Solve $2(y + 3) - xy \frac{dy}{dx} = 0$, given that $y(1) = -2$.

$x^2 (y + 3)^3 = e^{y + 2}$

The correct answer is Option (4) → $x^2 (y + 3)^3 = e^{y + 2}$ ##

Given that, $2(y + 3) - xy \frac{dy}{dx} = 0$

$\Rightarrow 2(y + 3) = xy \frac{dy}{dx}$

$\Rightarrow 2 \cdot \frac{dx}{x} = \left( \frac{y}{y + 3} \right) dy$

$\Rightarrow 2 \cdot \frac{dx}{x} = \left( \frac{y + 3 - 3}{y + 3} \right) dy$

$\Rightarrow 2 \cdot \frac{dx}{x} = \left( 1 - \frac{3}{y + 3} \right) dy$

[applying variable separable method]

On integrating both sides, we get:

$2 \log x = y - 3 \log(y + 3) + C \dots (i)$

When $x = 1$ and $y = -2$, then:

$2 \log 1 = -2 - 3 \log(-2 + 3) + C$

$\Rightarrow  2 \cdot 0 = -2 - 3 \cdot 0 + C \quad [∵\log 1 = 0]$

$\Rightarrow  C = 2$

On substituting the value of $C$ in Eq. (i), we get:

$2 \log x = y - 3 \log(y + 3) + 2$

$\Rightarrow  2 \log x + 3 \log(y + 3) = y + 2$

$\Rightarrow  \log x^2 + \log(y + 3)^3 = (y + 2) \quad [∵n \log m = \log m^n]$

$\Rightarrow  \log [x^2 (y + 3)^3] = y + 2 \quad [∵\log x + \log y = \log (xy)]$

$\Rightarrow  x^2 (y + 3)^3 = e^{y + 2}$