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.
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.