If `y=y(x)` is solution of differential equation `(5+e^(x))/(2+y) (dy)/(dx)=e^x` and `y(0)=4` than find `y(log_e 13)`

Answer

To solve the differential equation

\[ \frac{5 + e^x}{2 + y} \frac{dy}{dx} = e^x \]

with the initial condition \( y(0) = 4 \), we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to separate the variables \( y \) and \( x \):

\[ \frac{dy}{2 + y} = \frac{e^x}{5 + e^x} dx \]

### Step 2: Integrating Both Sides
Now, we integrate both sides. The left side requires the integration of \( \frac{1}{2 + y} \):

\[ \int \frac{dy}{2 + y} = \ln |2 + y| + C_1 \]

For the right side, we can use substitution. Let \( t = 5 + e^x \), then \( dt = e^x dx \). Thus, we rewrite the integral:

\[ \int \frac{e^x}{5 + e^x} dx = \int \frac{1}{t} dt = \ln |t| + C_2 = \ln |5 + e^x| + C_2 \]

### Step 3: Setting the Integrals Equal
Setting the two integrals equal gives us:

\[ \ln |2 + y| = \ln |5 + e^x| + C \]

where \( C = C_2 - C_1 \).

### Step 4: Exponentiating Both Sides
Exponentiating both sides, we have:

\[ 2 + y = k(5 + e^x) \]

where \( k = e^C \).

### Step 5: Applying the Initial Condition
Using the initial condition \( y(0) = 4 \):

\[ 2 + 4 = k(5 + e^0) \implies 6 = k(5 + 1) \implies 6 = 6k \implies k = 1 \]

### Step 6: Final Equation
Thus, we have:

\[ 2 + y = 5 + e^x \]

Rearranging gives:

\[ y = e^x + 3 \]

### Step 7: Finding \( y(\ln 13) \)
Now, we need to find \( y(\ln 13) \):

\[ y(\ln 13) = e^{\ln 13} + 3 = 13 + 3 = 16 \]

### Final Answer
Thus, the value of \( y(\ln 13) \) is

\[ \boxed{16} \]