The particular solution of the differential equation $\log(\frac{dy}{dx})=3x+4y$ satisfying $y = 0$ when $x = 0$ is:
Answer & explanation
Correct answer: option 3
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: $4e^{3x}+3e^{-4y}-7=0$