Observe and solve.

Solve the equation fully, using an integrating factor. $\begin{array}{rcl} \dfrac{dy}{dx} + \dfrac{dg}{dx}y & = & g\dfrac{dg}{dx} \\ e^g \dfrac{dy}{dx} + e^g \dfrac{dg}{dx} y & =& ge^g \dfrac{dg}{dx} \\ \dfrac{d}{dx}\big(e^g y\big) & = & ge^g \dfrac{dg}{dx} \; = \; \dfrac{d}{dx}\big((g-1)e^g\big) \\ e^g y & = & (g-1)e^g + c \\ y & = & g - 1 + ce^{-g} \end{array}$ Since $y(0) = g(0) = 0$ , we deduce that $c=1$ , and hence $y(x) \,=\, g(x) - 1 + e^{-g(x)}$ . Since $g(2) = 0$ , we therefore deduce that $y(2) = 0$ .

for ease of solving assume y(X) and g(X) as y and x respectively.now divide by g'(X) so now whole equation changes to first order diffrential equation as shown below. (dy/dx)+y=x .on solving we get y(e^x)=(x)(e^x)-(e^x)+C. replace y(X) and g(X) as y and x respectively and put X =0 to find C..and thus the required value :-)

Consider: $y'(x) = g(x)g'(x) - y(x)g'(x)$ Then factored out $g'(x)$ $y'(x) = g'(x)(g(x) - y(x))$ Then consider when $g(x) = Cy(x)$ . This gives: $y'(x) = Cg'(x)$ This would imply that $y(x)$ differs from $g(x)$ by some constant factor. Yet, the question asks for $y(2) = Cg(2) = C \cdot 0 = 0$

Trick is to choose suitable function for $g(x)$ which satisfy initial conditions . Hence I chose quadratic polynomial ..

Let $\displaystyle{g\left( x \right) =x(x-2)}$ , hence given Differential Equation becomes ..

$\displaystyle{\cfrac { dy }{ dx } +(2(x-1))y=2x(x-1)(x-2)}$ , It is Linear Differential equation , hence it's integrating factor is

$\displaystyle{IF={ e }^{ \int { 2(x-1)dx } }={ e }^{ { (x-1) }^{ 2 } }\\ y{ e }^{ { (x-1) }^{ 2 } }=\int { 2x(x-1)(x-2){ e }^{ { (x-1) }^{ 2 } } } dx\quad +c\\ let:\\ { (x-1) }^{ 2 }=t\quad \Rightarrow 2(x-1)dx=dt\\ y{ e }^{ { (x-1) }^{ 2 } }=\int { (\sqrt { t } +1)(\sqrt { t } -1){ e }^{ t } } dt\quad +c\\ y{ e }^{ { (x-1) }^{ 2 } }=\int { (t-1{ )e }^{ t } } dt\quad +c\\ y{ e }^{ { (x-1) }^{ 2 } }=(t-2{ )e }^{ t }\quad +c\\ y=({ (x-1) }^{ 2 }-2)+c{ e }^{ { (x-1) }^{ 2 } }\\ y(0)=0\quad \Rightarrow \boxed { c={ e }^{ -1 } } \\ \boxed { y=({ (x-1) }^{ 2 }-2)+{ e }^{ { (x-1) }^{ 2 }-1 } } \\ \boxed { y(2)=0 } }$