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Math Help - Differential Equation Question

  1. #1
    Super Member craig's Avatar
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    Differential Equation Question

    Hi again, not posted in a while but started my degree course now and getting back into the swing of things.

    Anyway, I was reading ahead in the differential equations module and had a go at a couple of questions. Not got any answers for them though so if someone would be kind enough to take a quick look would be appreciated

    Solve xy' + x +y = e^x

    First divide through by x, y' + x^{-1}y = x^{-1}e^x - 1

    Then using the integrating factor of e^{\int{\frac{1}{x}}}, I get the following equation:

    xy' + y = e^x - x

    Integrating both sides gives me:

    xy = e^x - \frac{x^2}{2} + C

    Which can be tidied up to give y = \frac{e^x}{x} - \frac{x}{2} + \frac{C}{x}

    Just realised that if I had spotted the implicit differentiation I could have saved myself a lot of time, however is my rather long winded attempt still ok?

    Thanks

    Craig
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by craig View Post
    Hi again, not posted in a while but started my degree course now and getting back into the swing of things.

    Anyway, I was reading ahead in the differential equations module and had a go at a couple of questions. Not got any answers for them though so if someone would be kind enough to take a quick look would be appreciated

    Solve xy' + x +y = e^x

    First divide through by x, y' + x^{-1}y = x^{-1}e^x - 1

    Then using the integrating factor of e^{\int{\frac{1}{x}}}, I get the following equation:

    xy' + y = e^x - x

    Integrating both sides gives me:

    xy = e^x - \frac{x^2}{2} + C

    Which can be tidied up to give y = \frac{e^x}{x} - \frac{x}{2} + \frac{C}{x}

    Just realised that if I had spotted the implicit differentiation I could have saved myself a lot of time, however is my rather long winded attempt still ok?

    Thanks

    Craig
    You're longer way is still correct, since this is a linear differential equation.
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You're longer way is still correct, since this is a linear differential equation.
    Thanks for replying.

    Craig
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