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Homework Help: Solving a linear ODE

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve 2*dy/dx - y = e^x at y(0) = 0

    3. The attempt at a solution

    So the integrating factor is e^-x

    Multiplying through by e^-x:

    2e^-x(dy/dx) - ye^-x = 1

    Now this is where I'm having a slight problem, isn't the left hand side meant to contract via the product rule to d/dx(...) ?

    Except that the derivative of 2e^-x is not -e^-x

    What have I done wrong?
  2. jcsd
  3. Sep 20, 2011 #2


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    You might want to check your integrating factor :wink:
  4. Sep 20, 2011 #3


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    The form of the differential equation needed before computing an integrating factor is
    y' + P(x)y = Q(x). This is done so the Product Rule can be applied: you are looking for a function I(x) such that

    [tex]I(x) \frac{dy}{dx} + I(x)P(x)y = I(x)Q(x) \Rightarrow \frac{d}{dx}[I(x)y(x)] = I(x)Q(x) ,[/tex]

    that is, so you will be able to re-make the left-hand side of the differential equation into the derivative of a simple product. After integration, you will have [itex]I(x)y(x) = \int I(x)Q(x) [/itex], after which you can simply divide I(x) back out again to obtain an expression for y (one that has a "closed form" if I(x)Q(x) can be integrated).
  5. Sep 20, 2011 #4


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    Then you must have the wrong integrating factor since this is exactly how "integrating factor" is defined!

  6. Sep 20, 2011 #5
    Ah...I see now, I should divide the equation through by 2 first, therefore the integrating factor becomes e^-1/2x

    and the equation eventually becomes y = e^x + e^-1/2x

    Thanks guys =)
  7. Sep 20, 2011 #6


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    Not quite: check what you did when you divided e-(1/2)x back out. Your solution doesn't work in the original DE...
    Last edited: Sep 20, 2011
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