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Math Help - 1st order ode

  1. #1
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    1st order ode

    I am trying to solve the following ODE in order to obtain a solution in explicit for:

    xdy=(x+y)dx

    WITH INITIAL CONDITION Y(X=0)=-2

    i have tried exact equations and attempted substituion but no luck.

    thanks
    Last edited by Jester; December 23rd 2010 at 11:08 AM. Reason: reduced font size
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  2. #2
    A Plied Mathematician
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    The equation is linear in y. Have you tried the integrating factor method?
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  3. #3
    A Plied Mathematician
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    Hmm. Come to think of it, the point x = 0 is a singularity of the DE. I'm not sure you can solve it as written. Are you sure the initial condition is applied at x = 0?

    Incidentally, could you please reduce the size of your font? It's a bit loud!
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  4. #4
    Super Member General's Avatar
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    It is homogeneous also.
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  5. #5
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    Quote Originally Posted by General View Post
    It is homogeneous also.
    @OP: To expand, the DE can be written \displaystyle \frac{dy}{dx} = 1 + \frac{y}{x} (provided x \neq 0). I suppose the given initial condition might be a limit, as in \displaystyle \lim_{x \to 0} y = -2 .... But that condition is false in fact false since \displaystyle \lim_{x \to 0} y = 0 ....
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    The substitution y=vx ( homogeneous equation ) provides the general solution:

    y=x(C+\log |x|)

    Possibly there is a typo and the initial condition is y(1)=-2 , so the particular solution would be:

    y=x(-2+\log |x|)

    Fernando Revilla
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