I recently learnt about solving partial differential equations using method of separation of variables.
The problem: Find a solution of the equation $\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial x} + 2u$ in the form $u = f(x)g(y)$. Solve the equation subject to the conditions $u = 0$ and $\frac{du}{dx} = 1 + e^{-3y}$ when $x = 0$ for all values of $y$.
Solution: On taking $u = f(x)g(y)$, I get $f(x) = c_1e^{2x} + c_2e^{-x}$. I assume $u = g(y)(c_1e^{2x} + c_2e^{-x})$ and then apply the boundary conditions. I end up getting couple of conditions: $c_1+c_2 = 0$ and $g(y) = \frac{1 + e^{-3y}}{2c_1-c_2}$ and then I am stuck.
How to calculate the constants and $g(y)$ here? No equation for $g(y)$ seems to be there. The given answer is: $u = \frac{1}{\sqrt{2}}sinh\sqrt{2}x + e^{-3y}sinx$.
You are almost done. Put $c_2=-c_1$ in $g$ and $u$ to get $g(y)=\frac{1+e^{-3y}}{3c_1}$ and $u(x,y)=g(y)c_1(e^{2x}-e^{-x})$, and therefore $$u(x,y)=\frac{1}{3}(1+e^{-3y})(e^{2x}-e^{-x})$$
Remark: Although this is a valid method to get a solution, it is pretty uncommon to use for these kind of problems, since there is no derivative wrt the variable $y$ in the equation. Treating this problem as ODE instead (by taking $c_1,c_2$ to be functions of $y$), will get you the same result, but will prove this is the only solution to the problem.