y2+(x−1y)dydx=0
y2=−(x−1y)dydx
y3=−−(xy−1)ydydx
y3=(1xy)dydx
∫11−xydx=∫1y3dy
Solving LHS→ ∫11−xydx
Substitute u=1−yx
⇒ ∫−1yudu=−1y∫duu
⇒ −1ydu=−1y∫−duu
as ∫1/udu=ln|u|
⇒ −1yln|u|=−1yln|1−yx|
⇒ −1yln|1−yx|+C1
solving RHS.∫1y3dy
⇒ ∫y−3dy
as ∫x2dx=xn+1n+1+C
∴ ∫y−3dy⇒ −12y−2+C2
Hence, −1yln|1−yx|+C1=−12y−2+C2
−1yln|1−yx|+12y−2+C