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Math Help - inhomogeneous linear differential equation

  1. #1
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    inhomogeneous linear differential equation

    b)
    Determine the general solution of the inhomogeneous linear differential equation

    y′ = [xy / (1+x^2)] + SQRT [ (1+x^2) / (1-x^2)]
    by the method of integrating factor.
    Last edited by rajr; February 1st 2009 at 11:13 AM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by rajr View Post
    b)
    Determine the general solution of the inhomogeneous linear differential equation

    y′ = [xy / (1+x^2)] + SQRT [ (1+x^2) / (1-x^2)]
    by the method of integrating factor.
    First, rewrite the DE as follows: \frac{\,dy}{\,dx}-\frac{x}{1+x^2}y=\sqrt{\frac{1+x^2}{1-x^2}}

    The integrating factor would be \varrho\!\left(x\right)=e^{-\int\frac{x}{1+x^2}\,dx}

    By u-substitution, it can be shown that \varrho\!\left(x\right)=\frac{1}{\sqrt{1+x^2}}

    Thus, the DE becomes \frac{\,d}{\,dx}\left[\frac{1}{\sqrt{1+x^2}}\cdot y\right]=\frac{1}{\sqrt{1-x^2}}

    It should be pretty straightforward from here.

    Does this make sense?
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  3. #3
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    hi thx for ur help

    should i then integrate both sides to get Y
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  4. #4
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    no you just integrate the left side and you should be left with an answer like y=(1+x^2)^-1/2[sin^-1(x)+C]
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  5. #5
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    thx for ur help
    Last edited by rajr; February 1st 2009 at 12:00 PM.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by anon18 View Post
    no you just integrate the left side and you should be left with an answer like y=(1+x^2)^-1/2[sin^-1(x)+C]
    No. You integrate both sides to get \frac{1}{\sqrt{1+x^2}}\cdot y=\sin^{-1}\!\left(x\right)+C\implies y=\sqrt{1+x^2}\left[\sin^{-1}\!\left(x\right)+C\right] \implies y=\sqrt{1+x^2}\sin^{-1}\!\left(x\right)+C\sqrt{1+x^2}.

    When writing the solution, though, we usually prefer to write it as y=\sqrt{1+x^2}\left[\sin^{-1}\!\left(x\right)+C\right]
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  7. #7
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    thx for ur help Mr Chris....
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