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Math Help - xyy' = yy' + x(y')^2

  1. #1
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    xyy' = yy' + x(y')^2

    Hi everyone,
    I need help trying to solve the following:

    xyy'' = yy' + x(y')^2

    Mathematica says the solution is y(x) = c_1e^{x^2} but I have no idea how to get there.
    I just noticed that the RHS can be written as (xyy')' but made no progress from there.
    Any help is appreciated. Thanks!
    Last edited by DJDorianGray; January 20th 2010 at 02:56 PM.
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  2. #2
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    Quote Originally Posted by DJDorianGray View Post
    Hi everyone,
    I need help trying to solve the following:

    xyy' = yy' + x(y')^2

    Mathematica says the solution is y(x) = c_1e^{x^2} but I have no idea how to get there.
    I just noticed that the RHS can be written as (xyy')' but made no progress from there.
    Any help is appreciated. Thanks!
    Check you ODE - seems like something isn't correct (since y' factors from what you have here).
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  3. #3
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    Corrected it, sorry

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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by DJDorianGray View Post
    Hi everyone,
    I need help trying to solve the following:

    xyy'' = yy' + x(y')^2

    Mathematica says the solution is y(x) = c_1e^{x^2} but I have no idea how to get there.
    I just noticed that the RHS can be written as (xyy')' but made no progress from there.
    Any help is appreciated. Thanks!
    I do not know a special method for solving equations of this type, but your equation can be solved so:

    xyy'' = yy' + x{\left( {y'} \right)^2} \Leftrightarrow \left( {yy'' - {{\left( {y'} \right)}^2}} \right)x = yy' \Leftrightarrow

    \Leftrightarrow \frac{{y''}}{{y'}} - \frac{{y'}}{y} = \frac{1}{x} \Leftrightarrow {\left( {\ln \left| {y'} \right|} \right)^\prime } - {\left( {\ln \left| y \right|} \right)^\prime } = \frac{1}{x} \Leftrightarrow

    \Leftrightarrow {\left( {\ln \left| {y'} \right| - \ln \left| y \right|} \right)^\prime } = \frac{1}{x} \Leftrightarrow {\left( {\ln \left| {\frac{{y'}}{y}} \right|} \right)^\prime } = \frac{1}{x} \Leftrightarrow

    \Leftrightarrow \ln \left| {\frac{{y'}}{y}} \right| = \ln \left| {{C_1}x} \right| \Leftrightarrow \frac{{y'}}{y} = {C_1}x \Leftrightarrow \int {\frac{{dy}}{y}}  = {C_1}\int {x\,dx}  \Leftrightarrow

    \Leftrightarrow \ln y = {C_1}{x^2} + {C_2} \Leftrightarrow y = {e^{{C_1}{x^2} + {C_2}}} = C_2e^{C_1x^2}.

    Do you understand?
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  5. #5
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    Try letting y = e^u and see what happens.
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