The general solution of the differential equation $\log_e\left(\frac{dy}{dx}\right) = ax + by$ is

Given: $\log_e\left(\frac{dy}{dx}\right) = ax + by$

---

Taking exponential (base $e$) on both sides:

$e^{\log_e\left(\frac{dy}{dx}\right)} = e^{ax + by}$

$\frac{dy}{dx} = e^{ax + by}$

---

Using $e^{A+B} = e^A \cdot e^B$:

$\frac{dy}{dx} = e^{ax} \cdot e^{by}$

---

Separating variables by dividing both sides by $e^{by}$:

$\frac{1}{e^{by}} \cdot dy = e^{ax} \cdot dx$

$e^{-by} \, dy = e^{ax} \, dx$

---

Integrating both sides:

$\int e^{-by} \, dy = \int e^{ax} \, dx$

Left side: $\int e^{-by} dy = \frac{e^{-by}}{-b} = -\frac{e^{-by}}{b}$

Right side: $\int e^{ax} dx = \frac{e^{ax}}{a}$

Therefore:

$-\frac{e^{-by}}{b} = \frac{e^{ax}}{a} + C_1$

---

Rearranging:

$\frac{e^{ax}}{a} + \frac{e^{-by}}{b} + C_1 = 0$

Replacing $C_1$ with $C$:

$\frac{e^{ax}}{a} + \frac{e^{-by}}{b} + C = 0$

where $C$ is the constant of integration.