Integration by substitution

integrate (x+2)g'(x)-2g(x)+2=0. we will get g(x)=C(x+2)^2+1, where C is a constant.---(1) by using g(0)=0 and the above equation we will get C=-1/4. now put x=2 and C=-1/4 in equation (1) we will get g(2)=-3

We can write the above as,

$\dfrac {d}{dx} g(x) - \dfrac {2}{(x+2)} g(x) = -\dfrac {2}{(x+2)}$

The integrating factor is $e^{\int -\frac{2}{(x+2)}} dx = \dfrac {1}{(x+2)^2}$

Multiply the above equation by Integrating Factor

$\dfrac {1}{(x+2)^2} \dfrac {d}{dx} g(x) - \dfrac {2}{(x+2)^3} g(x) =- \dfrac {2}{(x+2)^3}$

$\dfrac {1}{(x+2)^2} \dfrac {d}{dx} g(x) + \dfrac {d}{dx} \dfrac {1}{(x+2)^2} g(x) = -\dfrac {1}{(x+2)^3}$

This becomes,

$\dfrac {d}{dx} \begin{pmatrix} g(x)\dfrac {1}{(x+2)^2} \end{pmatrix} = -\dfrac {2}{(x+2)^3}$

Hence, $\dfrac {1}{(x+2)^2} g(x) = \displaystyle \int -\dfrac {2}{(x+2)^3} dx + C$

$\dfrac {g(x)}{(x+2)^2} = \dfrac {1}{(x+2)^2} + C$

We get, $g(x) = 1 + C(x+2)^2$

Put $g(0) = 0$ we get $C = -\dfrac 14$

Hence, our equation is ${\color{#20A900}{g(x) = 1 - \dfrac 14 (x+2)^2}}$

And $g(2) = 1 - \dfrac 14 (2+2)^2 = \boxed {-3}$