Checking some work on differential $y''-4y=\sinh x$ I want to check that I have done this correctly, because I feel like I am most of he way there but missing something important (or maybe just something obvious that I ought to know) 
We have the following differential equation: 
$$y''-4y=\sinh x$$
This should be simple enough, or so I thought. I see the characteristic equation for the linear homogeneous expression ($y''-4y=0$) should be $r^2 - 4 = 0$ so the roots are $r\pm2$. 
All-righty then, the solution $y_p$ ought to look like this: $y_p = c_1e^{2x}+c_2e^{-2x}$ because both roots are real and distinct. 
I take derivatives of $y_p$: 
$$y'_p = 2c_1e^{2x}-2c_2e^{-2x}$$
$$y''_p = 4c_1e^{2x}+4c_2e^{-2x}$$
and plugging it back into the original DE: 
$\sinh x = 4c_1e^{2x}+4c_2e^{-2x} - 4(c_1e^{2x}+c_2e^{-2x})$
But you can see the problem I end up with zero, and thus can't equate the coefficients of like powers (understand I had planned to just convert sinh into the Euler expression $\frac{e^x-e^{-x}}{2}$ and go from there).
So, I did something wrong, and it looks like I should have used variation of parameters here, but I admit that technique I am a bit fuzzy on. But if that's what I have to use... 
 A: It should be that the homogenous solution $y_h=c_1e^{2x}+c_2e^{-2x}$, this satisfies the homogenous part $y''_h-4y_h=0$.
Now the particular solution $y_p$ should satisfy $y_p''-4y_p=\sinh x$, then the full solution $y=y_c+y_p$ must satisfy the ODE and the BCs.
Here $y_p=A\cosh x+B\sinh x$, then
$y''_p=A\cosh x+ B\sinh x=y_p$, so we get:
$y''_p-4y_p=-3A\cosh x-3A\sinh x=\sinh x$, so $B=0,A=-1/3$, thus $y_p=-\frac{1}{3}\sinh x$.
Now $y = c_1e^{2x}+c_2e^{-2x}-\frac{1}{3}\sinh x$
if you had BC's you would implement them at this point.
A: Here's a technique that Oliver Heaviside used for these problems: Rewrite your equation as
\begin{align}
\left(1 - \frac{1}{4}\frac{\mathrm{d}^{2}}{\mathrm{d}x^2} \right)y = -\frac{1}{4}\sinh(x)
\end{align}
You've already found $y_{h}$ (which this method won't find), so let's focus on $y_{p}$: Divide both sides by $\left(1 - \frac{1}{4}\frac{\mathrm{d}^{2}}{\mathrm{d}x^2} \right)$ to obtain
\begin{align}
y_{p}(x) &=-\frac{1}{4} \frac{1}{\left(1 - \frac{1}{4}\frac{\mathrm{d}^{2}}{\mathrm{d}x^2} \right)}\sinh(x) \\
&= -\frac{1}{4}\sum_{j=0}^{\infty} \left(\frac{1}{4}\frac{\mathrm{d}^{2}}{\mathrm{d}x^2} \right)^{j} \sinh(x)  \\
&= -\frac{1}{4}\sum_{j=0}^{\infty} \frac{1}{4^{j}} \sinh(x)  \\
&= -\frac{1}{3}\sinh(x)
\end{align}
Kinda fun, I guess.
