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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?

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  • $\begingroup$ Where did $F\{1\}$ go? $\endgroup$
    – mjqxxxx
    Nov 3, 2021 at 18:31
  • $\begingroup$ Taking the inverse Fourier transform reduces that part back to just $1$. Hence the relevant inverse Fourier term will just involve the fraction expression $\endgroup$
    – variations
    Nov 3, 2021 at 18:40
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    $\begingroup$ 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. $\endgroup$
    – march
    Nov 3, 2021 at 19:09
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    $\begingroup$ 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. $\endgroup$
    – march
    Nov 3, 2021 at 19:20

1 Answer 1

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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.

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  • $\begingroup$ @LutzLehmann. That $x$ was a mistake. $\endgroup$
    – md2perpe
    Nov 3, 2021 at 21:06

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