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

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

Thanks in advance.

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  • $\begingroup$ What is $\delta(x)$? Is it the Dirac Delta function or something else? $\endgroup$ – Amzoti Jun 1 '13 at 23:30
  • $\begingroup$ Yes, it is Delta function. $\endgroup$ – alans Jun 1 '13 at 23:34
  • $\begingroup$ No, there is no initial conditions. I have to find general solution. $\endgroup$ – alans Jun 1 '13 at 23:46
  • $\begingroup$ No, I am not familiar with Laplace Transforms. $\endgroup$ – alans Jun 2 '13 at 0:02
  • $\begingroup$ I think that I could use Laplace transforms. I would say that solution is $C_1\cos{\sqrt{3}x}+C_2\sin{\sqrt{3}x}+\frac{\delta}{3}+\frac{1}{3}$. Am I right? Please, could you write down your solution or give some another hint. Thanks. $\endgroup$ – alans Jun 2 '13 at 7:35
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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 $H$.

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