y''-(2xy')/(1-x^2)+(2y)/(1-x^2)=0, we know that y1(x)=x is one solution, find the other(adsbygoogle = window.adsbygoogle || []).push({});

(starting at y2'' I'll let the multiplication and y1 be assumed)

so y2=u*y1

y2'=u'*y1+u*y1'

y2''=u''y+2u'y'+uy''

plug all that crap into the original and seperate the variables and it simplifies too

u''/u'=-[2y'/y-2x/(1-x^2)]

which is what we expect it too, so I'm pretty sure I'm right up to here, then we integrate both sides

lnu'=-2lny+ln(1-x^2)+c1, plug x back into y, and exponentiate both sides

u'=x^-2(1-x^2)*C

so integrate again

u=-c(x^2+1)/x

y2=u*y = -(x^2+1)*x (I set c = x to cancel out that one x in the denominator)

but this is totally not what the book got

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# Differential Equation help!

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