If `f'(x)=f(x)+int_(0)^(1)f(x)dx`,given `f(0)=1`, then the value of `f(log_(e)2)` is

Options

A. `1/(3+e)`
B. `(5-e)/(3-e)`
C. `(2+e)/(e-2)`
D. none of these

Explanation

To solve the problem, we start with the given information:

1. We have the differential equation:
\[ f'(x) = f(x) + \int_0^1 f(x) \, dx \]
2. We are also given the initial condition:
\[ f(0) = 1 \]

### Step 1: Rewrite the equation
Let \( C = \int_0^1 f(x) \, dx \). The equation simplifies to:
\[ f'(x) = f(x) + C \]

### Step 2: Solve the differential equation
This is a first-order linear ordinary differential equation. We can rearrange it as:
\[ f'(x) - f(x) = C \]
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{-\int 1 \, dx} = e^{-x} \]
Multiplying the entire equation by the integrating factor:
\[ e^{-x} f'(x) - e^{-x} f(x) = C e^{-x} \]
The left-hand side can be rewritten as:
\[ \frac{d}{dx}(e^{-x} f(x)) = C e^{-x} \]

### Step 3: Integrate both sides
Integrating both sides with respect to \( x \):
\[ e^{-x} f(x) = -C e^{-x} + K \]
where \( K \) is a constant of integration. Multiplying through by \( e^{x} \):
\[ f(x) = -C + K e^{x} \]

### Step 4: Apply the initial condition
Using the initial condition \( f(0) = 1 \):
\[ f(0) = -C + K = 1 \]
Thus, we have:
\[ K - C = 1 \quad \text{(1)} \]

### Step 5: Find \( C \)
Now, we need to find \( C \):
\[ C = \int_0^1 f(x) \, dx = \int_0^1 (-C + K e^{x}) \, dx \]
Calculating the integral:
\[ = \int_0^1 -C \, dx + \int_0^1 K e^{x} \, dx \]
\[ = -C + K[e^{x}]_0^1 = -C + K(e - 1) \]
So we have:
\[ C = -C + K(e - 1) \]
Rearranging gives:
\[ 2C = K(e - 1) \quad \text{(2)} \]

### Step 6: Solve equations (1) and (2)
From equation (1):
\[ K = C + 1 \]
Substituting into equation (2):
\[ 2C = (C + 1)(e - 1) \]
Expanding:
\[ 2C = C(e - 1) + (e - 1) \]
Rearranging gives:
\[ C(2 - (e - 1)) = e - 1 \]
\[ C(3 - e) = e - 1 \]
Thus:
\[ C = \frac{e - 1}{3 - e} \]

### Step 7: Find \( K \)
Substituting \( C \) back into \( K = C + 1 \):
\[ K = \frac{e - 1}{3 - e} + 1 = \frac{e - 1 + (3 - e)}{3 - e} = \frac{2}{3 - e} \]

### Step 8: Write the function \( f(x) \)
Now we have:
\[ f(x) = -C + K e^{x} = -\frac{e - 1}{3 - e} + \frac{2}{3 - e} e^{x} \]
\[ = \frac{- (e - 1) + 2 e^{x}}{3 - e} = \frac{2 e^{x} - e + 1}{3 - e} \]

### Step 9: Find \( f(\log_e 2) \)
Now we need to find \( f(\log_e 2) \):
\[ f(\log_e 2) = \frac{2 e^{\log_e 2} - e + 1}{3 - e} = \frac{2 \cdot 2 - e + 1}{3 - e} = \frac{4 - e + 1}{3 - e} = \frac{5 - e}{3 - e} \]

### Final Answer
Thus, the value of \( f(\log_e 2) \) is:
\[ \frac{5 - e}{3 - e} \]