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

Answer 1

Answer: (C)

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