Solution of the differential equation \(\log \left(\frac{dy}{dx}\right)=ax+by\) is
Answer & explanation
Correct answer: option 1
\(\frac{dy}{dx}=e^{ax}e^{ay}⇒\int e^{-by}dy=\int e^{ax}dx\)
$=-\frac{e^{-by}}{b}=\frac{e^{ax}}{a}+C$
so $be^{ax}+ae^{-by}=c'$