Find the particular solution of the following :
`(x+1)(dy)/(dx)=2e^(-y)-1,y(0)=0`.

Answer

To find the particular solution of the differential equation

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

we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables \(y\) and \(x\):

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

### Step 2: Separating Variables
Next, we can separate the variables by moving all terms involving \(y\) to one side and all terms involving \(x\) to the other side:

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

### Step 3: Integrating Both Sides
Now, we integrate both sides:

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

### Step 4: Substituting for Integration
To simplify the left-hand side, we can let \(t = 2 - e^{-y}\). Then, we have:

\[ e^{-y} = 2 - t \quad \text{and} \quad dy = -\frac{dt}{e^{-y}} = -\frac{dt}{2 - t}. \]

Substituting this into the integral gives:

\[ \int \frac{-dt}{t} = \int \frac{dx}{x + 1}. \]

### Step 5: Performing the Integrals
Now we perform the integrals:

\[ -\ln |t| = \ln |x + 1| + C, \]

where \(C\) is the constant of integration.

### Step 6: Exponentiating Both Sides
Exponentiating both sides results in:

\[ \frac{1}{t} = K(x + 1), \quad \text{where } K = e^{-C}. \]

### Step 7: Substituting Back for \(t\)
Now substituting back for \(t\):

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

### Step 8: Solving for \(e^{-y}\)
Rearranging gives:

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

Thus, we have:

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

### Step 9: Finding the Particular Solution
We need to find \(K\) using the initial condition \(y(0) = 0\):

\[ e^{0} = 2 - \frac{1}{K(0 + 1)} \implies 1 = 2 - \frac{1}{K} \implies \frac{1}{K} = 1 \implies K = 1. \]

### Step 10: Final Solution
Substituting \(K\) back into the equation gives:

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

Thus, the final particular solution is:

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