Is this general solution correct?

  • #1
2
0
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)
 

Answers and Replies

  • #2
34,887
6,626
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)
Or ##y = Ce^{-at} + \frac b a##
Did you check to see if your solution satisfies the diff. equation?
 
  • #3
2
0
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
34,887
6,626
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 ?
Substitute your solution in the original differential equation and you should get a true statement.
 

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