If `inte^(ax)cosbx dx=(e^(2x))/(29)f(x)+C` , then f'' (x)=

Options

A. 29 f (x)
B. `-29 f(x)`
C. 25 f(x)
D. `-25 f(x)`

Explanation

To solve the problem, we need to find \( f''(x) \) given that:

\[ \int e^{ax} \cos(bx) \, dx = \frac{e^{2x}}{29} f(x) + C \]

### Step 1: Identify the values of \( a \) and \( b \)

From the given integral, we can use the known formula for the integral of the form \( \int e^{ax} \cos(bx) \, dx \):

\[ \int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + C \]

Comparing this with the given expression \( \frac{e^{2x}}{29} f(x) + C \), we can see that:

- The coefficient of \( e^{ax} \) suggests that \( a = 2 \).
- The denominator \( a^2 + b^2 = 29 \).

### Step 2: Solve for \( b \)

Substituting \( a = 2 \) into the equation \( a^2 + b^2 = 29 \):

\[ 2^2 + b^2 = 29 \]
\[ 4 + b^2 = 29 \]
\[ b^2 = 29 - 4 = 25 \]
\[ b = \sqrt{25} = 5 \]

### Step 3: Write the function \( f(x) \)

Now that we have \( a = 2 \) and \( b = 5 \), we can express \( f(x) \):

\[ f(x) = a \cos(bx) + b \sin(bx) = 2 \cos(5x) + 5 \sin(5x) \]

### Step 4: Differentiate to find \( f'(x) \)

Now we will differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2 \cos(5x) + 5 \sin(5x)) \]
Using the chain rule:
\[ f'(x) = 2 \cdot (-5 \sin(5x)) + 5 \cdot (5 \cos(5x)) \]
\[ f'(x) = -10 \sin(5x) + 25 \cos(5x) \]

### Step 5: Differentiate again to find \( f''(x) \)

Now we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}(-10 \sin(5x) + 25 \cos(5x)) \]
Using the chain rule again:
\[ f''(x) = -10 \cdot 5 \cos(5x) - 25 \cdot 5 \sin(5x) \]
\[ f''(x) = -50 \cos(5x) - 125 \sin(5x) \]

### Step 6: Factor out common terms

We can factor out \(-25\):

\[ f''(x) = -25(2 \cos(5x) + 5 \sin(5x)) \]

Since we know \( f(x) = 2 \cos(5x) + 5 \sin(5x) \):

\[ f''(x) = -25 f(x) \]

### Final Answer

Thus, the final result is:

\[ f''(x) = -25 f(x) \]

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