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## Answers

(1/x dx -x dx)= y^2/1-y dy+ y/y-1dy

can b written as 1/x dx- x dx = (y^2+1-1)/1-y + (y+1-1)/1-y

integrating both sides

logx- x^2/2= (y+1)(y-1)/(1-y)dy +2/(1-y)dy +(y-1)/(1-y)dy

on solving we get

logx-x^2/2= -y^2/2 -y +2log(1-y)-y + c

c be any constant

logx-2log(1-y)=x^2/2-y^2/2-y+c

that is lhs = logx-log(1-y)^2= logx-log(y^2+2y-1)

using lhs log(x/y^2+2y-1) log a- log b =log (a/b)

hence x/y^2+2y-1= e^(x^2/2-y^2/2+c)

0rrr x= (y^2+2y-1)e^(x^2/2-y^2/2+c)

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Integrating on both sides we get,