$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$
$2[ \, 2y(x + 2) - 3 \, ] \, dx + (x + 2)^2 \, dy = 0$
$\dfrac{2[ \, 2y(x + 2) - 3 \, ] \, dx}{(x + 2)^2 \, dx} + \dfrac{(x + 2)^2 \, dy}{(x + 2)^2 \, dx} = 0$
$\dfrac{4y(x + 2) - 6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$
$\dfrac{4y(x + 2)}{(x + 2)^2} - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$
$\left( \dfrac{4}{x + 2} \right) y - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} + \left( \dfrac{4}{x + 2} \right) y = \dfrac{6}{(x + 2)^2}$ → linear in y
Hence,
$P = \dfrac{4}{x + 2}$
$Q = \dfrac{6}{(x + 2)^2}$
Integrating factor:
$\displaystyle e^{\int P \, dx} = e^{\int \frac{4}{x + 2} \, dx}$
$\displaystyle e^{\int P \, dx} = e^{4\int \frac{dx}{x + 2}}$
$\displaystyle e^{\int P \, dx} = e^{4\ln (x + 2)}$
$\displaystyle e^{\int P \, dx} = e^{\ln (x + 2)^4}$
$\displaystyle e^{\int P \, dx} = (x + 2)^4$
Thus,
$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx} \, dx + C$
$\displaystyle y(x + 2)^4 = \int \dfrac{6}{(x + 2)^2} \cdot (x + 2)^4 \, dx + c$
$\displaystyle y(x + 2)^4 = 6\int (x + 2)^2 \, dx + c$
$\displaystyle y(x + 2)^4 = 6 \cdot \dfrac{(x + 2)^3}{3} + c$
$\displaystyle y(x + 2)^4 = 2(x + 2)^3 + c$
Mulitply by (x + 2)-4
$\displaystyle y = 2(x + 2)^{-1} + c(x + 2)^{-4}$ answer
