This was a problem on my final last semester. I am trying to learn how this is done for the future. I thought that I figured it out and I emailed the instructor and he said there was a simpler solution without using Fourier transforms. I was wondering if anyone knows what to do as I am very stuck…
What I did:
We want a solution of the form $$LE=\delta$$ where $\delta$ is the delta function.
Then, $$u=E * \phi$$
So, we have
$$\partial E +aE = \delta$$ This is where I took the Fourier transform to get,
$$-i\xi\hat{E}+a\hat{E}=1$$
$$\Rightarrow (a-i\xi)\hat{E}=1$$
$$\Rightarrow \frac1{(a-i\xi)}+c\delta=\hat{E}$$ is a family of solutions. Then we can take $c=0$.
Then I have $$E=\int_{-\infty}^\infty \frac{e^{-i\xi x}}{a-i \xi} \frac{d\xi}{2\pi}$$
Again, I was told there was a much simpler solution. I am hoping someone can help me out.
Thanks.
Best Answer
Using the method of integrating factor: $$E'+aE = \delta$$ $$(Ee^{ax})' = \delta e^{ax} = \delta$$ $$Ee^{ax} = H$$ $$E = He^{-ax},$$ where $H$ is the Heavyside function.