Solve the following initial value problems and find the corresponding solution curves :
`(x+1)y'=2 e^(-y)-1, y(0)=0`.

Answer

To solve the initial value problem given by the differential equation

\[ (x+1)y' = 2e^{-y} - 1, \quad y(0) = 0, \]

we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation in a more manageable form. We can express it as:

\[ (x+1) \frac{dy}{dx} = 2e^{-y} - 1. \]

### Step 2: Separate variables
Next, we will separate the variables \(y\) and \(x\):

\[ \frac{dy}{2e^{-y} - 1} = \frac{dx}{x+1}. \]

### Step 3: Simplify the left-hand side
To simplify the left-hand side, we can rewrite it as:

\[ \frac{dy}{2e^{-y} - 1} = \frac{e^y}{2 - e^y} dy. \]

Thus, we have:

\[ \frac{e^y}{2 - e^y} dy = \frac{dx}{x+1}. \]

### Step 4: Integrate both sides
Now we integrate both sides:

\[ \int \frac{e^y}{2 - e^y} dy = \int \frac{dx}{x+1}. \]

For the left-hand side, we can use substitution. Let \(u = 2 - e^y\), then \(du = -e^y dy\), which gives us:

\[ -\int \frac{1}{u} du = \ln |u| + C = \ln |2 - e^y| + C. \]

For the right-hand side, we have:

\[ \int \frac{dx}{x+1} = \ln |x+1| + C_1. \]

### Step 5: Combine the results
Combining both integrals, we have:

\[ -\ln |2 - e^y| = \ln |x+1| + C. \]

### Step 6: Solve for \(C\) using the initial condition
Using the initial condition \(y(0) = 0\):

\[ -\ln |2 - e^0| = \ln |0 + 1| + C \implies -\ln |2 - 1| = \ln |1| + C \implies -\ln 1 = 0 + C \implies C = 0. \]

### Step 7: Rewrite the equation
Thus, we can rewrite the equation as:

\[ -\ln |2 - e^y| = \ln |x+1| \implies |2 - e^y| = \frac{1}{|x+1|}. \]

### Step 8: Solve for \(y\)
Now, we can solve for \(y\):

\[ 2 - e^y = \frac{1}{x+1} \implies e^y = 2 - \frac{1}{x+1} \implies y = \ln \left(2 - \frac{1}{x+1}\right). \]

### Final Solution
Thus, the solution to the initial value problem is:

\[ y = \ln \left(2 - \frac{1}{x+1}\right). \]