Value of Differential Equation Solution at Given Point

The given differential equation is:
$y'(x) + y(x)g'(x) = g(x)g'(x)$

This is a first-order linear differential equation of the form:
$\frac{dy}{dx} + P(x) y = Q(x)$
where $P(x) = g'(x)$ and $Q(x) = g(x)g'(x)$.

Step 1: Find the Integrating Factor ($\text{I.F.}$)
$\text{I.F.} = e^{\int P(x) dx} = e^{\int g'(x) dx} = e^{g(x)}$

Step 2: Solve the differential equation
Multiplying the differential equation by the integrating factor gives:
$\frac{d}{dx} \left[ y(x) e^{g(x)} \right] = g(x) g'(x) e^{g(x)}$

Integrating both sides with respect to $x$:
$y(x) e^{g(x)} = \int g(x) g'(x) e^{g(x)} dx$

To compute the integral on the right-hand side, substitute $u = g(x)$, which gives $du = g'(x) dx$:
$\int u e^u du = u e^u - e^u + C = (g(x) - 1)e^{g(x)} + C$

Therefore, the general solution is:
$y(x) e^{g(x)} = (g(x) - 1)e^{g(x)} + C$

Dividing both sides by $e^{g(x)}$:
$y(x) = g(x) - 1 + C e^{-g(x)}$

Step 3: Determine the constant $C$
We are given $y(0) = 0$ and $g(0) = 0$. Substituting $x = 0$ into the general solution:
$y(0) = g(0) - 1 + C e^{-g(0)}$
$0 = 0 - 1 + C e^0$
$0 = -1 + C \implies C = 1$

Thus, the particular solution is:
$y(x) = g(x) - 1 + e^{-g(x)}$

Step 4: Calculate $y(2)$
We are given $g(2) = 0$. Substituting $x = 2$:
$y(2) = g(2) - 1 + e^{-g(2)}$
$y(2) = 0 - 1 + e^0$
$y(2) = -1 + 1 = 0$

Hence, the value of $y(2)$ is $0$.