TITLE: Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$?
QUESTION [5 upvotes]: here's my question
If $y^2=ax^2+bx+c$ then prove that  $y^3[d^2y/dx^2]$  is a constant .
I have solved this using the conventional method, taking square root, differentiating w.r.t to x  and using chain and quotient rule
But can't think of some alternative smaller and more efficient method ??
Can you ?
Also can someone suggest me any alternative to quotient rule
I need something more non conventional, time saving and vastly
Applicable method .

REPLY [2 votes]: Here is a proof that uses only the fact that $u = ax^2 + bx + c$ is quadratic and hence $u''' = 0$.
We know that $y^2 = u$ and hence $2 y y' = u' , \, y^4 = u^2$.
Differentiate the second identity twice:
$$
4 y^3y' = 2 u u', \quad 4 y^3 y'' + 12 y^2 (y')^2 = 2uu'' + 2 (u')^2 \, .
$$
Since $4y^2(y')^2 = (u')^2$, this implies
$$
4 y^3 y'' =  2uu'' + 2 (u')^2 - 12 y^2 (y')^2 = 2 uu'' - (u')^2 \, .
$$
Differentiate this one more time and use the assumption $u''' = 0$ to obtain
$$
\frac{d}{dx} \left( 4 y^3 \frac{d^2y}{dx^2} \right) = 2 u u''' + 2 u' u''- 2u' u'' = 0 \, . 
$$
Hence $y^3 \frac{d^2y}{dx^2} $ is constant.