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

We start from the given differential equation

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

First divide every term by the factor $$5+e^{x}$$ so that the structure becomes clearer:

$$\frac1{2+y}\,\frac{dy}{dx}+\frac{e^{x}}{5+e^{x}}=0.$$

Recall the derivative of a natural logarithm. In general,

$$\frac{d}{dx}\bigl(\ln u\bigr)=\frac1u\,\frac{du}{dx}.$$

Using this, observe that

$$\frac{d}{dx}\left[\ln(2+y)\right]=\frac1{2+y}\,\frac{dy}{dx}$$

and

$$\frac{d}{dx}\left[\ln(5+e^{x})\right]=\frac{e^{x}}{5+e^{x}}.$$

Hence the left‐hand side of our transformed differential equation can be recognised as

$$\frac1{2+y}\,\frac{dy}{dx}+\frac{e^{x}}{5+e^{x}}=\frac{d}{dx}\Bigl[\ln(2+y)+\ln(5+e^{x})\Bigr]$$

or, after combining the logarithms,

$$\frac{d}{dx}\Bigl[\ln\!\bigl((2+y)(5+e^{x})\bigr)\Bigr].$$

Therefore the differential equation simply reads

$$\frac{d}{dx}\Bigl[\ln\!\bigl((2+y)(5+e^{x})\bigr)\Bigr]=0.$$

Whenever the derivative of a function is zero, that function must be a constant. So we can write

$$\ln\!\bigl((2+y)(5+e^{x})\bigr)=C,$$

where $$C$$ is a constant. Exponentiating both sides gives an equivalent relation without the logarithm:

$$(2+y)(5+e^{x})=K,$$

with $$K=e^{C}$$ (still a constant).

To evaluate $$K$$ we use the initial condition $$y(0)=1.$$ At $$x=0$$ we have $$e^{0}=1,$$ so

$$(2+1)(5+1)=K\;\Longrightarrow\;3\cdot6=K\;\Longrightarrow\;K=18.$$

Thus the solution connecting $$x$$ and $$y$$ is

$$(2+y)(5+e^{x})=18.$$

Now we substitute $$x=\log_e13$$. Since $$\log_e13=\ln13,$$ we get $$e^{x}=13,$$ and therefore

$$5+e^{x}=5+13=18.$$

Placing this into the relation gives

$$(2+y)\cdot18=18.$$

Dividing both sides by 18 we find

$$2+y=1\;\Longrightarrow\;y=-1.$$

Hence, the correct answer is Option B.