Question

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

Answer 1

solve this diff eq.?

Answer 2

yes :) can you?

Answer 3

I did this:
\[(1-x ^{2})y'-2xy ^{2}=xy\]
\[(1-x ^{2})y'=2xy ^{2}+xy\]
I divided this by x and got
\[(x ^{-1}-x)y'=2y ^{2}+y\]

and this is the type which separates variables. Is this ok?

Answer 4

but why you don't change y'=1 why do you need it like y' ??

Answer 5

y'=dy/dx

Answer 6

why y'=1?

Answer 7

never mind...

Answer 8

is this ok?

Answer 9

ahh... :@:@ ... I don't know , let me grab a paper , latex sucks !

Answer 10

... I'm not sure about this, but if you divided by x

then why y' isn't \[\Large \frac{y\;'}{x}\]

Answer 11

I know... it's multiply , Leave it ... seems right to me !.....

Answer 12

because \[(1-x ^{2})y'/x=[(1-x ^{2})/x]y'=(1/x-x)y'\]

Answer 13

ok?

Answer 14

\[\LARGE (1-x^2)y'=2xy^2+xy\]

divided by X...

\[\LARGE \frac{(1-x^2)y'}{x}=\frac{2xy^2+xy}{x}\]

\[\LARGE \left(\frac{1}{x}-\frac{x^2}{x}\right) \cdot y'=\frac{2xy^2}{x}+\frac{xy}{x}\]

\[\LARGE \left(x^{-1}-x\right) \cdot y'=2y^2+y\]

that's correct... well done !

Answer 15

:)