Consider the heat equation
$\color{blue}{\begin{align} u_t&=ku_{xx}-bt^2u,\quad-\infty
I am asked to solve it using the Fourier transform pair
$\begin{align} \color{blue}{F(\omega)}&\color{blue}{=\frac1{2\pi}\int_{-\infty}^\infty f(x)e^{i\omega x}dx,}\\ \color{blue}{f(x)}&\color{blue}{=\int_{-\infty}^\infty F(\omega)e^{-i\omega x}d\omega.} \end{align}$
This is what I ended up with:
$ U_t(\omega,t)=-k\omega^2U(\omega,t)-bt^2U(\omega,t), $
where the solution to this ODE is
$ U(\omega,t)=c(\omega)\exp\left[-\frac{bt^3}3-k\omega^2t\right]. $
Applying the initial condition to solve for $c$ yields
$ c(\omega)=\frac{\exp\left[-\frac{\omega^2}4\right]}{2\sqrt\pi}. $
Hence, the final solution is
$ U(\omega,t)=\frac{\exp\left[-\frac{bt^3}{3}-\frac{\omega^2}{4}(4kt+1)\right]}{2\sqrt\pi}. $
I wonder if this is correct. The reason why I ask this is that our professor hinted us that we use the following two transforms:
$\begin{align} \mathcal F\left(\exp\left[-x^2\right]\right)(\xi)&\stackrel{?}{=}(2\pi)^{\frac12}\exp\left[-\frac{\xi^2}2\right],\\ \mathcal F(f(ax))(\xi)&=a^{-1}\mathcal F\left(\frac\xi a\right), \end{align}$
which make no sense because I am not sure of the validity of the first, and I see no place where the second one could be of any help.
Is this perhaps a typo? Thanks in advance.