Find the function $$ F $$ that satisfies the following differential equations and initial conditions. $$ F^{\prime \prime}(x)=\cos x, F^{\prime}(0)=9, F(\pi)=9 $$ $$ F(x)=\square $$ (Type an exact answer in terms of $$ \pi $$.)

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After solving the differential equation $$ F''(x) = \cos x $$ with the given initial conditions $$ F'(0) = 9 $$ and $$ F(\pi) = 9 $$, we find the function $$ F(x) $$ as follows:

1. **Integrate the second derivative:**
   $$
   F'(x) = \int \cos x \, dx = \sin x + C
   $$
   Applying the condition $$ F'(0) = 9 $$:
   $$
   F'(0) = \sin(0) + C = 0 + C = 9 \implies C = 9
   $$
   So,
   $$
   F'(x) = \sin x + 9
   $$

2. **Integrate the first derivative to find $$ F(x) $$:**
   $$
   F(x) = \int (\sin x + 9) \, dx = -\cos x + 9x + D
   $$
   Applying the condition $$ F(\pi) = 9 $$:
   $$
   F(\pi) = -\cos(\pi) + 9\pi + D = 1 + 9\pi + D = 9 \implies D = 8 - 9\pi
   $$
   
3. **Combine the constants to write the final function:**
   $$
   F(x) = -\cos x + 9x + 8 - 9\pi
   $$
   
Final Answer:
After simplifying, F (x ) is nine x minus cosine x plus eight minus nine π. Thus,

F(x) = 9 x − cos x + 8 − 9 π
$$ F(x) = 9x - \cos x + 8 - 9\pi $$

Find the function \( F \) that satisfies the following differential equations and initial conditions. \( F^{\prime \prime}(x)=\cos x, F^{\prime}(0)=9, F(\pi)=9 \) \( F(x)=\square \) (Type an exact answer in terms of \( \pi \).)

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