The differential equation $\frac{dy}{dx} +\frac{y^2}{x^2} =\frac{y}{x}$ has the solution
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : $x=y (log x + C)$
We have,
$\frac{dy}{dx} +\frac{y^2}{x^2}=\frac{y}{x}$
Putting $y = vx $ and $\frac{dy}{dx} = v + x\frac{dv}{dx}, $ we get
$v + x\frac{dv}{dx} + v^2 = v $
$⇒x\frac{dv}{dx} = -v^2$
$⇒-\frac{1}{v^2}dv=\frac{1}{x}dx$
On integrating, we get
$\frac{1}{v}= log x + C $
$⇒x = y (log x + C)$