Let \(y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x) y(0)=0, x \in R\), where \(f^{\prime}(x)\) denotes \(\frac{d f(x)}{d x}\) and \(g(x)\) is a given non-constant differentiable function on \(R\) with \(g(0)=g(2)=0\). Then, the value of \(y\) (2) is...

\(\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) g^{\prime}(x)\)
\[
\mathrm{IF}=e^{\int g^{\prime}(x) d x}=e^{g(x)}
\]
\(\therefore\) Solution is

\[
y\left(e^{g(x)}\right)=\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+C
\]
Put \(g(x)=t, g^{\prime}(x) d x=d t\)
\[
\begin{aligned}
& y\left(e^{g(x)}\right)=\int t \cdot e^t d t+C \\
= & t \cdot e^t-\int 1 \cdot e^t d t+C=t \cdot e^t-e^t+C \\
& y e^{g(x)}=(g(x)-1) e^{g(x)}+C \quad \ldots(\mathrm{i})
\end{aligned}
\]
Given, \(y(0)=0, g(0)=g(2)=0\)

\(\therefore\) Eq. (i) becomes,
\[
\begin{aligned}
& \quad y(0) \cdot e^{g(0)}=(g(0)-1) \cdot e^{g(0)}+C \\
& \Rightarrow \quad 0=(-1) \cdot 1+C \Rightarrow C=1 \\
& \therefore \quad y(x) \cdot e^{g(x)}=(g(x)-1) e^{g(x)}+1 \\
& \Rightarrow \quad y(2) \cdot e^{g(2)}=(g(2)-1) e^{g(2)}+1 \\
& \text { where, } \quad g(2)=0 \\
& \Rightarrow \quad y(2) \cdot 1=(-1) \cdot 1+1 \Rightarrow y(2)=0
\end{aligned}
\]