The general solution of the differential equa | vedclass

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(B) Given the differential equation: $\log \left( \frac{dy}{dx} \right) = x + y$By the definition of logarithm,we can write: $\frac{dy}{dx} = e^{x+y}$

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${e^x} + {e^{-y}} = c$

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(B) Given the differential equation: $\log \left( \frac{dy}{dx} \right) = x + y$By the definition of logarithm,we can write: $\frac{dy}{dx} = e^{x+y}$Using the property of exponents,this becomes: $\frac{dy}{dx} = e^x \cdot e^y$Now,separate the variables: $\frac{dy}{e^y} = e^x \, dx$Which is equivalent to: $e^{-y} \, dy = e^x \, dx$Integrating both sides: $\int e^{-y} \, dy = \int e^x \, dx$$-e^{-y} = e^x + C$Rearranging the terms,we get: $e^x + e^{-y} = -C$Since $-C$ is an arbitrary constant,we can write it as $c$: $e^x + e^{-y} = c$.

The general solution of the differential equation $\log \left( \frac{dy}{dx} \right) = x + y$ is

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