Question

Differential equation whose general solution is $y = c_{1}x + \frac{c_{2}}{x}$ for all values of $c_{1}$ and $c_{2}$ is

• A. $\frac{d^{2}y}{dx^{2}} + \frac{x^{2}}{y} + \frac{dy}{dx} = 0$
• B. $\frac{d^{2}y}{dx^{2}} + \frac{y}{x^{2}} - \frac{dy}{dx} = 0$
• C. $\frac{d^{2}y}{dx^{2}} - \frac{1}{2x}\frac{dy}{dx} = 0$
• D. $\frac{d^{2}y}{dx^{2}} + \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^{2}} = 0$

Answer 1

Answer: (D)

$\frac{d^{2}y}{dx^{2}} + \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^{2}} = 0$

Answer 2

$y = c_{1}x + \frac{c_{2}}{x}$ .....(i)

There are two arbitrary constants. To eliminate these constants, we need to differentiate (i) twice.

Differentiating (i) with respect to x,

$\frac{dy}{dx} = c_{1} - \frac{c_{2}}{x^{2}}$ .....(ii)

Again differentiating with respect to x,

$\frac{d^{2}y}{dx^{2}} = \frac{2c_{2}}{x^{3}}$ ......(iii)

From (iii), $c_{2} = \frac{x^{3}}{2}\frac{d^{2}y}{dx^{2}}$ and from (ii), $c_{1} = \frac{dy}{dx} + \frac{c_{2}}{x^{2}}$;

∴ $c_{1} = \frac{dy}{dx} + \frac{x}{2}\frac{d^{2}y}{dx^{2}}$

From (i), $y = \left( \frac{dy}{dx} + \frac{x}{2} \cdot \frac{d^{2}y}{dx^{2}} \right)x + \frac{x^{2}}{2} \cdot \frac{d^{2}y}{dx^{2}}$

⇒ $y = x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx}$

∴ $\frac{d^{2}y}{dx^{2}} + \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^{2}} = 0$