\(\begin{aligned}
& \frac{d y}{d x}+y g^{\prime}(x)=g(x) \cdot g^{\prime}(x) \\
& \therefore P=g^{\prime}(x) \text { and } Q=g(x) \cdot g^{\prime}(x) \\
& \text {IF }=e^{\int p d x} \\
& =e^{\int g^{\prime}(x) d x} \\
& \mathrm{IF}=e^{g(x)} \\
\end{aligned}\)
General Solution is
\(\begin{aligned}
y(\mathrm{IF}) & =\int Q(\mathrm{IF}) d x+c \\
y\left(e^{g(x)}\right) & =\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+c
\end{aligned}\)
Put, \(g(x)=t\)
\(\begin{gathered}
g^{\prime}(x) d x=d t \\
y e^{g(x)}=\int t e^t d t+e^c \\
y e^{g(x)}=e^t(t-1)+e^c \\
y e^{g(x)}=e^t \cdot t-e^t+e^c \\
y e^{g(x)}=e^{g(x)} \cdot g(x)-e^{g(x)}+e^c \\
e^{g(x)}[y+1-g(x)]=e^c
\end{gathered}\)
Taking \(\log\) on both sides,
\(\begin{aligned}
\log _e e^{g(x)}[y+1-g(x)] & =\log _e e^c \\
g(x)+[y+1-g(x)] & =C
\end{aligned}\)
Hence, option (b) is correct.