Question

Solve :

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

Answer 1

gotcha

Answer 2

sorry..glitch

Answer 3

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....

Answer 4

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

Answer 5

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

Answer 6

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...

Answer 7

ggabnore i think u have understood the way to solve this type of DE

Answer 8

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.