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

$\frac{5+e^{x}}{2+y} \cdot \frac{d y}{d x}=-e^{x}$

$\int \frac{d y}{2+y}=-\int \frac{e^{x}}{5+e^{x}} d x$

$\Rightarrow \log _{e}|2+y| \cdot \log _{e}\left|5+e^{x}\right|=\log _{e} C$

$\Rightarrow\left|(2+y)\left(5+e^{x}\right)\right|=C \quad \because y(0)=1$

$C=18$

$\therefore(2+y) \cdot\left(5+e^{x}\right)=18$

When $x=\log _{e} 13$ then $(2+y) \cdot 18=18$

$\Rightarrow 2+y=\pm 1$

$\therefore y=-1,-3$

$\therefore y(\ln 13)=-1$