Differential equation - distributions

Question

How to find solution to the following problem (in $D'(R)$):

$$u''+3 u=1+\delta (x)\text{ ?}$$

Thanks in advance.

Answer 1

In $\delta(x)$, does $x$ mean a particular number at which the delta function is located, or is it the independent variable? If the latter, then it's not clear at what point the delta function is located. I'll assume it's at $0$.

Let's try the variation of parameters. The homogeneous equation has solutions $u_1(x)=\cos \sqrt{3}x$ and $u_2(x)=\sin \sqrt{3}x$, with the Wronskian $W=\sqrt{3}$. The general solution is $$ u(x)=-\frac{u_1(x)}{W}\int \sin \sqrt{3}x (1+\delta(x))\,dx + \frac{u_2(x)}{W}\int \cos \sqrt{3}x (1+\delta(x))\,dx $$ With my interpretation that $\delta$ is located at $0$, the contribution of $\delta$ to the first integral is $0$ (because $\sin \sqrt{3}x=0$ there). The contribution of $\delta$ to the second integral is antiderivative of $\delta$, which is the Heaviside function $H$.