The differential equation of the curve xc−1+yc+1=1 is given by

[c] xc−1+yc+1=1                            ? (i) ⇒1c−1+y′c+1=0                           ? (ii) ⇒y′1=c+11−c⇒y′−1y′+1=c Put value of c in (1) ⇒xy′−1y′+1−1+yy′−1y′+1+1=1 ⇒x(y′)+1−2+y(y′+1)2y′=1 ⇒(y′+1)2(yy′−x)=1 ⇒(1+dydx)(y−xdydx)=2dydx