The solution of log left( frac{dy}{dx} right) | vedclass

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(B) Given the differential equation: $\log \left( \frac{dy}{dx} \right) = ax + by$ Taking the exponential of both sides,we get: $\frac{dy}{dx} = e^{ax

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

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(B) Given the differential equation: $\log \left( \frac{dy}{dx} \right) = ax + by$ Taking the exponential of both sides,we get: $\frac{dy}{dx} = e^{ax + by}$ Using the property of exponents $e^{m+n} = e^m \cdot e^n$,we can write: $\frac{dy}{dx} = e^{ax} \cdot e^{by}$ Separating the variables $x$ and $y$: $e^{-by} \, dy = e^{ax} \, dx$ Integrating both sides: $\int e^{-by} \, dy = \int e^{ax} \, dx$ Performing the integration: $\frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + c$ Thus,the solution is $\frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + c$.

The solution of $\log \left( \frac{dy}{dx} \right) = ax + by$ is

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