 
and its roots are
$$\lambda =\frac{\sqrt{-4}}{2}=\frac{i\sqrt{4}}{2}=\pm i$$ 
So $$x_h(t)=c_1 \cos(t)+c_2 \sin(t)$$
Second step is to find $x_p(t):$
My guess will be,
$$x_p(t)=A \cos(\omega t)+B \sin(\omega t)$$
Now I take the derivative of my guess.
$$x_p'(t)=-A \sin(\omega t) \cdot \omega+B \cos(\omega t) \cdot \omega$$
$$x_p''(t)= -A \cos(\omega t) \cdot \omega^2-B \sin(\omega t) \cdot \omega^2$$
The I have replace it in the equation
$$-A \cos(\omega t) \cdot \omega^2-B \sin(\omega t) \cdot \omega^2 + A \cos(\omega t)+B \sin(\omega t) = 3 \cos(\omega t)$$
Since we have no $\sin(\omega t)$ on the RHS the B must be $0$ on the LHS. Then
$$-A \cos(\omega t) \cdot \omega^2 - A \cos(\omega t) = 3 \cos(\omega t)$$
And isolate A
$$A\big(- \cos(\omega t) \cdot \omega^2-\cos(\omega t)\big)=3 \cos(\omega t)$$
$$A=\frac{3 \cos(\omega t)}{- \cos(\omega t) \cdot \omega^2-\cos(\omega t)}$$
$$A=\frac{3 \cos(\omega t)}{\cos(\omega t)\big(-\omega^2-1 \big)}$$
$$A=\frac{3}{-\omega^2-1}$$
Then replace this into our guess
$$x_p(t)=\frac{3}{-\omega^2-1} \cos(\omega t)$$
$$x_p(t)=\frac{3 \cos(\omega t)}{-\omega^2-1}$$
Last the general solution is 
$$x(t)=c_1 \cos(t)+c_2 \sin(t)-\frac{3 \cos(\omega t)}{\omega^2-1}$$
 A: The homogeneous solution is
$$ x_h(t) = c_1\cos t + c_2\sin t $$
There are two cases:
If $\omega \ne 1$, the particular solution is
$$ x_p(t) = A\cos(\omega t) + B\sin(\omega t) $$
where $A$ and $B$ are constants (not $3$, you don't know what they are yet).
If $\omega = 1$, our previous guess becomes $x_p(t) = A\cos t + B\sin t$, which is the same as the homogeneous solution. When you plug in this guess, you will get $0$ on the RHS. In order to fix it, we have to add a factor of $t$
$$ x_p(t) = At\cos t + Bt\sin t$$
A: In general a second order equation of the type
\begin{equation*}
\partial _{t}^{2}x(t)+a\partial _{t}x(t)+bx(t)=f(t)
\end{equation*}
can be handled by rewriting it as a coupled set of first order equations by
setting
\begin{equation*}
x_{1}=x,\;x_{2}=\partial _{t}x
\end{equation*}
Then, denoting
\begin{equation*}
\mathbf{x}(t)=\left(
\begin{array}{c}
x_{1}(t) \\
x_{2}(t)
\end{array}
\right)
\end{equation*}
one obtains the equation
\begin{equation*}
\partial _{t}\mathbf{x}(t)=\mathsf{A}\cdot \mathbf{x}(t)+\mathbf{y}(t)
\end{equation*}
where $\mathsf{A}$ is a $t$-independent $2\times 2$ matrix. In your case
\begin{equation*}
\partial _{t}^{2}x(t)+x(t)=3\cos (\omega t)
\end{equation*}
we have
\begin{equation*}
\mathsf{A}=\left(
\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}
\right) ,\;\mathbf{y}(t)=\left(
\begin{array}{c}
0 \\
3\cos (\omega t)
\end{array}
\right)
\end{equation*}
and the solution is
\begin{equation*}
\mathbf{x}(t)=\exp [\mathsf{A}t]\cdot \mathbf{x}(0)+\int_{0}^{t}ds\exp [%
\mathsf{A}(t-s)]\cdot \mathbf{y}(s)
\end{equation*}
Since
\begin{equation*}
\mathsf{A}^{2}=-\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) =-\mathsf{I}
\end{equation*}
you have
\begin{equation*}
\exp [\mathsf{A}t]=\mathsf{I}\cos t+\mathsf{A}\sin t
\end{equation*}
Given
\begin{equation*}
\mathbf{y}(s)=\left(
\begin{array}{c}
0 \\
3\cos (\omega s)
\end{array}
\right)
\end{equation*}
you can do the $s$-integral and obtain the required result.
