# Solving $y''+4y'+4y=1$ using the Fourier transform

Assuming the function $$y(t)$$ has a Fourier-transform, we can use this method to find a particular solution for this equation. The homogenous solution will not appear ( completely ) because non-decaying exponentials do not have Fourier-transforms. Taking Fourier-transform of both sides $$F\{y''+4y'+4y\}=F\{1\}$$ $$-\omega^2Y(\omega)+4i\omega Y(\omega)+4Y(\omega)=F\{1\}$$$$Y(\omega)(ki+2)^2=F\{1\}$$ $$Y(\omega)=\frac{1}{(ki+2)^2}\cdot F\{1\}$$

where $$Y(\omega)$$ is the Fourier-transform of $$y(t)$$. Now, if we take the inverse Fourier-transform, we end up with $$y(t)=F^{-1}\{(\frac{1}{(ki+2)}\cdot\frac{1}{(ki+2)}\}$$

It seems like we should apply the convolution-theorem here. Doing so gives $$y(t)=\int_{-\infty}^{\infty}e^{-2\omega}e^{-2(t-\omega)}d\omega=\int_{-\infty}^{\infty}e^{-2t}d\omega$$

Note that the functions inside the integral are actually step functions, equal to zero when $$t<0$$. Still, I don't think the convolution theorem is used correctly here. The result will not be a solution of the equation. What mistake have I done here?

• Where did $F\{1\}$ go? Nov 3, 2021 at 18:31
• Taking the inverse Fourier transform reduces that part back to just $1$. Hence the relevant inverse Fourier term will just involve the fraction expression Nov 3, 2021 at 18:40
• Note that the Fourier transform of 1 is a Dirac delta, and so in taking the inverse Fourier transform, you should have just evaluated the integral at $\omega=0$, yielding a constant. But I think you want to write the right-hand side as $\Theta(t)$ (the Heavisede theta function) rather than 1. That way, the source term "turns on" at $t=0$, and what you actually get on the right-hand side is a function of $\omega$ plus a Dirac-delta. The point is, you've got to be careful with that right-hand side. Nov 3, 2021 at 19:09
• Furthermore, if you want to use the convolution theorem, you need to include the $\mathcal{F}(\mbox{right-hand side})$, because it's not just a constant. Nov 3, 2021 at 19:20

Your assumption that $$y$$ has a Fourier transform is only partly valid.
The general solution to the differential equation is $$y(t) = (At+B)e^{-2t} + \frac{1}{4}.$$
The first part of this, $$(At+B)e^{-2t},$$ is not Fourier transformable, not even as a distribution. Only $$\frac{1}{4}$$ is; it has transform $$C\delta(\omega)$$ for some constant $$C$$ dependent of exact definition of the transform.
• @LutzLehmann. That $x$ was a mistake. Nov 3, 2021 at 21:06