The general solution of the differential equation $\frac{dy}{dx}= e^{ax+by}$ is: (Here C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
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:
\[ e^{-by} + \frac{b}{a} e^{ax} = K \]
\(a{e^{-by} + be^{ax} = C}\)