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Differential equation: y'-y=(1/2)x-1

  1. Mar 15, 2008 #1
    I know this looks really easy, but trying to solve this is amazingly difficult. I couldnt do it, i kept getting wrong answers. Any body got any idea how to solve this?
  2. jcsd
  3. Mar 15, 2008 #2
    When methods have you leart because there is more then one way to solve the above equation.
  4. Mar 15, 2008 #3
    I just used the basic linear form, got an integrating factor, and then just took the antiderivative of what was left
  5. Mar 15, 2008 #4
    If no one else helps you out I'll try to solve it with that method tomorrow. First though you have to show me the steps you tried.
  6. Mar 15, 2008 #5
    y'-y=g(x)....linear form
    d(x)y'-s(x)y=(d(x)y)'.....product rule
    only works if d(x)'=s(x)
    the only function that is its own integral, that i could think of, is e[tex]^{x}[/tex]
    but since the y is negative, it would have to be e[tex]^{-x}[/tex]

    so then u could say,


    (e[tex]^{-x}[/tex]y)'=e[tex]^{-x}[/tex]g(x)....use the product rule
    e[tex]^{-x}[/tex]y=antiderivative(e[tex]^{-x}[/tex]g(x))........took antiderivative
    y=antiderivative(e[tex]^{-x}[/tex]((1/2)x-1))/e[tex]^{-x}[/tex].......divide both sides by e[tex]^{-x}[/tex]

    I think im screwing up in the simplifying part. But, the integration becomes tricky
  7. Mar 15, 2008 #6
    Hey I just figured it out. The first time i did it i made a really small error in the beginning that made it unsolvable. Thanks anyways. In case your wondering though,

  8. Mar 15, 2008 #7
    You know that the solution isn't unique, right? You can add e^x times any constant to your answer and still have a valid solution.
  9. Apr 12, 2008 #8
    Yeah, the solution will only be unique if you specify an initial condition such as y(0). Otherwise your integrating factor method should still leave you with a term Ce^x for some undetermined constant C.
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