Question

The general solution of the differential equation [2$\sqrt{xy}$ – x] dy + y dx = 0 is –

• A. log x + $\sqrt{\frac{y}{x}}$ = c
• B. log y – $\sqrt{\frac{x}{y}}$ = c
• C. log y +$\sqrt{\frac{x}{y}}$ = c
• D. None of these

Answer 1

Answer: (C)

log y +$\sqrt{\frac{x}{y}}$ = c

Answer 2

We have, $\frac{dy}{dx}$ =$\frac{y}{x - 2\sqrt{xy}}$ which is homogeneous.

Put y = Vx so that $\frac{dy}{dx}$ = x $\frac{dV}{dx}$ + V

Ž x$\frac{dV}{dx}$ = $\frac{V}{1 - 2\sqrt{V}}$– V = $\frac{2V^{3/2}}{1 - 2\sqrt{V}}$

Ž $\frac{dx}{x}$ = $\frac{1 - 2\sqrt{V}}{2V^{3/2}}$ dV = $\left( \frac{1}{2V^{3/2}} - \frac{1}{V} \right)$ dV

Integrating, we get

– c + log x = – V^–1/2 – log V = – $\sqrt{\frac{x}{y}}$ – log y + log x

Ž log y + $\sqrt{\frac{x}{y}}$ = c.

Hence (3) is the correct answer