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anonymous

  • 5 years ago

Solve : y' = (xy+2)/(1-x^2)

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  1. anonymous
    • 5 years ago
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    gotcha

  2. anonymous
    • 5 years ago
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    sorry..glitch

  3. anonymous
    • 5 years ago
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    see this will be a series solution... let \[y=x ^{m}\sum_{n=0}^{\infty}a _{n}x ^{n}\] \[y'=(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+n-1}\] \[(1-x ^{2})(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+n-1}=\sum_{n=0}^{\infty}a _{n}x ^{m+n+1}+2\] now in both side you have to equate the power of x and have to evaluate coefficient a0,a1,a2 amd hence the series....

  4. dumbcow
    • 5 years ago
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    i took this as a normal DE, found integrating factor of sqrt(1-x^2) found solution \[y = \frac{2\sin^{-1} x}{\sqrt{1-x ^{2}}}\]

  5. anonymous
    • 5 years ago
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    \[(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+n-1}-(m+n)\sum_{n=o}^{\infty}a _{n}x ^{m+n+1}=\sum_{n=0}^{\infty}a _{n}x ^{m+n+1} +2\] \[ma _{0}x ^{m-1}+(m+1)a _{1}x ^{m}+(m+n+2)\sum_{n=0}^{\infty}a _{n+2}x ^{m+n+1}-(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+m+1}\]=\[\sum_{n=0}^{\infty}a _{n}x ^{m+n+1}+2\]from here obviously ma0=0, (m+1)a1=0, and so on...

  6. anonymous
    • 5 years ago
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    see what dumbcow has got , it also can be represented by a series and the coefficient will be what you find from this equation... the general way to do this is for series solution......it always works...

  7. anonymous
    • 5 years ago
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    ggabnore i think u have understood the way to solve this type of DE

  8. anonymous
    • 5 years ago
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    Yeah.. I really don't think I'm expected to know this for my course, the questions aren't this extreme. But thanks everyone for the replies.

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