` (x+1)(dy)/(dx) -1 = 2e^(-y) , y=0, " when " x=1`

Answer

To solve the differential equation \((x+1)\frac{dy}{dx} - 1 = 2e^{-y}\) with the initial condition \(y(1) = 0\), we will follow these steps:

### Step 1: Rearrange the Equation
We start by rearranging the given equation to isolate \(\frac{dy}{dx}\):
\[ (x+1)\frac{dy}{dx} = 2e^{-y} + 1 \]
\[ \frac{dy}{dx} = \frac{2e^{-y} + 1}{x+1} \]

### Step 2: Separate Variables
Next, we separate the variables \(y\) and \(x\):
\[ dy = \left(\frac{2e^{-y} + 1}{x+1}\right) dx \]
To facilitate integration, we can multiply both sides by \(e^y\):
\[ e^y dy = \frac{(2 + e^y)}{x+1} dx \]

### Step 3: Integrate Both Sides
Now we will integrate both sides. The left side becomes:
\[ \int e^y dy = e^y + C_1 \]
For the right side, we can rewrite it as:
\[ \int \frac{(2 + e^y)}{x+1} dx = \int \frac{2}{x+1} dx + \int \frac{e^y}{x+1} dx \]
The first integral is:
\[ 2 \ln|x+1| + C_2 \]
Thus, we have:
\[ e^y = 2 \ln|x+1| + C \]

### Step 4: Apply Initial Condition
Now we will apply the initial condition \(y(1) = 0\):
\[ e^0 = 2 \ln|1+1| + C \]
This simplifies to:
\[ 1 = 2 \ln(2) + C \]
Solving for \(C\):
\[ C = 1 - 2 \ln(2) \]

### Step 5: Final Solution
Substituting \(C\) back into our equation gives:
\[ e^y = 2 \ln|x+1| + 1 - 2 \ln(2) \]
This can be simplified to:
\[ e^y = 2 \ln\left(\frac{x+1}{2}\right) + 1 \]

### Step 6: Solve for \(y\)
Taking the natural logarithm of both sides:
\[ y = \ln\left(2 \ln\left(\frac{x+1}{2}\right) + 1\right) \]

### Final Answer
Thus, the solution to the differential equation with the given initial condition is:
\[ y = \ln\left(2 \ln\left(\frac{x+1}{2}\right) + 1\right) \]