Solution of the differential equation $log(\frac{dy}{dx})=3x+4y $ given that $y=0$ when $x=0 $ is :
Answer & explanation
Correct answer: option 3
$\log(\frac{dy}{dx})=3x+4y⇒\frac{dy}{dx}=e^{3x}e^{4y}$
so $\int e^{-4y}dy=\int e^{3x}dx$
$\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+c$
so $4e^{3x}+3e^{-4y}+c'=0$