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Math Help - 2nd order non-linear ODE, no 'x' term

  1. #1
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    2nd order non-linear ODE, no 'x' term

    I'm trying to solve the ODE

    yy"=4(y')^2

    I'm "close" to the answer provided, but I can't see where I'm going wrong. This is what I've done:

    let z=y', so y" = \frac{dz}{dx} = \frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}

    Then the ODE becomes yz\frac{dz}{dy}=4z^2, which is separable and gives me

    z = Cy^4

    Now integrating wrt x to recover y,

    y = Cxy^4+K

    However... the answer provided is y = (Cx+K)^{-1/3}, and I don't think manipulating what I have will get me here

    So what have I done wrong?
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  2. #2
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    Divide both sides by yy' and integrate.
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  3. #3
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    Quote Originally Posted by HD09 View Post
    I'm trying to solve the ODE

    yy"=4(y')^2

    I'm "close" to the answer provided, but I can't see where I'm going wrong. This is what I've done:

    let z=y', so y" = \frac{dz}{dx} = \frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}

    Then the ODE becomes yz\frac{dz}{dy}=4z^2, which is separable and gives me

    z = Cy^4

    Now integrating wrt x to recover y,

    y = Cxy^4+K

    However... the answer provided is y = (Cx+K)^{-1/3}, and I don't think manipulating what I have will get me here

    So what have I done wrong?
    If z = cy^4 then \dfrac{dy}{dx} = cy^4. Now separate \dfrac{dy}{y^4} = cdx, integrate and solve for y.
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