The solution curve of the differential equation \(\log_e\left(\displaystyle\frac{dy}{dx}\right)=3x+4y\), passing through the origin is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4e^{3x}+3e^{-4y}-7=0\)
Removing the logarithm:
\( \displaystyle \frac{dy}{dx} = e^{3x+4y} = e^{3x}\,e^{4y} \)
The variables separate:
\( \displaystyle \int e^{-4y}\,dy = \int e^{3x}\,dx \)
\( \displaystyle -\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} + C \)
The curve passes through the origin, so put x = 0, y = 0:
\( \displaystyle -\frac{1}{4} = \frac{1}{3} + C \)
\( \displaystyle C = -\frac{7}{12} \)
Hence
\( \displaystyle -\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} - \frac{7}{12} \)
Multiplying by −12:
\(3e^{-4y} = -4e^{3x} + 7\)
\(4e^{3x}+3e^{-4y}-7=0\)
NCERT (2026-27 edition):
Differential Equations
