# Find the general eqn of this(1-x^2)(1-y)dx=xy(1+y)dy

2
by ayushk9919

• Brainly User
2014-08-26T19:40:56+05:30
(1-x^2)/x dx=y(y+1)/(1-y)
(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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Thank u but ans is x(1-y^2)=c exp(x^2-y^2-4y)/2
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2014-08-27T07:19:46+05:30

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Integrating on both sides we get,

thanks and u r welcome !