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Math Help - [SOLVED] First order homogeneous ODE

  1. #1
    Super Member Matt Westwood's Avatar
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    [SOLVED] First order homogeneous ODE

    Another one from G.F.Simmons' "Differential Equations": Chapter 2 Miscellaneous Exercises no. 20:

    \left({y^2 - 3 x y - 2 x^2}\right) dx = \left({x^2 - xy}\right) dy

    This can be put in the format:
    \frac {dy} {dx} = \frac {y^2 - 3 x y - 2 x^2} {x^2 - xy} = \frac {\left({\frac y x}\right)^2 - 3\left({\frac y x}\right) - 2} {1 - \left({\frac y x}\right)}

    Substituting z = \frac y x, from which x \frac {dz}{dx} + z = \frac {dy} {dx}

    which leads to:
    x \frac {dz}{dx} = \frac {z^2 - 3 z - 2} {1 - z} - z = \frac {z^2 - 3 z - 2} {1 - z} - \frac {z - z^2} {1 - z} = \frac {2 z^2 - 4 z - 2} {1 - z}

    This leads to the messy:
    \int \frac {1 - z} {2 z^2 - 4 z - 2} dz = \int \frac {dx} x

    Flogging through the calculus and after tidying up:
    \sqrt 2 \ln \left({\frac {z + 1 - \sqrt 2} {z + 1 + \sqrt 2}}\right) = \ln \left({2 z^2 - 4 z - 2}\right) = 4 \ln x

    which is miles away from the given answer:
    x^2 y^2 - 2 x^3 y - x^4 = C
    (although that x^4 looks promising)

    If I try solving it by doing:
    \frac {dx} {dy} = \frac {x^2 - xy} {y^2 - 3 x y - 2 x^2}

    and dividing T&B by x^2 and substuting z = \frac x y it's even worse.

    I must be doing something fundamentally wrong in the first few lines but for the life of me I can't work out what.
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  2. #2
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    Your mistake is at the point where you say 'This leads to the messy:'.
    The integral on the LHS is of the standard form
     \int \frac{f'(x)}{f(x)} dx = \ln{f(x)} + C
    (after you have multiplied top and bottom of the integrand by -4 ).
    You seem to have lost the negative sign in your integration. With that, the RHS becomes
     -4\ln x which then creates a  \frac{C}{x^4}.
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  3. #3
    Super Member Matt Westwood's Avatar
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    That's excellent. Many thanks.

    Yes I thoroughly accept the fact that my tidying up was inaccurate! I really need to practice more.

    I suspect that if I try it again and take more care over my signs, the messy glob of \sqrt 2 \ln ... stuff will magically vanish. It ought to, the same result should be obtainable whatever the technique.
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