The solution of the differential equation $\log_e \left(\frac{dy}{dx}\right)=5x+2y$ is given by

$2e^{5x}+5e^{-2y}+C=0$: C is an arbitrary constant

The correct answer is Option (2) → $2e^{5x}+5e^{-2y}+C=0$: C is an arbitrary constant

Given: $\log_e\left(\frac{dy}{dx}\right) = 5x + 2y$

Take exponential both sides:

$\frac{dy}{dx} = e^{5x + 2y} = e^{5x} \cdot e^{2y}$

Separate variables:

$\frac{dy}{e^{2y}} = e^{5x} dx$

Integrate both sides:

$\int e^{-2y} dy = \int e^{5x} dx$

Left side: $\int e^{-2y} dy = \frac{-1}{2} e^{-2y}$

Right side: $\int e^{5x} dx = \frac{1}{5} e^{5x}$

So:

$\frac{-1}{2} e^{-2y} = \frac{1}{5} e^{5x} + C$

Multiply both sides by 10:

$-5e^{-2y} = 2e^{5x} + 10C$

Bring all terms to one side:

$2e^{5x} + 5e^{-2y} + C = 0$ (absorbing $10C$ into arbitrary constant $C$)