1. Given differential equation

$\frac{5+e^x}{2+y}\frac{dy}{dx}+e^x=0$

with initial condition

$y(0)=1.$

We need to find $y(\ln 13)$.

2. Rewrite the equation

Multiply both sides by $(2+y)$:

$(5+e^x)\frac{dy}{dx}+e^x(2+y)=0$

So,

$\frac{dy}{dx}=-\frac{e^x(2+y)}{5+e^x}.$

Now separate variables:

$\frac{dy}{2+y}=-\frac{e^x}{5+e^x}\,dx.$

3. Integrate both sides

$\int \frac{dy}{2+y}= -\int \frac{e^x}{5+e^x}\,dx.$

Left side:

$\int \frac{dy}{2+y}=\ln|2+y|.$

Right side: let

$u=5+e^x \Rightarrow du=e^x dx.$

Hence,

$-\int \frac{e^x}{5+e^x}\,dx=-\int \frac{du}{u}=-\ln|u|=-\ln(5+e^x).$

Therefore,

$\ln|2+y|=-\ln(5+e^x)+C.$

So,

$\ln|2+y|+\ln(5+e^x)=C$

or

$\ln\big(|2+y|(5+e^x)\big)=C.$

Thus,

$|2+y|(5+e^x)=K.$

Since the initial condition will give a positive value, we can write

$2+y=\frac{C}{5+e^x}.$

4. Use the initial condition

At $x=0$, $y=1$ and $e^0=1$.

So,

$2+1=\frac{C}{5+1}$

$3=\frac{C}{6}$

$C=18.$

Hence the solution is

$2+y=\frac{18}{5+e^x}$

which gives

$y=\frac{18}{5+e^x}-2.$

5. Find $y(\ln 13)$

Since

$e^{\ln 13}=13,$

we get

$y(\ln 13)=\frac{18}{5+13}-2=\frac{18}{18}-2=1-2=-1.$

6. Check options

The value is

$-1$

So the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They agree.