Question

In each of the exercises $1$ to $3$, form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b$
1. $\frac{x}{a}+\frac{y}{b}=1$
2. $y^2=a(b^2-x^2)$
3. $y=a{e}^{3x}+b{e}^{2x}$

Answer 1

Given,

1.

$\frac{x}{a}+\frac{y}{b}=1$

On differentiating on both sides, we get,

$\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{b}{a}$

Again differentaiting on both sides, we get,

$\frac{d^{2}y}{d{x}^2}=0$

Hence, this is the required equation.

2.

$y^2=a(b^2-x^2)$

$y^2=a{b}^2-a{x}^2$

On differentiating on both sides, we get,

$2y\frac{dy}{dx}=0-2ax$

$y\frac{dy}{dx}=-ax$

$\frac{dy}{dx}=-a\frac{x}{y}$

Again differentiating on both sides, we get

$\frac{d^{2}y}{d{x}^2}=-a\frac{y-x\frac{dy}{dx}}{y^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{ax}{y^2}\frac{dy}{dx}-\frac{a}{y}$

$\frac{d^{2}y}{d{x}^2}=\frac{x}{b^2-x^2}\frac{dy}{dx}-\frac{a}{y}$

Hence, this is the required equation.

3.

$y=a{e}^{3x}+b{e}^{2x}$

On differentiating on both sides, we get,

$\frac{dy}{dx}=3a{e}^{3x}+2b{e}^{2x}$

Again differentiating on both sides, we get,

$\frac{d^{2}y}{d{x}^2}=9a{e}^{3x}+4b{e}^{2x}$

$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}=9a{e}^{3x}+4be2x-3a{e}^{3x}-2b{e}^{2x}$

$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}=6a{e}^{3x}+2b{e}^{2x}$

Hence, this is the required equation.