And a nice light one to end off the day (maybe, I'm tempted to do more solutions later tonight)! We are simply asked to solve a fairly simple differential equation, specifically this one, representing
the wavefunction that has minimum position-momentum uncertainty. After coming to a few realizations about how this minimal uncertainty case must occur when the Cauchy-Schwarz inequality becomes an equility, we come
to the following differential equation:
$$\Big( \frac{\hbar}{i}\frac{\text{d}}{\text{dx}} \ - \ \langle p \rangle \Big)\Psi \ = \ ia(x \ - \ \langle x \rangle)\Psi$$
We are asked to solve this equation, distributing out the $\Psi$ on the left side of the equation, we get:
$$\Big( \frac{\hbar}{i}\frac{\text{d}}{\text{dx}} \ - \ \langle p \rangle \Big)\Psi \ = \ ia(x \ - \ \langle x \rangle)\Psi \ \Rightarrow \ \frac{\hbar}{i}\frac{\text{d}\Psi}{\text{dx}} \ - \ \langle p \rangle\Psi \ = \ ia(x \ - \ \langle x \rangle)\Psi$$
$$\Rightarrow \ \frac{1}{\Psi} \ \frac{\hbar}{i} \ \frac{\text{d}\Psi}{\text{dx}} \ - \ \langle p \rangle \ = \ ia(x \ - \ \langle x \rangle) \ \Rightarrow \ \frac{1}{\Psi} \ \frac{\text{d}\Psi}{\text{dx}} \ = \ -\frac{a(x \ - \ \langle x \rangle)}{\hbar} \ + \ \frac{i\langle p \rangle}{\hbar}$$
$$\Rightarrow \ \frac{1}{\Psi} \ \text{d}\Psi \ = \ \Big( -\frac{a(x \ - \ \langle x \rangle)}{\hbar} \ + \ \frac{i\langle p \rangle}{\hbar} \Big) \ \text{dx} \ \Rightarrow \ \displaystyle\int \ \frac{1}{\Psi} \ \text{d}\Psi \ = \ \displaystyle\int \ \Big( -\frac{a(x \ - \ \langle x \rangle)}{\hbar} \ + \ \frac{i\langle p \rangle}{\hbar} \Big) \ \text{dx}$$
$$\Rightarrow \ \ln \ \Psi \ = \ \Big( -\frac{a(x \ - \ \langle x \rangle)^2}{2\hbar} \ + \ \frac{i\langle p \rangle x}{\hbar} \Big) \ \Rightarrow \ \Psi \ = \ e^{-\frac{a(x \ - \ \langle x \rangle)^2}{2\hbar} \ + \ \frac{i\langle p \rangle x}{\hbar}}$$
That's it! Like I said, fairly straight-forward!