Let y=f(x) is a solution of differential equation `e^(y)((dy)/(dx)-1)=e^(x)` and f(0)=0 then f(1) is equal to

Answer

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

### Step 1: Rewrite the differential equation
We start with the equation:
\[ e^y \left( \frac{dy}{dx} - 1 \right) = e^x \]
This can be rearranged to:
\[ e^y \frac{dy}{dx} - e^y = e^x \]
Thus, we have:
\[ e^y \frac{dy}{dx} = e^x + e^y \]

### Step 2: Substitute \( e^y = t \)
Let \( t = e^y \). Then, differentiating \( y \) with respect to \( x \) gives:
\[ \frac{dy}{dx} = \frac{dt}{dx} \cdot \frac{1}{t} \]
Substituting this into our equation gives:
\[ t \cdot \frac{dt}{dx} \cdot \frac{1}{t} = e^x + t \]
This simplifies to:
\[ \frac{dt}{dx} = e^x + t \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \frac{dt}{dx} - t = e^x \]

### Step 4: Finding the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int -1 \, dx} = e^{-x} \]

### Step 5: Multiply through by the integrating factor
Multiplying the entire equation by the integrating factor:
\[ e^{-x} \frac{dt}{dx} - e^{-x} t = e^{-x} e^x \]
This simplifies to:
\[ e^{-x} \frac{dt}{dx} - e^{-x} t = 1 \]

### Step 6: Left-hand side as a derivative
The left-hand side can be expressed as:
\[ \frac{d}{dx}(e^{-x} t) = 1 \]

### Step 7: Integrate both sides
Integrating both sides with respect to \( x \):
\[ e^{-x} t = \int 1 \, dx = x + C \]
Thus,
\[ t = e^x (x + C) \]

### Step 8: Substitute back for \( y \)
Recalling that \( t = e^y \), we have:
\[ e^y = e^x (x + C) \]
Taking the natural logarithm gives:
\[ y = x + \ln(x + C) \]

### Step 9: Apply the initial condition
Using the initial condition \( f(0) = 0 \):
\[ 0 = 0 + \ln(0 + C) \implies \ln(C) = 0 \implies C = 1 \]
Thus, the equation simplifies to:
\[ y = x + \ln(x + 1) \]

### Step 10: Find \( f(1) \)
Now, we need to find \( f(1) \):
\[ f(1) = 1 + \ln(1 + 1) = 1 + \ln(2) \]

### Final Answer
Thus, \( f(1) = 1 + \ln(2) \).

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