The particular solution of the differential equation $\log(\frac{dy}{dx})=3x+4y$ satisfying $y = 0$ when $x = 0$ is:

$4e^{3x}+3e^{-4y}-7=0$

The correct answer is Option (3) → $4e^{3x}+3e^{-4y}-7=0$

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

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

\(e^{-4y} \frac{dy}{dx} = e^{3x}\)

\(\frac{d}{dx}\left(-\frac{1}{4} e^{-4y}\right) = e^{3x}\)

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

Using \(y(0)=0\): \(-\frac14 e^{0} = \frac13 e^{0} + C\)

$(-\frac14 = \frac13 + C$

$C = -\frac{7}{12})$

\(-\frac14 e^{-4y} = \frac13 e^{3x} - \frac{7}{12}\)

\(e^{-4y} = -\frac43 e^{3x} + \frac73\)

Particular solution: $4e^{3x}+3e^{-4y}-7=0$