The Fourier transform of the function $\frac{d}{dx}f(x)+xf(x)$ is $i[\frac{d}{dk}\widetilde{f}(k)+k\widetilde{f}(k)]$, so if a function $f(x)$ satisfies the ode $\frac{d}{dx}f(x)+xf(x)=0$, then the Fourier transform will satisfy the same ode, that means that $f(x)$ and $\widetilde{f}(k)$ are described by the same functions.
Show that in this case $f(x)=e^{-x^2}$, $\widetilde{f}(k)=Ae^{-k^2}$. Calculate the value of the constant $A$. $$$$ I have done the following so far:
$$\frac{d}{dx}f(x)+xf(x)=0 \Rightarrow \frac{d}{dx}f(x) \frac{1}{f(x)}=-x \Rightarrow ln|f(x)|=-\frac{x^2}{2}+c \text{ let } c=0, \Rightarrow ln|f(x)=-\frac{x^2}{2} \Rightarrow f(x)=e^{-\frac{x^2}{2}}$$
But I have to show that $f(x)=e^{-x^2}$. What have I done wrong? :/
Nothing. Obviously, $f(x) = \exp(-x^2)$ is not a solution as you can check easily.