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Please check my answer?

  1. Aug 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of y'+ay=0.


    2. Relevant equations
    dy/dx=-ay
    dy/y=-a dx
    ∫ dy/y=∫ -a dx
    ln(y)=-ax+C
    y=e^(-ax)



    3. The attempt at a solution
    What I want to know is that is y=e^(-ax) the answer or y=ce^(-ax) is the answer?
     
  2. jcsd
  3. Aug 3, 2014 #2

    verty

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    Homework Helper

    Find which values of c solve the equation. If it is only true for c = 1, THEN y = e^(-ax) will be the answer.
     
  4. Aug 3, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, and now you solve for y by taking the exponential of both sides.

    ??? What happened to the "C"?



     
  5. Aug 4, 2014 #4
    Thank you, Hallsoflvy.
     
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