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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.

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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 $0$.

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    Which Delta property is being used here? I don't seem to follow. – 2017-02-06
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    @KoltPenny If $a$\int_a^b f(x) \delta(x) dx$ is either $f(0)$ if $a<0$0$ otherwise. This is not really a *property* but rather the definition. – 2017-02-06
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    @Ian Oh yeah, I understand the integral but not the left side of the equation anf how does it relate to H(t). Can I also make a tend to -∞? – 2017-02-06
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    @KoltPenny The right side is essentially $H(t)$ (modulo the meaning at $t=0$). – 2017-02-06