The curve satisfying the equation dydx=yx+y3xy3−x and passing through the point (4,−2) is

Given differential equation is dydx=yx+y3xy3−x xy3dy−y4dx=xydx+x2dyxy3dy−y4dx=x(ydx+xdy)xy3dy−y4dxx=(ydx+xdy)⇒xxy3dy−y4dxx2=d(xy)⇒xy3xdy−ydxx2=d(xy)⇒xy3dyx=d(xy)x2y2⋅yxdyx=d(xy)⇒yxdyx=1(xy)2d(xy)⇒12dyx2=d(xy)−1⇒12yx2=−1xy+c As it passes through (4,−2)⇒12−242=−14(−2)+c⇒1214=18+c⇒c=012y2x2=1xy⇒y2=−2x2xy⇒y2=−2xy⇒y3=−2x