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Separable differential equation

  1. May 1, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{du}{dt}[/itex] = [itex]e^{5u + 7t}[/itex]

    Solve the separable differential equation for u:

    Use the following initial condition: u(0) = 6.

    3. The attempt at a solution

    I tried to take the natural log of each side but now I'm stuck. How can I separate the equation when both the u and t terms are an exponent? And how can I use u(0) = 6? Sorry if I'm clueless, I missed a bunch of lectures on separable differential equations.
  2. jcsd
  3. May 1, 2014 #2


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    ##a^{x+y}=a^x\cdot a^y ##
  4. May 1, 2014 #3


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    What do you remember about working with powers? Can you write an expression like ##e^{a+b}## a different way?

    Edit: Sammy told you the answer while I was typng :smile:

    The general solution to the differential equation will contain an arbitrary constant. You can eliminate that using the initial condition u(0) = 6.
  5. May 1, 2014 #4
    Ok so I can rewrite as:

    [itex]\frac{du}{dt}[/itex] = [itex]e^{5u}[/itex] * [itex]e^{7t}[/itex]

    [itex]e^{-5u}[/itex] du = [itex]e^{7t}[/itex] dt

    After finding the antiderivatives I cross-multiplied:

    5[itex]e^{7t}[/itex] = -7[itex]e^{-5u}[/itex]

    [itex]\frac{-7}{5}[/itex] = [itex]\frac{e^{7t}}{e^{-5u}}[/itex]

    [itex]\frac{-7}{5}[/itex] = [itex]e^{7t + 5u}[/itex]

    This is now where I'm running into a problem. When I try to natural log both sides to get rid of the e and bring down the 7t + 5u exponent, this means I have ln([itex]\frac{-7}{5}[/itex]) on the other side of the equation, which is unreal. Did I do something wrong or is there some sort of way to work around this?
  6. May 1, 2014 #5


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    You forgot the +C part when you integrated.
  7. May 2, 2014 #6
    Do I add a constant to both sides of the equation?

    [itex]e^{-5u}[/itex] + [itex]C_{u}[/itex] = [itex]\frac{-5e^{7t}}{7}[/itex] + [itex]C_{t}[/itex]
  8. May 2, 2014 #7


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    That would be silly.

    You could then replace [itex]C_{t}-C_{u}[/itex] with [itex]C[/itex]
  9. May 2, 2014 #8
    Got it, thanks guys!
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