# Solving $y-y'' = \exp(-|t|)$ with the Fourier transform

I would like to solve the differential equation $$y-y'' = \exp(-|t|),\ \ \ t \in \mathbb{R}$$ by using the Fourier transform. I have calculated the Fourier transform $\mathcal{F}(\exp(-|t|))(\xi) = \frac{2}{\sqrt{2\pi}(1+\xi^2)}$ already and went on to using the fourier transform on both sides of the equation above. Using the linearity of the Fourier operator and differentiation rules, I end up with the equation $$\int_\mathbb{R} e^{-i \xi x} y(x) \ d\lambda(x)= \frac{2}{(1+\xi^2)^2}$$ This is where I am kind of stuck... It seems like there must be a rather simple way of solving this equation for $y$, but I am not quite sure what the most straight forward way to continue would be here.. I might be able to use the inversion theorem and then solve the equation, but the calculations seem a little bit tedious.. Any help would be greatly appreciated!

• You should apply the inverse transform. The convolution theorem may help. – eranreches Nov 23 '17 at 16:57

I have tried something, there might be something useful in it:

$$y-y''=\exp(-|t|)$$ $$\mathcal{F}(y-y'')=\mathcal{F}(\exp(-|t|))$$ $$\mathcal{F}(y)-\mathcal{F}(y'')=\frac{2}{1+\omega^2}$$ $$F(\omega)-(i\omega)^2F(\omega)=\frac{2}{1+\omega^2}$$ $$(1+\omega^2)F(\omega)=\frac{2}{1+\omega^2}$$ $$F(\omega)=\frac{2}{(1+\omega^2)^2}$$ We know that:

$\mathcal{F}(y')=i\omega\mathcal{F}(y)$

$\mathcal{F}(t\cdot y)=i \frac{\mathrm{d}}{\mathrm{d}\omega} \mathcal{F}(y)$

$\mathcal{F}(\exp(-|t|))=\frac{2}{1+\omega^2}$

$$\mathcal{F}(t\exp(-|t|))=i\frac{\mathrm{d}}{\mathrm{d}\omega}\frac{2}{1+\omega^2}=-i\frac{4\omega}{(1+\omega^2)^2}=-2 i \omega \frac{1}{(1+\omega^2)^2}=-2 i \omega F(\omega)$$ $$\mathcal{F}(y')=(i \omega) F(\omega)$$ So: $$-2\mathcal{F}(y')=\mathcal{F}(t\exp(-|t|))$$ $$-2y'=t\exp(-|t|)$$ $$y=-\frac{1}{2}\int t\exp(-|t|) \mathrm{d}t$$ If t>0: $$y=\frac{1}{2} e^{-t}(1+t)+C$$ $$y''=\frac{1}{2}e^{-t}(t-1)$$ $$y-y''=C+e^{-t}=e^{-t}$$ So $C=0$ here.

If t<0: $$y=\frac{1}{2}e^t(1-t)+C$$ $$y''=-\frac{1}{2}e^{t}(t+1)$$ $$y-y''=C+e^t=e^t$$ So $C=0$ here.

• Thanks, this is very helpful. I just realised though, that I dont understand how the step $\mathcal{F}(t\exp(-|t|)) = -2i\omega F(\omega)$ works... I suppose you use the convolution theorem $\mathcal{F}(f\cdot g) = \mathcal{F}(f) * \mathcal{F}(g)$? – Markus Peschl Nov 23 '17 at 17:31
• @GreenLogic $\mathcal{F}(t^n*f(t))=i^n*\frac{\mathrm{d}^n F(\omega)}{\mathrm{d} \omega^n}$. I will edit it a bit more soon. But this is only the particular solution for the diff. eq., I don't know how could we get the complementary solution. – Botond Nov 23 '17 at 17:45
• @GreenLogic I think I've corrected 1 big mistake: The missing constant. – Botond Nov 23 '17 at 18:20
• Thanks a lot again! It all makes sense now. – Markus Peschl Nov 23 '17 at 19:14