# Is this general solution correct?

• I
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)

## Answers and Replies

Mentor
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?

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 ?

Mentor
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.
The text passage contains a mathematical question (a differential equation) and a proposed solution, followed by a discussion on how to verify the solution.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Is this general solution correct?

• I
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)

## Answers and Replies

Mentor
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?

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 ?

Mentor
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.
user
The text passage contains a mathematical question (a differential equation) and a proposed solution, followed by a discussion on how to verify the solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER