If 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) is equal to 1,  then the value of $y\left({\log }_{e}13\right)$ is

The given differential equation is $\frac{dy}{2+y}+\frac{e^x}{5+e^x}dx=0$

 Integrating both sides we get       $\log \left(y+2\right)+\log \left(5+e^x\right)=c$

                                          $\Rightarrow$$\left(y+2\right)\left(5+e^x\right)=C$

plug in y(0) =1 , we get $C=18$

Therefore, the curve is $\left(y+2\right)\left(5+e^x\right)=18$

 $put x=\ln 13$

$(y+2)(5+e^{{\log }_e 13})=18$

$$\begin{gathered} \left(y+2 \right) 18=18 \\ \Rightarrow y+2=1\Rightarrow y=-1 \end{gathered}$$