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Math Help - Solving non-linear ODE

  1. #1
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    Solving non-linear ODE

    Hi, I would appreciate any help in solving this:

    y''(x) \cdot y(x) = (y'(x))^2 - y'(x)

    With initial conditions y(0) = 1, y'(0) = 2

    It may be helpful to note the fact that  (y(x) \cdot y'(x))' = (y'(x))^2 + y(x) \cdot y''(x)

    or that \left( \frac{y'(x)}{y(x)} \right) ' = \frac{y''(x) \cdot y(x) - (y'(x))^2}{(y(x))^2} (we get this if we divide both sides of the equations by y^2)

    But I was not able to solve this using either.

    Thanks for any help.
    Last edited by Defunkt; March 8th 2010 at 01:59 PM.
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  2. #2
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    Quote Originally Posted by Defunkt View Post
    Hi, I would appreciate any help in solving this:

    y''(x) \cdot y(x) = (y'(x))^2 - y'(x)

    With initial conditions y(0) = 1, y'(0) = 2

    It may be helpful to note the fact that  (y(x) \cdot y'(x))' = (y'(x))^2 + y(x) \cdot y''(x)

    or that \left( \frac{y'(x)}{y(x)} \right) ' = \frac{y''(x) \cdot y(x) - (y'(x))^2}{(y(x))^2} (we get this if we divide both sides of the equations by y^2)

    But I was not able to solve this using either.

    Thanks for any help.
    The last hint is very useful.

     <br />
\frac{y y'' - y'^2}{y^2} = - \frac{y'}{y^2}<br />
or \left(\frac{y'}{y}\right)' = \left(\frac{1}{y} \right)'

    so

     <br />
\frac{y'}{y} = \frac{1}{y} + c_1<br />

    or y' - c_1 y = 1 (this is separable)
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  3. #3
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    Quote Originally Posted by Danny View Post
    The last hint is very useful.

     <br />
\frac{y y'' - y'^2}{y^2} = - \frac{y'}{y^2}<br />
or \left(\frac{y'}{y}\right)' = \left(\frac{1}{y} \right)'

    so

     <br />
\frac{y'}{y} = \frac{1}{y} + c_1<br />

    or y' - c_1 y = 1 (this is separable)
    Yep, for some reason I forgot to add the constant to the integral :\

    thanks
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