Question

The solution of the differential equation $x^{2}dy+y(x+y)dx=0$ is:
(where $c$ is integration constant)

• A. $\left|\frac{xy}{2x+y}\right|=c(c>0)$
• B. $\left|\frac{x^{2}y}{2x+y}\right|=c(c>0)$
• C. $\left|\frac{x^{3}y}{2x-y}\right|=c(c>0)$
• D. $\left|\frac{x^2{y}^2}{2x+y}\right|=c(c>0)$

Answer 1

The correct option is B $\left|\frac{x^{2}y}{2x+y}\right|=c(c>0)$

$x^{2}dy+y(x+y)dx=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y(x+y)}{x^2}$
$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}-\frac{y^2}{x^2}$
Putting $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Given equation transforms to
$v+x\frac{dv}{dx}=-v-v^2$
$\Rightarrow \int \frac{1}{v^2+2v}dv=-\int \frac{dx}{x}$
$\Rightarrow \frac{1}{2}\int [\frac{1}{v}-\frac{1}{v+2}]dv=-\int \frac{dx}{x}$
$\Rightarrow \ln \left|v\right|-\ln |v+2|=-2\ln \left|x\right|+\ln c(c>0)$
$\Rightarrow \left|\frac{v{x}^2}{v+2}\right|=c$
$\Rightarrow \left|\frac{x^{2}y}{2x+y}\right|=c(c>0)$