Question

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

• A. $b{e}^{ax}+a{e}^{-by}=c$
• B. $b{e}^{ax}-a{e}^{-hy}=c$
• C. $b{e}^{-ax}-a{e}^{-by}=c$
• D. $b{e}^{ax}+a{e}^{by}=c$

Answer 1

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

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

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

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

$e^{-by}dy=e^{ax}dx$

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

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