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Math Help - Solve the ode

  1. #1
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    Solve the ode

    x^3y' + 2y = x^3 + 2x y(1) = e+1

    Which method would you use to solve this?
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  2. #2
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    Quote Originally Posted by sarah232 View Post
    x^3y' + 2y = x^3 + 2x y(1) = e+1

    Which method would you use to solve this?
    The equation is linear so first put it in standard form

    y' + \frac{2y}{x^3} = 1 + \frac{2}{x^2}

    The integrating factor is

    \mu = e^{-\frac{1}{x^2}} so

    \frac{d}{dx} e^{-\frac{1}{x^2}} \cdot y = \left( 1 + \frac{2}{x^2}\right) e^{-\frac{1}{x^2}} which integrates giving

     e^{-\frac{1}{x^2}} \cdot y = x e^{-\frac{1}{x^2}} + c

    I think you can take it from here.
    Last edited by Jester; April 18th 2009 at 08:09 AM. Reason: typo
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    The equation is linear so first put it in standard form

    y' + \frac{2y}{x^3} = 1 + \frac{2}{x^3}

    The integrating factor is

    \mu = e^{-\frac{1}{x^2}} so

    \frac{d}{dx} e^{-\frac{1}{x^2}} \cdot y = \left( 1 + \frac{2}{x^2}\right) e^{-\frac{1}{x^2}} which integrates giving

     e^{-\frac{1}{x^2}} \cdot y = x e^{-\frac{1}{x^2}} + c

    I think you can take it from here.
    Your standard form is wrong.

    If x^3\frac{dy}{dx} + 2y = x^3 + 2x then

    \frac{dy}{dx} + 2x^{-3}y = 1 + 2x^{-2}.
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