Solve $y'' + 4y = e^{-x^2}$ using Fourier transforms I need to solve the equation $y'' + 4y = e^{-x^2}$ using Fourier transforms.  I was able to take the Fourier transform of both sides and solve for $\hat y$.  I have $\hat y = \frac{e^{-k^2/4}}{\sqrt{2}(4-k^2)}$.  I presumably need to take the inverse Fourier transform of both sides.  How exactly would I go about doing this?
 A: You want to use the Fourier convolution property. You have that
$$\hat{y} = C e^{-\alpha k^2} \cdot \left(\frac{1}{\beta + \gamma k^2}\right) = \hat{f} \hat{g}$$
for various numbers $C,\alpha, \beta, \gamma$ that it looks like you've figured out. You can inverse Fourier transform each of the factors in the above, so then it will follow that $y = f \ast g$, where $\ast$ indicates convolution. Depending on which Fourier transform you're using, it might be $y = \sqrt{2 \pi}^{-1} f \ast g$. Just make sure that you use one consistent convention for the Fourier transform.
A: Note that the inverse FT of $(4-k^2)^{-1}$ is $(1/4) \sin{(2 x)} \mathrm{sgn}(x)$, where $\mathrm{sgn}(x)$ is the signum function, or $|x|/x$ when $x \ne 0$.
By the convolution theorem:
$$\begin{align}y(x) &= \frac{1}{2 \pi} \frac{1}{4} \int_{-\infty}^{\infty} dx' \: e^{-(x-x')^2} \sin{(2 x)} \mathrm{sgn}(x)\\ &= \frac{1}{8 \pi}\left [ -\int_{-\infty}^{0} dx' \: e^{-(x-x')^2} \sin{(2 x)} + \int_{0}^{\infty} dx' \: e^{-(x-x')^2} \sin{(2 x)}\right ]\\ &= \frac{1}{4 \pi} \Re{[F(1+i x)]} \\\end{align}$$
where 
$$F(z) = e^{-z^2} \int_0^z dt e^{t^2}$$
is Dawson's integral.
