Question

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

• A. $b{e}^{ax}+a{e}^{-by}=k$
• B. $e^{ax}+e^{-by}=c$
• C. $e^{ax+by}=c$
• D. $(ax+by)=cxy$

Answer 1

The correct option is A $b{e}^{ax}+a{e}^{-by}=k$

$log\left(\frac{dy}{dx}\right)=ax+by\frac{dy}{dx}=e^{ax+by}\frac{dy}{dx}=e^{ax}.e^{by}\int e^{-by}dy=\int e^{ax}dx+c\Rightarrow \frac{e^{-by}}{-b}=\frac{e^{ax}}{a}+c$

$\Rightarrow c=\frac{e^{ax}}{a}+\frac{e^{-by}}{b}$

$\Rightarrow b{e}^{ax}+a{e}^{-by}=k$