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If $y^2=a x^2+b x+c$, where $a, b, c$ are constants, then $y^3 \frac{d^2 y}{d x^2}$ is equal to
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Verified Answer
The correct answer is:
function of $x$
$$
y^2=a x^2+b x+c
$$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& 2 y \frac{d y}{d x}=2 a x+b \quad \Rightarrow 2 y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2=2 a \\
& \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=a \\
& \therefore y^3 \frac{d^2 y}{d x^2}=\left(a x^2+b x+c\right)\left[\left(\frac{2 a x+b}{2}\right)^2-a\right]
\end{aligned}
$$
R.H.S. of eq. (1) is a function of ' $x$ ' only.
y^2=a x^2+b x+c
$$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& 2 y \frac{d y}{d x}=2 a x+b \quad \Rightarrow 2 y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2=2 a \\
& \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=a \\
& \therefore y^3 \frac{d^2 y}{d x^2}=\left(a x^2+b x+c\right)\left[\left(\frac{2 a x+b}{2}\right)^2-a\right]
\end{aligned}
$$
R.H.S. of eq. (1) is a function of ' $x$ ' only.
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