Verify that convolution with $\exp(-|z|)$ gives a solution of $y''-y=f$ I am currently trying to find a solution to the ODE $y''(x)-y(x)=f(x)$. Now, I calculated the solution, which was given to be:
$$y(x)=-\frac{1}{2}\int_{-\infty}^\infty f(x-z)e^{-|z|}dz$$
I am trying to now show that the integral in fact solves this ODE. First I must find $y''(x)$. Moving this into the integral yields:
$$y''(x) = -\frac{1}{2}\int_{-\infty}^\infty \frac{\partial^2 f(x-z)}{\partial x^2} e^{-|z|} dz$$
Now a hint says to divide this integral up into parts where $y\geq x$ and $y<x$, and integrate twice over each region. How does this help, and how am I supposed to proceed? I don't really see why this hint is there. 
 A: First, observe that 
$$
\frac{\partial^2 f(x-z)}{\partial x^2} = \frac{\partial^2 f(x-z)}{\partial z^2}
$$
Then integrate by parts twice, throwing the derivative onto the exponential term. The latter is not differentiable at $z=0$, so the integral should be split:
$$-2y''(x) = \int_{-\infty}^0 \frac{\partial^2 f(x-z)}{\partial z^2} e^{z} dz +
\int_{-0}^\infty \frac{\partial^2 f(x-z)}{\partial z^2} e^{-z} dz \\ =  
 \frac{\partial f(x-z)}{\partial z} e^{z} \bigg|_{-\infty}^0 
 - \int_{-\infty}^0 \frac{\partial f(x-z)}{\partial z} e^{z} dz +
\frac{\partial f(x-z)}{\partial z} e^{-z} \bigg|_{0}^\infty
 + \int_{-0}^\infty \frac{\partial f(x-z)}{\partial z} e^{-z} dz
$$
The boundary values at $\infty$ are zero, and at $0$ they cancel out. Integrate by parts again:
$$
- f(x-z)e^z\bigg|_{-\infty}^0 
 + \int_{-\infty}^0 f(x-z) e^{z} dz 
+ f(x-z)e^{-z}\bigg|_0^{\infty}
 + \int_{-0}^\infty f(x-z) e^{-z} dz
$$
This simplifies to
$$
-2f(x) + \int_{-0}^\infty f(x-z) e^{-|z|} dz = -2f(x) - 2y(x)
$$
as expected.
A: You can also use directly the property of the convolution product under differentiation $(f*g)'=f'*g=f*g'$ to conclude that $y''=(f*g)''=f*g''$ and with $g(x)=-\frac12e^{-|x|}$
\begin{align}
g'(x)&=(u(x)-\frac12)e^{-|x|}\\
g''(x)&=\delta(x)e^{-|x|}-\frac12(2u(x)-1)^2e^{-|x|}\\
&=\delta(x)+g(x)
\end{align}
where $u(x)$ is the unit jump at $x=0$, $u(x)=0$ for $x<0$, $u(x)=1$ for $x\ge 0$ so that $(\frac{d}{dx}|x|)^2=(2u(x)-1)^2=1$ (a.e.).
Now the claim follows as $\delta * f=f$.
