Question

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

• A. $y-2x=k{x}^{2}y$
• B. $y+2x=k{x}^{2}y$
• C. $2y-x=k{x}^{2}y$
• D. $2y+x=k{x}^{2}y$

Answer 1

The correct option is A $y-2x=k{x}^{2}y$

$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow x\frac{dv}{dx}+2v=v^2\Rightarrow \frac{dv}{v^2-2v}=\frac{dx}{x}$
$\Rightarrow \frac{v-(v-2)}{v(v-2)}dv=\frac{2dx}{x}$
$\Rightarrow \frac{dv}{v-2}-\frac{dv}{v}=\frac{2dx}{x}$
Integrating we get, $\log \left(\frac{v-2}{v}\right)=2\log x+\log k$
$\Rightarrow 1-\frac{2}{v}=x^{2}k\Rightarrow y-2x=k{x}^{2}y$