If y=y(x) is the solution of the differential | vedclass

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(A) Given the differential equation: $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx} + e^x = 0$. Separating the variables,we get: $\frac{dy}{2+y} = -\fr

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(A) Given the differential equation: $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx} + e^x = 0$. Separating the variables,we get: $\frac{dy}{2+y} = -\frac{e^x}{5+e^x} dx$. Integrating both sides: $\int \frac{dy}{2+y} = -\int \frac{e^x}{5+e^x} dx$. This gives: $\log |2+y| = -\log |5+e^x| + C$. Using the initial condition $y(0)=1$: $\log |2+1| = -\log |5+e^0| + C \Rightarrow \log 3 = -\log 6 + C \Rightarrow C = \log 3 + \log 6 = \log 18$. Thus,$\log |2+y| = \log \left|\frac{18}{5+e^x}\right|$,which implies $2+y = \frac{18}{5+e^x}$. So,$y(x) = \frac{18}{5+e^x} - 2$. For $x = \log 13$,$y(\log 13) = \frac{18}{5+e^{\log 13}} - 2 = \frac{18}{5+13} - 2 = \frac{18}{18} - 2 = 1 - 2 = -1$.

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx}+e^x=0$ satisfying $y(0)=1$,then a value of $y(\log 13)$ is

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