Question

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

Answer 1

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

Let $y=mx$

$\frac{dy}{dx}=m+x\frac{dm}{dx}$

Substituting in the first equation, we get

$m+x\frac{dm}{dx}=\frac{mx(x-mx)}{x(x+mx)}$

$\Rightarrow m+x\frac{dm}{dx}=\frac{x^{2}m(1-m)}{x^2(1+m)}$

$\Rightarrow x\frac{dm}{dx}=\frac{m-m^2}{1+m}-m$

$\Rightarrow x\frac{dm}{dx}=\frac{m-m^2-m-m^2}{1+m}$

$\Rightarrow x\frac{dm}{dx}=\frac{-2m^2}{1+m}$

$\Rightarrow \left(\frac{1+m}{2m^2}\right)dm=-\frac{dx}{x}$

Integrating we get,

$\Rightarrow \frac{1}{2}\int \frac{1}{m^2}dm+\frac{1}{2}\int \frac{1}{m}dm=-\int \frac{dx}{x}$

$\Rightarrow -\frac{1}{2m}+\frac{1}{2}\ln m=-\ln x+C$

$\Rightarrow -\frac{x}{2y}+\frac{1}{2}\ln \frac{y}{x}=-\ln x+C$