Given PDE:
$$u_t=u_{xx}-\gamma u+g(x), -\infty \lt x \lt \infty, t \gt0, \gamma \gt 0$$ with $$u(x,0)=0, g(x)=xe^{-x^2/2}.$$
Use Fourier transform to find the solution of the given PDE.
**My Approach **
I took the Fourier transform and got the corresponding equation $$U_t(k,t)=-k^2U(k,t)-\gamma U(k,t)-ik \sqrt{2\pi}e^{-k^2/2}$$
My question is how to revert back into the original space using the inverse Fourier transformation.
After correcting the mistake of the Fourier Transform of $g(x)$, I got this: $$U_t(k,t)=-k^2U(k,t)-\gamma U(k,t)-ik \sqrt{2\pi}e^{-k^2/2}$$ $$U_t(k,t)+(k^2+\gamma) U(k,t)=-ik \sqrt{2\pi}e^{-k^2/2}$$ Solving the DE with Integrating Factor: $$(U(k,t)e^{(k^2+\gamma)t} )'=-ik \sqrt{2\pi}e^{-k^2/2}e^{(k^2+\gamma)t} $$ Integrate: $$U(k,t)e^{(k^2+\gamma)t} =-ik \sqrt{2\pi}e^{-k^2/2}\int e^{(k^2+\gamma)t}dt +C(k)$$ $$U(k,t)e^{(k^2+\gamma)t} =-ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2} e^{(k^2+\gamma)t} +C(k)$$ $$U(k,t) =-ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2} +C(k)e^{-(k^2+\gamma)t}$$ $$U(k,0)=0 \implies C(k) =ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2}$$ $$U(k,t) =-ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2}(1 -e^{-(k^2+\gamma)t})$$
Edit:
You have the condition that should simplify a little the equation: $$U(k,0)=0$$ Note that you have this result: $$\mathcal{F^{-1}}\{ikF(k)\}=f'(x)$$ And you also have: $$\mathcal{F^{-1}} \{\frac {2a}{a^2+k^2}\}=e^{-a|x|}$$