TITLE: Solving distributional differential equation
QUESTION [7 upvotes]: How to solve differential equation in $\mathcal D'(R)$:
$$u''+u=\delta'(x),$$ where $\delta$ is Dirac Delta function?
Solution of homogeneous problem is $C_1\cos{x}+C_2\sin{x}$, so using the variation of parameters, I got that the final solution to the problem should be:
$$-\cos{x}\int\delta'(x)\sin{x}\mathrm dx+\sin{x}\int\delta'(x)\cos{x}\mathrm dx,$$ 
but I don't know how to evaluate integrals $\int\delta'(x)\sin{x} \mathrm dx$, $\int\delta'(x)\cos{x}\mathrm dx$. Any help is appreciated. Thanks in advance.

REPLY [8 votes]: Integrating by parts, we have 
$$\int\delta'(x)\sin x \ dx 
= \delta(x)\sin x - \int\delta(x)\cos x\ dx
= 0 - \theta(x) = -\theta(x),$$ 
where $\theta(x)$ is the Heaviside step function. 
(The distributional product of $\delta(x)$ with $\sin x$ is zero.)
I leave it to you to show that 
$\int\delta'(x)\cos x\ dx = \delta(x)$. 
It is a good exercise to show that the resulting particular solution 
$$u(x) = \theta(x)\cos x$$
satisfies the inhomogeneous differential equation.