Solve the following differential equation: $x^2 dy + y(x + y) dx = 0$

$x^2 y = C(y + 2x)$

The correct answer is Option (1) → $x^2 y = C^2(y + 2x)$ ##

We have, $x^2 dy + y(x + y) dx = 0$

$\frac{x^2 dy}{y dx} + (x + y) = 0$

$\frac{x^2 dy}{y^2 dx} \cdot \frac{x}{y} + \frac{x}{y} + 1 = 0 \quad \dots(i)$

Put $y = vx ⇒\frac{dy}{dx} = v + x \frac{dv}{dx}$

From eq (i), we get:

$\frac{1}{v^2} \left( v + x \frac{dv}{dx} \right) + \frac{1}{v} + 1 = 0$

$\Rightarrow x \frac{dv}{dx} = -(2v + v^2)$

$\Rightarrow \frac{1}{2} \left[ \frac{1}{v} - \frac{1}{v + 2} \right] dv = -\frac{dx}{x}$

On integrating both sides, we get:

$\Rightarrow \frac{1}{2} (\log v - \log(v + 2)) = -\log x + C_1$

$\frac{1}{2} \log \left( \frac{v}{v + 2} \right) = -\log x + C_1$

$\Rightarrow \log \left( \frac{v}{v + 2} \right) = -2 \log x + \log k$

where $k = e^{2C_1}$

$\Rightarrow \log \left( \frac{v}{v + 2} \right) = \log \left( \frac{k}{x^2} \right)$

$\Rightarrow \frac{v}{v + 2} = \frac{k}{x^2}$

Substituting $v = \frac{y}{x}$, we get:

$\frac{y}{2x + y} = \frac{k}{x^2}$

$\Rightarrow x^2 y = C^2(2x + y) \text{ where } k = C^2$