x+ydydxy−xdydx=x2+2y2+y4x2 , the solution of the differential equation is kyx−1x2+y2=C, then k=
Given equation can be rewritten as
xdx+ydyydx−xdy=x4+2x2y2+y4x2⇒xdx+ydyx2+y22=ydx−xdyy2×y2x2⇒∫dx2+y2x2+y22=2∫1(x/y)2⋅dxy⇒−1x2+y2=−2x/y+c⇒2yx−1x2+y2=C