The general solution of the differential equation $\frac{dy}{dx}= e^{ax+by}$ is: (Here C is an arbitrary constant)

$ae^{-by}+be^{ax} = C$

The correct answer is Option (3) → $ae^{-by}+be^{ax} = C$

Given differential equation:

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

Rewrite as:

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

Separate variables:

\[ \frac{dy}{e^{by}} = e^{ax} dx \]

Rewrite left side:

\[ e^{-by} dy = e^{ax} dx \]

Integrate both sides:

\[ \int e^{-by} dy = \int e^{ax} dx \]

Integrate left side:

\[ \int e^{-by} dy = -\frac{1}{b} e^{-by} + C_1 \]

Integrate right side:

\[ \int e^{ax} dx = \frac{1}{a} e^{ax} + C_2 \]

Combine constants \(C = C_2 - C_1\), so:

\[ -\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C \]

Multiply both sides by \(-b\):

\[ e^{-by} = -\frac{b}{a} e^{ax} - bC \]

Rewrite constant \(-bC\) as \(K\), arbitrary constant:

\[ e^{-by} + \frac{b}{a} e^{ax} = K \]

\(a{e^{-by} + be^{ax} = C}\)