Explanation

(b)
$\because\ \log{\left( \ \frac{dy}{dx} \right) = ax + by}$
$\Rightarrow \ \frac{dy}{dx} = e^{ax + by} = e^{ax}e^{by}$
$\Rightarrow e^{- by}dy = e^{ax}dx$
On integrating both sides, we get
$\int_{}^{}{e^{- by}dy =}\int_{}^{}{e^{ax}dx}$
$\Rightarrow \ \frac{e^{- by}}{- b} = \frac{e^{ax}}{a} + c$