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Math Help - how to solve this?

  1. #1
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    how to solve this?

    Solve:
    y''=yy'/x
    And show that y=c3 is a solution of the problem too.
    where y'=dy/dx

    How to solve it?
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  2. #2
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    Re: how to solve this?

    The second question is easy: if y is a constant, then both y' and y'' are 0 so the equation becomes 0= 0.
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: how to solve this?

    Quote Originally Posted by blocnt View Post
    Solve:
    y''=yy'/x
    And show that y=c3 is a solution of the problem too.
    where y'=dy/dx

    How to solve it?
    Well I have a stab at it that might lead somewhere. But I can't finish it. Perhaps it will give someone an idea.

    Do you have initial conditions on this?

    Note that \frac{1}{2} \frac{d}{dx}(y^2) = y y' and that x y'' = \frac{d}{dx}(xy') - y'.

    Then
    y'' = \frac{y y'}{x}

    x y'' = y y'

    \int x y'' dx = \int y y' dx

    \int \left ( \frac{d}{dx}(x y') - y' \right ) dx = \frac{1}{2}y^2

    xy' - \int y' dx = \frac{1}{2}y^2

    xy' - y = \frac{1}{2}y^2 + a
    where a is the accumulated integration constant.

    My hope was for a Bernoulli equation, but we have that extra constant that seems to screw things up. I can't find a way to proceed from here.

    -Dan

    Edit: I was hoping that, since y = constant was a solution (leading to a =0 in the above equation) would help, but it leads to a Bernoulli equation where y = constant is not a solution to the Bernoulli, so this doesn't help.
    Last edited by topsquark; October 30th 2013 at 08:54 AM.
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  4. #4
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    Re: how to solve this?

    I know a little late but the ODE is now separable.
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