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if in the form
y' + P(x) y = Q(x)
the integrating factor is e^F
where F= integral P(x) dx

\[(1-x^2)\frac{dy}{dx} -xy = x\]

integral -x dx/(1-x^2) = 1/2 * ln(1-x^2) (toss in abs signs to be sure it's positive)

exp (ln(1-x^2)^(1/2))= sqrt(1-x^2)

integral du/u = ln(u)
in this case u = 1-x^2 du = -2x dx

ok - i see it now . I was confused with the sign when integrating
thanx

i need more practice in FODEs .......

are you saying you can't finish ?

no i think i've got it now
I get the solution to be
y = A(1 - x^2)^(-1/2) - 1

that looks good

I got it down to
|dw:1359820446243:dw|

from here its easy
thanks again

though you could have tossed in a minus sign on both sides before integrating ?

hmm i'm confused
how come the integral is negative of square root bit?

i'm allergic to integration!!!

except you have a^2 - x^2 not a^2 + x^2

maybe the Greeks had the right idea, stick with positive rational numbers at all times

then none of these stray minus signs to keep track of.

yes its the signs which have confused me here

if this was easy, anybody could do it.

right