The solution of the differential equation $log \left(\frac{dy}{dx}\right)=ax+by $ is :

$\frac{-1}{b}e^{-by}=\frac{1}{a}e^{ax}+C$ Where C is an arbitrary constant

Given: $\ln\!\left(\frac{dy}{dx}\right)=ax+by$

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

Separate variables: $e^{-by}\,dy=e^{ax}\,dx$

Integrate both sides:

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

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