Is this general solution correct?

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Discussion Overview

The discussion revolves around the general solution of the differential equation dy/dt = -ay + b. Participants are exploring the steps involved in deriving the solution and verifying its correctness.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a solution involving the transformation of the differential equation and integration steps.
  • Another participant reiterates the same solution and asks whether it satisfies the original differential equation.
  • Some participants inquire about the method to verify the solution, suggesting differentiation or substitution into the original equation.
  • Concerns are raised regarding the assumption that eCe^{-at} can be simplified to Ce^{-at}.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the solution or the verification method, as participants are still discussing the steps and assumptions involved.

Contextual Notes

Participants have not fully resolved the implications of their assumptions, particularly regarding the simplification of terms and the verification process for the solution.

xavier777
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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)
 
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xavier777 said:
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?
 
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 ?
 
xavier777 said:
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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