Question

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

• A. $\frac{e^{by}}{b}=\frac{e^{ax}}{a}+c$
• B. $\frac{e^{-by}}{-b}=\frac{e^{ax}}{a}+c$
  
• C. $\frac{e^{-by}}{a}=\frac{e^{ax}}{a}+c$
• D. None of these

Answer 1

The correct option is B $\frac{e^{-by}}{-b}=\frac{e^{ax}}{a}+c$


$\frac{dy}{dx}=e^{ax+by}=e^{ax}{e}^{by}$
or $e^{-by}dy=e^{ax}dx$
$\therefore -\frac{1}{b}{e}^{-by}=\frac{1}{a}{e}^{ax}+c$