The correct option is D y=2√cx+c2
Write p=dydx and differentiating w.r.t.x, we have
p=2p+2xdpdx+2xp4+4p4+4p3x2dpdx⇒0=p(1+2xp3)+2x(1+2p3x)⇒p+2xdpdx=0⇒2dpdx=−dxx⇒2logp+logx=const⇒p2x=c or p=√cx
Substituting this value in the given equation, we get y=2√cx+c2