Suppose `f(x)=e^(ax)+e^(bx),` where `aneb and f''(x)-2f'(x)-15f(x)=0` for all x, then the value of `|a+b|` is equal to......

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the function
We start with the function given in the problem:
\[ f(x) = e^{ax} + e^{bx} \]

### Step 2: Find the first derivative
Next, we differentiate \(f(x)\) to find the first derivative \(f'(x)\):
\[ f'(x) = \frac{d}{dx}(e^{ax}) + \frac{d}{dx}(e^{bx}) = ae^{ax} + be^{bx} \]

### Step 3: Find the second derivative
Now, we differentiate \(f'(x)\) to find the second derivative \(f''(x)\):
\[ f''(x) = \frac{d}{dx}(ae^{ax}) + \frac{d}{dx}(be^{bx}) = a^2e^{ax} + b^2e^{bx} \]

### Step 4: Substitute into the differential equation
We substitute \(f(x)\), \(f'(x)\), and \(f''(x)\) into the differential equation given in the problem:
\[ f''(x) - 2f'(x) - 15f(x) = 0 \]
Substituting the expressions we found:
\[ (a^2e^{ax} + b^2e^{bx}) - 2(ae^{ax} + be^{bx}) - 15(e^{ax} + e^{bx}) = 0 \]

### Step 5: Combine like terms
Now, we can combine the terms:
\[ (a^2 - 2a - 15)e^{ax} + (b^2 - 2b - 15)e^{bx} = 0 \]
Since \(e^{ax}\) and \(e^{bx}\) are never zero, we can set the coefficients to zero:
1. \(a^2 - 2a - 15 = 0\)
2. \(b^2 - 2b - 15 = 0\)

### Step 6: Solve the quadratic equations
We will solve each quadratic equation separately.

#### For \(a\):
\[ a^2 - 2a - 15 = 0 \]
Factoring gives:
\[ (a - 5)(a + 3) = 0 \]
Thus, \(a = 5\) or \(a = -3\).

#### For \(b\):
\[ b^2 - 2b - 15 = 0 \]
Factoring gives:
\[ (b - 5)(b + 3) = 0 \]
Thus, \(b = 5\) or \(b = -3\).

### Step 7: Determine valid pairs for \(a\) and \(b\)
Since it is given that \(a \neq b\), we have the following valid pairs:
- \(a = 5\) and \(b = -3\)
- \(a = -3\) and \(b = 5\)

### Step 8: Calculate \(|a + b|\)
Now we calculate \(|a + b|\):
\[ |a + b| = |5 + (-3)| = |2| = 2 \]

### Final Answer
The value of \(|a + b|\) is:
\[ \boxed{2} \]