Question

Solve the initial value probelm of y'-y=e^x, y(1)=0

Answer 1

$\text{The integrating factor is } e^{-\int dx}=e^{-x}\\\text{Multiply the given differential equation by } e^{-x} \\\implies e^{-x}y'-e^{-x}y=e^{-x}e^x\\\int d(e^{-x}y)=\int1dx\\e^{-x}y=x+c\\y=xe^x+ce^{x}\\\text{Recall that y(1)=0}\\\implies0=e^1 +ce^1\\\text{Therefore c = -1, hence the general solution is }y =xe^x-e^x$