Solution of the differential equation $x^2\big(y-x\frac{dy}{dx}\big)=y\big(\frac{dy}{dx}\big)^2$ which does not contain singular solution is
A) $x^2(y-xc)=yc^2$
B) $y=cx+c^2$
C) $y^2=cx^2+c^2$
D) $xy=cx^2+c$
My Attempt:
I put $\frac{dy}{dx}=p$ and tried differentiating w.r.t $x$, but couldn't conclude.
I also tried to rearrange the equation but couldn't form the Clairaut's equation.
Best Answer
D'Alembert's differential equation $$x^2\Big(y-x\frac{dy}{dx}\Big)=y\Big(\frac{dy}{dx}\Big)^2$$ Substitute $u=x^2$ $$y-2u\frac{dy}{du}=4y\Big(\frac{dy}{du}\Big)^2$$ $$y(1-4y'^2)=2uy'$$ $$y=u\dfrac {2y'}{(1-4y'^2)}$$ This is D'Alembert's differential equation $$y=uf(y')+g(y')$$ With $g=0$
Clairaut's differential equation
I also tried to rearrange the equation but couldn't form the Clairaut's equation.
$$x^2\Big(y-x\frac{dy}{dx}\Big)=y\Big(\frac{dy}{dx}\Big)^2$$ Multiply by $y$ $$x^2\Big(y^2-xy\frac{dy}{dx}\Big)=y^2\Big(\frac{dy}{dx}\Big)^2$$ Substitute $w=y^2$ $$w-\dfrac x2w'=\dfrac 1 {4x^2}(w')^2$$ Substitute $u=x^2$ $$w-uw'=(w')^2$$ This is Clairaut's differential equation: $$w=uw'+w'^2$$ The general solution is: $$w=uC+C^2$$ $$\boxed {y^2=x^2C+C^2}$$