Question

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

• A. $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}$
• B. $\left(x^{2}-y^{2}\right) \frac{d^{2} y}{d x^{2}}-2 x y \frac{d y}{d x}-y=0$
• C. $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$
• D. $x^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}-2 y=0$

Answer 1

Answer: (C)

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
On differentiating,
$\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x}=0 $
$\Rightarrow \frac{y}{x} \frac{d y}{d x}=\frac{-b^{2}}{a^{2}}$
Again differentiating w.r.t. $x$
$\frac{y}{x} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}\left[\frac{x \frac{d y}{d x}-y}{x^{2}}\right]=0$
$\Rightarrow x y \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}\left[x \frac{d y}{d x}-y\right]=0$
$ \Rightarrow 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$