$y''-y=e^x$ initial conditions $y(0)=0$, $y'(0) = \frac{3}{2}$

Question

This is the solution to the reference:

How did the blue line come out, or set it out? This is really hard for me to think about, what is the general method?

Answer 1

Since $e^x$ is a solution of $y''-y=0$, then a particular solution has the form $$ y_p=Axe^x. $$ Putting this into the equation gives $A=\frac12$. Let the general solution be $$ y=C_1e^x+C_2e^{-x}+\frac12xe^x. $$ Then use the initial condition to solve $C_1,C_2$. You can try yourself.