- #1

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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)

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- Thread starter xavier777
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- #1

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My solution :

dy/dt = -a(y-b/a)

(dy/dt)/(y-(b/a)) = -a

ln | y-(b/a) | = -at + C

e

Ce

- #2

Mark44

Mentor

- 34,887

- 6,626

Or ##y = Ce^{-at} + \frac b a##

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)

Did you check to see if your solution satisfies the diff. equation?

- #3

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- #4

Mark44

Mentor

- 34,887

- 6,626

Substitute your solution in the original differential equation and you should get a true statement.^{C}e^{-at}= Ce^{-at}?

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