Can particular solution be found using Laplace transform without initial condition given? I was getting stuck with the following second order differential equation problem.
The problem is to use Laplace transform to find a particular solution for:
$$x'' - x = \cosh (t)$$
where one recalls
$$\cosh (t) = \frac{e^t+e^{-t}}{2}$$
and
$$\sinh (t) = \frac{e^t-e^{-t}}{2}$$
For hours I have tried to keep on using Laplace transform with the both initial conditions kept as unknowns. However, I could not figure out the partial fraction part. I read the textbook and searched the web for tutorials and still could not find any answers or hints on how to solve the question.
Can anyone be kind enough to tell me what to do or how to do this? Or is this impossible to do as there aren't any initial conditions to start with?
 A: Call $x(0)=a;\;x'(0)=b$ and $X$ the Laplace transform of $x(t)$ 
You get, from tables
$$-a s-b+s^2 X-X=\frac{s}{s^2-1}$$
and solving wrt $X$
$$X=\frac{a s^3+bs^2+(1-a) s-b}{\left(s^2-1\right)^2}$$
Partial fractions
$$\frac{P}{s+1}+\frac{Q}{(s+1)^2}+\frac{R}{s-1}+\frac{S}{(s-1)^2}=(*)\\=\frac{s^2 (-P+Q+R+S)+s (-P-2 Q-R+2 S)+s^3 (P+R)+P+Q-R+S}{(s-1)^2 (s+1)^2}$$
We get the system
$\left\{
\begin{array}{l}
 P+Q-R+S=-b \\
 -P-2 Q-R+2 S=1-a \\
 -P+Q+R+S=b \\
 P+R=a \\
\end{array}
\right.
$
And solutions are $$P= \frac{a-b}{2},Q= -\frac{1}{4},R= \frac{a+b}{2},S= \frac{1}{4}$$
So $(*)$ can be written as
$$\frac{a-b}{2 (s+1)}+\frac{a+b}{2 (s-1)}+\frac{1}{4 (s-1)^2}-\frac{1}{4 (s+1)^2}$$
and looking back to the tables we get
$$x(t)=\frac{1}{2} e^{-t} (a-b)+\frac{1}{2} e^t (a+b) +\frac{e^t t}{4}-\frac{e^{-t} t}{4} $$
and finally
$$x(t)=a \cosh t+b \sinh t+\frac{1}{2} t \sinh t$$
A: Let $x(0)=a$ and $x'(0)=b$, then with Laplace of both sides
\begin{align}
s^2{\cal L}(x)-sx(0)-x'(0)-{\cal L}(x)&=\dfrac{s}{s^2-1}\\
(s^2-1){\cal L}(x)&=\dfrac{s}{s^2-1}+as+b\\
{\cal L}(x)&=\dfrac14\dfrac{1}{(s-1)^2}-\dfrac14\dfrac{1}{(s+1)^2}+a\dfrac{s}{s^2-1}+b\dfrac{1}{s^2-1}\\
x(t)&=\dfrac14te^t-\dfrac14te^{-t}+a\cosh t+b\sin ht
\end{align}
A: Only a particular solution is asked for, and this is independent of any initial conditions.  So you may as well choose $x(0)=0$, $x'(0)=0$ then proceed as in other people's answers (but with much less work) to obtain
$$X=\frac{s}{\left(s^2-1\right)^2}$$
and hence
$$x=\tfrac{1}{2} t \sinh t\ .$$
