The general solution of the differential equation $\log_e (\frac{dy}{dx})=ax+by$ is

$\frac{e^{ax}}{a}+\frac{e^{-by}}{b} +C=0$, where C is constant of integration

The correct answer is Option (3) → $\frac{e^{ax}}{a}+\frac{e^{-by}}{b} +C=0$, where C is constant of integration

Given

$\log_e\left(\frac{dy}{dx}\right)=ax+by$

$\frac{dy}{dx}=e^{ax+by}=e^{ax}e^{by}$

$e^{-by}dy=e^{ax}dx$

Integrate both sides

$\int e^{-by}dy=\int e^{ax}dx$

$-\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+C$

$\frac{e^{ax}}{a}+\frac{e^{-by}}{b}+C=0$

The general solution is $\frac{e^{ax}}{a}+\frac{e^{-by}}{b}+C=0$.