# I Is this general solution correct?

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1. Jun 20, 2016

### xavier777

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)

2. Jun 20, 2016

### Staff: Mentor

Or $y = Ce^{-at} + \frac b a$
Did you check to see if your solution satisfies the diff. equation?

3. Jun 20, 2016

### 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 ?

4. Jun 21, 2016

### Staff: Mentor

Substitute your solution in the original differential equation and you should get a true statement.