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.

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    $\begingroup$ Yes, distributional derivative of the Dirac delta distribution. $\endgroup$ – alans Jun 2 '13 at 19:58

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.

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  • $\begingroup$ Could you explain why distribution product of $\delta$ and $\sin⁡x$ is zero? and how did the integral equal to Heaviside function? Shouldn't the integral be cos(0)=1? $\endgroup$ – Sahiba Arora Feb 8 '17 at 23:23
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    $\begingroup$ @SahibaArora: Note that $\int f(x)\delta(x)\sin x\,dx = f(0)\sin 0 = 0$. Also, $$\int^x\delta(t)\cos t\,dt = \begin{cases}\cos 0=1,&x>0 \\0,&\textrm{else.}\end{cases}$$ $\endgroup$ – user26872 Feb 9 '17 at 1:29
  • $\begingroup$ @SahibaArora: Glad to help. $\endgroup$ – user26872 Feb 9 '17 at 17:18

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