Solve a differential equation by Fourier transform $ y''+6y'+5y=\delta (t).e^{-t}$

Question

Solve a differential equation by Fourier transform: $ y''+6y'+5y=\delta (t).e^{-t}$

I know Fourier transform of $\delta (t)$ and $H(t).e^{-t} $ but I can't determine Fourier transform of $\delta (t).e^{-t}$.

Could you give me some hints? Thanks for helping

Answer 1

Hint: $$\delta(x-x_{0})f(x)=\delta(x-x_{0})f(x_{0})$$

Answer 2

You don't need to memorize Fourier transforms, just how to integrate delta! $$ \mathcal{F}(\delta(t)e^{-t})(\xi)= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-2\pi i \xi t}e^{-t}\delta(t)\mathrm dt\\ =\frac{1}{\sqrt{2\pi}}e^{-2\pi i \xi t}e^{-t}\vert_{t=0}=\frac{1}{\sqrt{2\pi}} $$