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If $a$ and $b$ are arbitrary constants, then the differential equation having $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its general solution is
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The correct answer is:
$x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
On differentiating,
$$
\frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x}=0 \quad \Rightarrow \quad \frac{y}{x} \frac{d y}{d x}=\frac{-b^2}{a^2}
$$
Again differentiating w.r.t. $x$
$$
\begin{aligned}
& \frac{y}{x} \frac{d^2 y}{d x^2}+\frac{d y}{d x}\left[\frac{x \frac{d y}{d x}-y}{x^2}\right]=0 \\
& \Rightarrow \quad x y \frac{d^2 y}{d x^2}+\frac{d y}{d x}\left[x \frac{d y}{d x}-y\right]=0 \\
& \Rightarrow \quad x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0 .
\end{aligned}
$$
On differentiating,
$$
\frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x}=0 \quad \Rightarrow \quad \frac{y}{x} \frac{d y}{d x}=\frac{-b^2}{a^2}
$$
Again differentiating w.r.t. $x$
$$
\begin{aligned}
& \frac{y}{x} \frac{d^2 y}{d x^2}+\frac{d y}{d x}\left[\frac{x \frac{d y}{d x}-y}{x^2}\right]=0 \\
& \Rightarrow \quad x y \frac{d^2 y}{d x^2}+\frac{d y}{d x}\left[x \frac{d y}{d x}-y\right]=0 \\
& \Rightarrow \quad x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0 .
\end{aligned}
$$
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