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

Given:

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

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

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

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Separating variables:

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

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

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Integrating both sides:

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

For the left side, using $\int e^{ku} \, du = \frac{e^{ku}}{k}$ with $k = -b$:

$\int e^{-by} \, dy = \frac{e^{-by}}{-b}$

$= -\frac{e^{-by}}{b}$

For the right side, with $k = a$:

$\int e^{ax} \, dx = \frac{e^{ax}}{a}$

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Combining results:

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

Multiplying through by $-ab$:

$ae^{-by} = -be^{ax} - abC'$

$ae^{-by} + be^{ax} = -abC'$

Since $-abC'$ is an arbitrary constant, replacing it with $C$:

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