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

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

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

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

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

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

On integrating both sides,

we get
$2\log x = y - 3\log (y + 3) + C$
……(i)

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

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

$\Rightarrow$ $2 \cdot 0 = - 2 - 3 \cdot 0 + C$

$\Rightarrow$ $C = 2$

On substituting the value of $C$ in Eq.

(i), we get $2\log x = y - 3\log (y + 3) + 2$

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

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

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

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