I'm trying to solve a differential equation similar to the one shown as a first solution here:
Inverse Fourier transform of $ \frac{1}{a+jw} $
The DE is the following:
$y' + 2y = \delta(x)$
I'm trying to solve without using Laplace but got stuck here:
$y=e^{-2x}\int^{}e^{2x}\delta(x)dx$
There should be some way to prove that $\int^{}e^{2x}\delta(x) = H(x)$ but I just can't see it. As far as I know $\int^{}e^{2x}\delta(x) = 1$, so there should be a way to shift the delta inside the integral as $\delta(x-a)$ when $a \neq 0$ so $H(x)$ makes sense.
The equation can be rewritten s $(e^{2x}y)'=e^{2x}\delta(x)$. Fix $a<0$ and integrate between $a$ an $x$: $$ e^{2x}y(x)-e^{2a}y(a)=\int_a^xe^{2t}\delta(t)\,dt. $$ If $x>0$ the integral on the right hand side is $1$; if $x<0$ then the integral is $0$.