The solution of the differential equation $\log_e(\frac{dy}{dx})=3x+4y$ is given by

$4e^{3x}+3e^{-4y}+ C = 0$, where C is constant of integration

The correct answer is Option (1) → $4e^{3x}+3e^{-4y}+ C = 0$, where C is constant of integration

Given:

$\log\left(\frac{dy}{dx}\right)=3x+4y$

Exponentiate both sides:

$\frac{dy}{dx}=e^{3x+4y}=e^{3x}e^{4y}$

Separate variables:

$e^{-4y}\,dy=e^{3x}\,dx$

Integrate:

$\int e^{-4y}\,dy=\int e^{3x}\,dx$

$-\frac{1}{4}e^{-4y}=\frac{1}{3}e^{3x}+C$

Multiply by 12 to simplify:

$-3e^{-4y}=4e^{3x}+12C$

Rewrite constant: $12C = C'$

$4e^{3x}+3e^{-4y}+C'=0$

Final answer: $4e^{3x}+3e^{-4y}+C=0$