Recent content by xavier777

How do I do that? Differentiate it to get back to the original diff eq ? If yes, then what about the assumption eCe-at = Ce-at ?

Question : General solution of dy/dt = -ay + b My solution : dy/dt = -a(y-b/a) (dy/dt)/(y-(b/a)) = -a Integrating both sides : ln | y-(b/a) | = -at + C e(-at+C) = y-(b/a) Ce(-at) = y-(b/a)