Solve the following differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x}, x > 0$

$y + \sqrt{x^2 + y^2} = Cx^2$

The correct answer is Option (1) → $y + \sqrt{x^2 + y^2} = Cx^2$ ##

Given differential equation is

$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x}, x > 0 \quad \dots(i)$

$\Rightarrow \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = f\left(\frac{y}{x}\right)$

Hence, homogeneous. Put $y = vx ⇒\frac{dy}{dx} = v + x \frac{dv}{dx} \quad \dots(ii)$

On comparing (i) & (ii), the differential equation becomes:

$v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}$

$\text{or } \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$

On integrating, we get:

$\log |v + \sqrt{1 + v^2}| = \log |x| + \log c$

$\Rightarrow \log |v + \sqrt{1 + v^2}| = \log |xc|$

$\Rightarrow v + \sqrt{1 + v^2} = \pm cx$

$\Rightarrow \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = \pm cx$

$\Rightarrow y + \sqrt{x^2 + y^2} = cx^2$

which is the required solution.