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

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

• What definition of the fourier transform are you using? – John Doe Dec 31 '17 at 14:25

Hint: $$\delta(x-x_{0})f(x)=\delta(x-x_{0})f(x_{0})$$
• @qbert He says he knows the transform of the delta function. He can use this to simplify $$\delta(t)e^{-t}=\:?$$ – eranreches Dec 31 '17 at 14:30
• @qbert this says that $\delta(t)e^{-t}=\delta(t)$, so $\mathcal F[\delta(t)e^{-t}]=\mathcal F[\delta(t)]$ – John Doe Dec 31 '17 at 14:32
You don't need to memorize Fourier transforms, just how to integrate delta! $$\mathcal{F}(\delta(t)e^{-t})(\xi)= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-2\pi i \xi t}e^{-t}\delta(t)\mathrm dt\\ =\frac{1}{\sqrt{2\pi}}e^{-2\pi i \xi t}e^{-t}\vert_{t=0}=\frac{1}{\sqrt{2\pi}}$$