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Seperable Equation

  1. Sep 18, 2007 #1
    [SOLVED] Seperable Equation

    1. Instruction: Solve the equation.

    2. Equations:
    dy/dx = g(x)p(y)
    h(y) = 1/p(y)
    h(y)dy = g(x)dx
    [tex]\int[/tex]h(y)dy = [tex]\int[/tex]g(x)dx
    H(y) = G(x) + C

    3. [​IMG]

    I tried to do it on the right side, but.....I got stuck there. If I add to the right, then I would be left with -C + e[tex]^{-y}[/tex] = ex + -ye[tex]^{-y}[/tex]. Can I say that -C = C? But what about e[tex]^{-y}[/tex] . Did I inegrate it right?

    Thank you in advance.
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 18, 2007 #2


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    Nope. Didn't integrate it right. Try taking d/dy on the left side. What went wrong?
  4. Sep 19, 2007 #3
    so I should be integrating

    [tex]\underline{d}[/tex]= [tex]\underline{d(e^{y})}[/tex]
    dx(e[tex]^{x}[/tex]) dy (y-1)
  5. Sep 19, 2007 #4


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    You were OK up to the next to the last step. How do you integrate y·exp(-y) ?
  6. Sep 19, 2007 #5
    I multiplied ye[tex]^{-y}[/tex]dy - e[tex]^{-y}[/tex]dy, then integrate
  7. Sep 19, 2007 #6


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    Right, and you integrated the *second* term correctly. What integration technique must you use on the term ye[tex]^{-y}[/tex] ?
  8. Sep 19, 2007 #7
    du dv right?

    I think I got it!!! thank you so much!!!
  9. Sep 19, 2007 #8


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    If you mean by that, "integration by parts", we are in agreement. I hope that works out for you...
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