Step-by-step Solution:

1. Identify the type of differential equation. The given differential equation is: $y'\left( x \right) + y\left( x \right)g'\left( x \right) = g\left( x \right)g'\left( x \right)$ This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where:

   • $P(x) = g'(x)$
   • $Q(x) = g(x)g'(x)$

2. Find the integrating factor (IF). The integrating factor is given by the formula $I.F. = e^{\int P(x)dx}$. $I.F. = e^{\int g'(x)dx} = e^{g(x)}$ (We can omit the constant of integration here.)

3. Solve the differential equation. Multiply the entire differential equation by the integrating factor $e^{g(x)}$: $e^{g(x)}y'(x) + y(x)g'(x)e^{g(x)} = g(x)g'(x)e^{g(x)}$ The left-hand side (LHS) of the equation is the derivative of the product of $y(x)$ and the integrating factor: $\frac{d}{dx} \left[ y(x) e^{g(x)} \right] = g(x)g'(x)e^{g(x)}$

4. Integrate both sides with respect to x. $\int \frac{d}{dx} \left[ y(x) e^{g(x)} \right] dx = \int g(x)g'(x)e^{g(x)} dx$ $y(x) e^{g(x)} = \int g(x)g'(x)e^{g(x)} dx + C$ where C is the constant of integration.

5. Evaluate the integral on the right-hand side (RHS). Let's solve the integral $I = \int g(x)g'(x)e^{g(x)} dx$. Use the substitution method. Let $t = g(x)$. Then $dt = g'(x)dx$. $I = \int t e^t dt$ Now, use integration by parts: $\int u dv = uv - \int v du$. Let $u = t$ and $dv = e^t dt$. Then $du = dt$ and $v = e^t$. $I = t e^t - \int e^t dt = t e^t - e^t = (t-1)e^t$ Substitute back $t = g(x)$: $I = (g(x)-1)e^{g(x)}$

6. Write the general solution. Substituting the result of the integral back into the equation from Step 4: $y(x) e^{g(x)} = (g(x)-1)e^{g(x)} + C$ Divide by $e^{g(x)}$ to isolate $y(x)$: $y(x) = g(x) - 1 + C e^{-g(x)}$

7. Use the initial condition to find the constant C. We are given the initial condition $y(0) = 0$ and also that $g(0) = 0$. Substitute $x=0$ into the general solution: $y(0) = g(0) - 1 + C e^{-g(0)}$ $0 = 0 - 1 + C e^{-0}$ $0 = -1 + C(1)$ $C = 1$

8. Write the particular solution. With $C=1$, the particular solution is: $y(x) = g(x) - 1 + e^{-g(x)}$

9. Calculate the value of y(2). We need to find $y(2)$. We are given that $g(2) = 0$. Substitute $x=2$ into the particular solution: $y(2) = g(2) - 1 + e^{-g(2)}$ $y(2) = 0 - 1 + e^{-0}$ $y(2) = -1 + 1$ $y(2) = 0$

Alternative Method (Using Definite Integration)

Starting from Step 3: $\frac{d}{dx} \left[ y(x) e^{g(x)} \right] = g(x)g'(x)e^{g(x)}$ Integrate both sides from $x=0$ to $x=2$: $\int_{0}^{2} \frac{d}{dx} \left[ y(x) e^{g(x)} \right] dx = \int_{0}^{2} g(x)g'(x)e^{g(x)} dx$ By the Fundamental Theorem of Calculus, the LHS is: $\left[ y(x) e^{g(x)} \right]_0^2 = y(2)e^{g(2)} - y(0)e^{g(0)}$ Given $y(0)=0$, $g(0)=0$, and $g(2)=0$: $LHS = y(2)e^{0} - 0 \cdot e^{0} = y(2)$ For the RHS, let $t = g(x)$, so $dt = g'(x)dx$. The limits of integration change from $x=0, 2$ to $t=g(0)=0, g(2)=0$. $RHS = \int_{g(0)}^{g(2)} t e^t dt = \int_{0}^{0} t e^t dt = 0$ Equating LHS and RHS: $y(2) = 0$

The final answer is 0.