Let `f(x)` be a polynomial satisfying f(0)=2 , `f'(0)=3` and `f''(x)=f(x)` then f(4) equals

Options

A. `(5(e^(8)+1))/(2e^(4))`
B. `(5(e^(8)-1))/(2e^(4))`
C. `(2e^(4))/(5(e^(8)-1))`
D. `(2e^(4))/(5(e^(8)+1))`

Answer

To solve the problem step by step, we need to find the polynomial \( f(x) \) that satisfies the given conditions.

### Step 1: Understand the conditions
We are given:
1. \( f(0) = 2 \)
2. \( f'(0) = 3 \)
3. \( f''(x) = f(x) \)

### Step 2: Identify the general form of the polynomial
The condition \( f''(x) = f(x) \) suggests that \( f(x) \) can be expressed in terms of exponential functions. The general solution to the differential equation \( f''(x) = f(x) \) is:
\[ f(x) = A e^x + B e^{-x} \]
where \( A \) and \( B \) are constants to be determined.

### Step 3: Calculate the first derivative
Now, we find the first derivative of \( f(x) \):
\[ f'(x) = A e^x - B e^{-x} \]

### Step 4: Apply the initial conditions
Using the first initial condition \( f(0) = 2 \):
\[ f(0) = A e^0 + B e^0 = A + B = 2 \quad \text{(1)} \]

Using the second initial condition \( f'(0) = 3 \):
\[ f'(0) = A e^0 - B e^0 = A - B = 3 \quad \text{(2)} \]

### Step 5: Solve the system of equations
We now have a system of two equations:
1. \( A + B = 2 \)
2. \( A - B = 3 \)

Adding these two equations:
\[ (A + B) + (A - B) = 2 + 3 \implies 2A = 5 \implies A = \frac{5}{2} \]

Substituting \( A \) back into equation (1):
\[ \frac{5}{2} + B = 2 \implies B = 2 - \frac{5}{2} = -\frac{1}{2} \]

### Step 6: Write the polynomial
Now we have:
\[ A = \frac{5}{2}, \quad B = -\frac{1}{2} \]
Thus, the polynomial is:
\[ f(x) = \frac{5}{2} e^x - \frac{1}{2} e^{-x} \]

### Step 7: Calculate \( f(4) \)
Now we need to find \( f(4) \):
\[ f(4) = \frac{5}{2} e^4 - \frac{1}{2} e^{-4} \]

### Step 8: Simplify \( f(4) \)
To simplify:
\[ f(4) = \frac{5}{2} e^4 - \frac{1}{2 e^4} \]
This can be expressed as:
\[ f(4) = \frac{5 e^4 - 1}{2 e^4} \]

### Final Answer
Thus, the value of \( f(4) \) is:
\[ f(4) = \frac{5 e^4 - 1}{2 e^4} \]