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anonymous
 5 years ago
Solve :
y' = (xy+2)/(1x^2)
anonymous
 5 years ago
Solve : y' = (xy+2)/(1x^2)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see this will be a series solution... let \[y=x ^{m}\sum_{n=0}^{\infty}a _{n}x ^{n}\] \[y'=(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+n1}\] \[(1x ^{2})(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+n1}=\sum_{n=0}^{\infty}a _{n}x ^{m+n+1}+2\] now in both side you have to equate the power of x and have to evaluate coefficient a0,a1,a2 amd hence the series....

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0i took this as a normal DE, found integrating factor of sqrt(1x^2) found solution \[y = \frac{2\sin^{1} x}{\sqrt{1x ^{2}}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+n1}(m+n)\sum_{n=o}^{\infty}a _{n}x ^{m+n+1}=\sum_{n=0}^{\infty}a _{n}x ^{m+n+1} +2\] \[ma _{0}x ^{m1}+(m+1)a _{1}x ^{m}+(m+n+2)\sum_{n=0}^{\infty}a _{n+2}x ^{m+n+1}(m+n)\sum_{n=0}^{\infty}a _{n}x ^{m+m+1}\]=\[\sum_{n=0}^{\infty}a _{n}x ^{m+n+1}+2\]from here obviously ma0=0, (m+1)a1=0, and so on...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see 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...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ggabnore i think u have understood the way to solve this type of DE

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah.. 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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