Question

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

Answer 1

(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)

Answer 2

$(1-x^2)\ (1-y) \ dx = x\ y\ (1+ y ) \ dy \\ \\ \frac{(1-x^2)\ dx}{x} = \frac{y(1+y)\ dy}{1-y} \\ \\ Let\ \ 1 - y = t ,\ so\ \ -dy = dt ,\ \ substitute\ in\ the\ above\ equation \\ \\ \frac{1}{x} dx - x dx =\ \frac{(1-t)(1+1-t)}{t}(-dt) = -\frac{2-3t+t^2}{t}dt = \frac{-2}{t} dt + 3 dt - t dt$

Integrating on both sides we get,

$Ln\ x\ -\ \frac{x^2}{2} = - 2 Ln t + 3 t - \frac{t^2}{2}+ K$

$Ln\ x + 2 Ln (1-y) = \frac{x^2 - (1-y)^2 + 6 (1-y)}{2} + K \\ \\ Ln\ x (1-y)^2 = \frac{1}{2} * [x^2 - y^2 + 5 - 4 y + 2K] \\ \\ x ( 1- y )^2 = e^{\frac{1}{2}[x^2 - y^2 - 4 y + 2K + 5]]}$