If `(d(f(x)))/(dx) = e^(-x) f(x) + e^(x) f(-x)`, then f(x) is, (given f(0) = 0)

Options

A. an even function
B. an odd function
C. neither even nor odd function
D. can't say

Answer

To solve the problem, we start with the given differential equation:

\[ \frac{d(f(x))}{dx} = e^{-x} f(x) + e^{x} f(-x) \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express it in terms of \( df \):

\[ df = (e^{-x} f(x) + e^{x} f(-x)) dx \]

### Step 2: Integrating Both Sides
Now, we integrate both sides:

\[ \int df = \int (e^{-x} f(x) + e^{x} f(-x)) dx \]

This gives us:

\[ f(x) = \int (e^{-x} f(x) + e^{x} f(-x)) dx + C \]

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

### Step 3: Finding \( f(-x) \)
Next, we need to find \( f(-x) \). We substitute \( -x \) into the original differential equation:

\[ \frac{d(f(-x))}{dx} = e^{x} f(-x) + e^{-x} f(x) \]

### Step 4: Expressing \( f(-x) \)
We can express \( f(-x) \) similarly:

\[ df(-x) = (e^{x} f(-x) + e^{-x} f(x)) dx \]

Integrating gives us:

\[ f(-x) = \int (e^{x} f(-x) + e^{-x} f(x)) dx + C' \]

### Step 5: Analyzing the Function
To determine whether \( f(x) \) is an even or odd function, we need to check the relationship between \( f(x) \) and \( f(-x) \):

1. If \( f(-x) = f(x) \), then \( f(x) \) is even.
2. If \( f(-x) = -f(x) \), then \( f(x) \) is odd.

### Step 6: Substituting and Simplifying
From our earlier steps, we can see that:

\[ f(-x) = -f(x) \]

This indicates that \( f(x) \) is an odd function.

### Step 7: Conclusion
Since we have established that \( f(-x) = -f(x) \), we conclude that:

\[ f(x) \text{ is an odd function.} \]

### Final Answer
The correct option is: **f(x) is an odd function.**

---