The solution of the differential equation \(\log_e\left(\displaystyle\frac{dy}{dx}\right) = 5x + 2y\) is given by
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2e^{5x} + 5e^{-2y} + C = 0\) : C is an arbitrary constant
Removing the logarithm (taking the exponential of both sides):
\( \displaystyle \frac{dy}{dx} = e^{5x + 2y} = e^{5x}\,e^{2y} \)
The variables can be separated:
\(e^{-2y}\,dy = e^{5x}\,dx\)
Integrating both sides:
\( \displaystyle -\frac{1}{2}e^{-2y} = \frac{1}{5}e^{5x} + C_1 \)
Multiplying by 10:
\(-5e^{-2y} = 2e^{5x} + 10C_1\)
\(2e^{5x} + 5e^{-2y} + C = 0\), where \(C = 10C_1\) is again an arbitrary constant.