Dear student We have,logdydx=ax+by⇒dydx=eax+by⇒dydx=eaxeby⇒dy=eaxeby dx⇒e-by dy=eaxdxIntegrating both sides, we gete-by-b=eaxa+C Regards

ANSWER ASAP

Dear student

We have, \log (\frac{dy}{dx}) = ax + by \Rightarrow \frac{dy}{dx} = e^{ax + by} \Rightarrow \frac{dy}{dx} = e^{ax} e^{by} \Rightarrow dy = e^{ax} e^{by} dx \Rightarrow e^{- by} dy = e^{ax} dx Integrating both sides, we get \frac{e^{- by}}{- b} = \frac{e^{ax}}{a} + C

Regards