The solution of the differential equation \(\log_e\left(\displaystyle\frac{dy}{dx}\right)=3x+4y\) is given by
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4e^{3x}+3e^{-4y}+ C = 0\), where C is constant of integration
Removing the logarithm gives an equation in variables separable form:
\( \displaystyle \frac{dy}{dx} = e^{3x + 4y} = e^{3x}\,e^{4y} \)
\(e^{-4y}\,dy = e^{3x}\,dx\)
Integrating both sides:
\( \displaystyle \int e^{-4y}\,dy = \int e^{3x}\,dx \)
\( \displaystyle -\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} + C_1 \)
Multiplying by 12 and writing \(C = 12C_1\):
\(-3e^{-4y} = 4e^{3x} + 12C_1\)
\(4e^{3x} + 3e^{-4y} + C = 0\)
NCERT (2026-27 edition):
Differential Equations
