General solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=a x+b y\) is

Given differential equation is
\(\begin{array}{ll}
& \log \left(\frac{d y}{d x}\right)=a x+b y \\
\therefore \quad & \frac{d y}{d x}=e^{a x+b y} \\
\therefore \quad & \frac{d y}{d x}=e^{a x} \cdot e^{b y}
\end{array}\)
\(\begin{aligned} & \therefore \quad \frac{\mathrm{d} y}{\mathrm{e}^{\mathrm{by}}}=\mathrm{e}^{\mathrm{ax}} \cdot \mathrm{d} x \\ & \mathrm{e}^{-\mathrm{by}} \mathrm{d} y-\mathrm{e}^{\mathrm{ax}} \mathrm{d} x=0 \\ & \text { Integrating both sides, we get } \\ & \int \mathrm{e}^{-\mathrm{by}} \mathrm{d} y-\int \mathrm{e}^{\mathrm{ax}} \mathrm{d} x=0 \\ & \frac{e^{-b y}}{-b}-\frac{e^{a x}}{a}+c=0 \\ & \text { i.e., } \frac{e^{-b y}}{b}+\frac{e^{x x}}{a}=c \\ & a e^{-b y}+b e^{a x}=a b c \\ & a e^{-b y}+b e^{a x}=c_1 \text {, where } c_1=a b c \\ & \end{aligned}\)