Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$? 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 .
 A: Differentiating both sides once, we have
$$
2y\frac{\mathrm{d}y}{\mathrm{d}x}=2ax+b,
$$
and differentiating twice, we reach
$$
2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+2y\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=2a.
$$
Substituting, we find
$$
\left(\frac{2ax+b}{2y}\right)^2+y\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=a
$$
$$
(2ax+b)^2+4y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=4ay^2
$$
$$
4y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=4a(ax^2+bx+c)-(2ax+b)^2
$$
$$
4y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=(4a^2x^2+4abx+4ac)-(4a^2x^2+4abx+b^2)
$$
$$
y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=\frac{4ac-b^2}{4}.
$$
We still use the chain rule but not the quotient rule.
A: Using implicit derivation:
$$
(y^2)^\prime = 2yy^\prime = 2ax+b \Rightarrow y^\prime=\frac{2ax+b}{2y}
$$
$$
(y^2)^{\prime\prime} = 2(y^\prime)^2+2yy^{\prime\prime} = 2a \Rightarrow y^{\prime\prime}=\frac{a-(y^\prime)^2}{y}
$$
Thus
\begin{align}
y^3y^{\prime\prime}=y^2(a-(y^\prime)^2) &= y^2(a-\left(\frac{2ax+b}{2y}\right)^2) \\ 
&= ay^2-\frac{1}{4}(2ax+b)^2 \\
&= (a^2x^2+abx+ac)-\frac{1}{4}(4a^2x^2+4abx+b^2) \\
&=  (a^2x^2+abx+ac)-(a^2x^2+abx+\frac{b^2}{4}) \\
&= ac - \frac{b^2}{4} \ \ \ \text{(which is constant)}
\end{align}
A: Using differentials, it is easy to see that
$$
2y dy = (2ax+ b) dx \tag{1}
$$
from which we can write
$$
u = \frac{dy}{dx} = \frac{2ax+b}{2y}
$$
Using again differentials and product rule,
\begin{eqnarray}
du 
&=& \frac{2a}{2y} dx - \frac{2ax+b}{2y^2} dy 
= \left[ \frac{a}{y} - \frac{(2ax+b)^2}{4y^3} \right] dx 
\end{eqnarray}
where we have used (1).
Finally
\begin{eqnarray}
y^3\frac{du}{dx} 
= y^3\frac{d^2y}{dx^2} 
&=& 
\left[ ay^2 - \frac{(2ax+b)^2}{4} \right]
= 
\left[ a(ax^2+bx+c) - \frac{(2ax+b)^2}{4} \right] 
=
ac-\frac{b^2}{4}
\end{eqnarray}
A: 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.
