Question

Solve$\frac{dy}{dx}=\frac{y(x-y)}{x(x+y)}$

• A. $kxy=-e^{x/y}$
• B. $kxy=e^{y/x}$
• C. $kxy=e^{x/y}$
• D. $kxy=e^{-y/x}$

Answer 1

The correct option is C $kxy=e^{x/y}$

$\frac{dy}{dx}=\frac{y(x-y)}{x(x+y)}$
$\Rightarrow \frac{dy}{dx}=\frac{yx-y^2}{x^2+xy}$ ....(1)
which is a homogeneous differential eqn
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, eqn (1) becomes
$v+x\frac{dv}{dx}=\frac{v-v^2}{1+v}$
$\Rightarrow x\frac{dv}{dx}=-\frac{2v^2}{1+v}$
$\Rightarrow \left(\frac{1+v}{v^2}\right)dv=-2\frac{1}{x}dx$
Integrating both sides,
$\int (\frac{1}{v^2}+\frac{1}{v})dv=-2\log x-\log k$
$\Rightarrow -\frac{1}{v}+\log v=-2\log x-\log k$
$\Rightarrow -\frac{x}{y}+\log \left(\frac{y}{x}\right)+2\log x+\log k=0$
$\Rightarrow \log kxy=\frac{x}{y}$
$\Rightarrow kxy=e^{x/y}$