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Math Help - [SOLVED] 1st order homogeneous diffeq from G.F.Simmons

  1. #1
    Super Member Matt Westwood's Avatar
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    [SOLVED] 1st order homogeneous diffeq from G.F.Simmons

    Chapter 2, miscellaneous exercises no. 4:

    x y' = \sqrt {x^2 + y^2}

    I made it homogeneous:
    \frac {dy} {dx} = \sqrt {1 + \left({\frac y x}\right)^2}

    Then I substituted z = \frac x y which makes it separable:
    z + x \frac {dz}{dx} = \sqrt {1 + z^2}

    and so
    \int \frac {dz} {\sqrt {1 + z^2} - z} = \int \frac {dx} x

    Now the nightmare begins. The integral on the LHS seems to be tractable only if I try the substitution z = \sinh w

    This leads to:
    \int \frac {\cosh w dw} {\cosh w - \sinh w}

    Converting the hyp funcs to exps:
    \int \frac {e^w + e^{-w}} {e^{-2w}}

    which gets us to
    \int \left({e^{3w} + e^w}\right) dw = \frac {e^{3w}} 3 + e^w + C

    To get back to z and thence y/x, I substitute back using the formula for w = \sinh^{-1} z = \ln \left({z + \sqrt{z^2 + 1}}\right) from which I end up with the answer to my diffeq:

    \left({\frac y x + \sqrt{\left({\frac y x}\right)^2 + 1}}\right)^3 + 3 \left({\frac y x + \sqrt{\left({\frac y x}\right)^2 + 1}}\right) = 3 \ln x + C

    So far so good, but the answer as given in the book is:
    y \sqrt {x^2 + y^2} + x^2 \ln \left({y + \sqrt {x^2 + y^2}}\right) + y^2 = 3 x^2 \ln x + c x^2

    Can these two results be reconciled? I fear not. Can anyone work it through and see whether (a) I've made a mistake somewhere, or (b) I haven't followed through far enough?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Matt Westwood View Post
    [snip]

    Chapter 2, miscellaneous exercises no. 4:

    x y' = \sqrt {x^2 + y^2}

    I made it homogeneous:
    \frac {dy} {dx} = \sqrt {1 + \left({\frac y x}\right)^2}

    Then I substituted z = \frac x y which makes it separable:
    z + x \frac {dz}{dx} = \sqrt {1 + z^2}

    and so
    \int \frac {dz} {\sqrt {1 + z^2} - z} = \int \frac {dx} x

    [/snip]
    Here's the trick (for LHS):

    \int\frac{\,dz}{\sqrt{1+z^2}-z}=\int\frac{1}{\sqrt{1+z^2}-z}\cdot\frac{\sqrt{1+z^2}+z}{\sqrt{1+z^2}+z}\,dz =\int\frac{\sqrt{1+z^2}+z}{1+z^2-z^2}\,dz=\int\sqrt{1+z^2}+z\,dz.

    Now all you need to do is apply a trig sub to \int\sqrt{1+z^2}\,dz; it should be a walk in the park from there.
    Last edited by Chris L T521; August 17th 2009 at 01:55 PM. Reason: fixed tiny typo... xD
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  3. #3
    Super Member Matt Westwood's Avatar
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    Ahh, gotcha. Genius. I'd completely forgotten about that sweet little manipulation.

    I like your DE tutorial, btw.
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