The general solution of the differential equa | vedclass

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(A) Given the differential equation: $\log_{e}\left(\frac{dy}{dx}\right) = x + y$. By the definition of logarithm, we can write this as: $\frac{dy}{dx

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the differential equation: $\log_{e}\left(\frac{dy}{dx}\right) = x + y$. By the definition of logarithm, we can write this as: $\frac{dy}{dx} = e^{x+y}$. Using the property of exponents, we have: $\frac{dy}{dx} = e^x \cdot e^y$. Separating the variables, we get: $e^{-y} dy = e^x dx$. Integrating both sides: $\int e^{-y} dy = \int e^x dx$. This yields: $-e^{-y} = e^x + C_1$. Rearranging the terms, we get: $e^x + e^{-y} = -C_1$. Letting $-C_1 = C$, the general solution is: $e^x + e^{-y} = C$.

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

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