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

Options

A. ln2
B. 2+ln2
C. 1+ln2
D. 3+ln2

Answer

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

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ e^{y} \left( \frac{dy}{dx} - 1 \right) = e^{x} \]
We can rearrange this to isolate \( \frac{dy}{dx} \):
\[ e^{y} \frac{dy}{dx} - e^{y} = e^{x} \]
Adding \( e^{y} \) to both sides gives:
\[ e^{y} \frac{dy}{dx} = e^{x} + e^{y} \]
Now, we can divide both sides by \( e^{y} \):
\[ \frac{dy}{dx} = e^{x - y} + 1 \]

### Step 2: Substitute \( e^{y} \)
Let \( e^{y} = t \). Then, \( y = \ln(t) \) and \( \frac{dy}{dx} = \frac{1}{t} \frac{dt}{dx} \). Substituting these into the equation gives:
\[ \frac{1}{t} \frac{dt}{dx} = e^{x - \ln(t)} + 1 \]
This simplifies to:
\[ \frac{1}{t} \frac{dt}{dx} = \frac{e^{x}}{t} + 1 \]
Multiplying through by \( t \) results in:
\[ \frac{dt}{dx} = e^{x} + t \]

### Step 3: Solve the Differential Equation
This is a linear first-order differential equation. We can rearrange it:
\[ \frac{dt}{dx} - t = e^{x} \]
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{-\int 1 \, dx} = e^{-x} \]
Multiplying the entire equation by the integrating factor:
\[ e^{-x} \frac{dt}{dx} - e^{-x} t = 1 \]
The left side can be rewritten as:
\[ \frac{d}{dx}(e^{-x} t) = 1 \]
Integrating both sides:
\[ e^{-x} t = x + C \]
Thus,
\[ t = e^{x}(x + C) \]

### Step 4: Substitute Back for \( y \)
Recall that \( t = e^{y} \), so:
\[ e^{y} = e^{x}(x + C) \]
Taking the natural logarithm of both sides:
\[ y = x + \ln(x + C) \]

### Step 5: Apply Initial Condition
We know \( f(0) = 0 \), so substituting \( x = 0 \):
\[ 0 = 0 + \ln(0 + C) \implies \ln(C) = 0 \implies C = 1 \]
Thus, the function simplifies to:
\[ y = x + \ln(x + 1) \]

### Step 6: 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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