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How to solve this kind of differential equation

  1. Oct 9, 2012 #1
    Can anyone give some clues on how to solve this differential equation:

    f '(x) = 3 * f(x)

    Thanks!
     
  2. jcsd
  3. Oct 9, 2012 #2
    Let me see if I can help...

    If the problem is asking to solve for f(x)...
    Then, I'd first ask myself, what function is equal to its derivative.

    I only know of one function: e^x
    Therefore, the answer must be of be form: e^(kx), where k is some constant.

    Does that help?
     
  4. Oct 9, 2012 #3
    f'(x) = 3f(x)
    lets say y = f(x) for funsies

    y' = 3y
    (dy/dx) = 3y
    dy/(3y) = dx

    Integrate with respect to both sides...

    right side = X+ C
    Left side, , it equals (1/3)*(lny) (technically absolute value of y, but whatevs)

    so
    ln(y) = 3x + 3c ... which we can also say 3x + C, since C is an arbitrary constant
    ln(y) = 3x+C
    y = e^(3x+C)
    y = (e^(3x))(e^C)
    e^C is a constant, so we can say thats C
    so...

    y = Ce^(3x)
     
  5. Oct 9, 2012 #4
    Thank you guys, got it now.
     
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