Q118 / 160 · TS EAMCET 2022 (18 Jul Shift 2) · Mathematics

The differential equation corresponding to the family of curves given by $a x^{2} + b y^{2} = 1$ where $a$ and $b$ are arbitrary constants is

A
$x \frac{d^{2} y}{d x^{2}} = \frac{d y}{d x}$
Correct answer
B
$x y \frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} = 0$
C
$x y \frac{d^{2} y}{d x^{2}} + y {(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} = 0$
D
$x y \frac{d^{2} y}{d x^{2}} - x {(\frac{d y}{d x})}^{2} + y \frac{d y}{d x} = 0$

Solution

$a x^{2} + b y^{2} = 1$
Differentiating, we get
$2 a x + 2 b y \frac{d y}{d x} = 0 \Rightarrow b y \frac{d y}{d x} = - a x$
Again differentiating, we get
$b y \frac{d^{2} y}{d x^{2}} + b {(\frac{d y}{d x})}^{2} = - a \Rightarrow y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = - \frac{a}{b}$
i.e. $y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = \frac{y}{x} \frac{d y}{d x} \Rightarrow x y \frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} = 0$

Hence, the correct answer is option: 'B'.

The differential equation corresponding to the family of curves given by ax2+by2=1 where a and b are arbitrary constants is

Solution

Relevant Questions

The differential equation corresponding to the family of curves given by $a x^{2} + b y^{2} = 1$ where $a$ and $b$ are arbitrary constants is