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?
Best Answer
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.