To solve for $f(x)$ given the conditions:

1. $f''(x) = \sin x + \cos x$
2. $f'(0) = 5$
3. $f(0) = 5$

we'll proceed by integrating step-by-step.

Step 1: Find $f'(x)$ by integrating $f''(x)$.

Given: $f''(x) = \sin x + \cos x$ Integrate $f''(x)$ with respect to $x$: $f'(x) = \int (\sin x + \cos x) \, dx$ $f'(x) = -\cos x + \sin x + C_1$ where $C_1$ is the constant of integration.

Step 2: Use $f'(0) = 5$ to solve for $C_1$.

Substitute $x = 0$ into $f'(x)$: $f'(0) = -\cos(0) + \sin(0) + C_1 = -1 + 0 + C_1 = 5$ $C_1 = 6$ So, $f'(x) = -\cos x + \sin x + 6$

Step 3: Find $f(x)$ by integrating $f'(x)$.

Now integrate $f'(x)$: $f(x) = \int (-\cos x + \sin x + 6) \, dx$ $f(x) = -\sin x - \cos x + 6x + C_2$ where $C_2$ is another constant of integration.

Step 4: Use $f(0) = 5$ to solve for $C_2$.

Substitute $x = 0$ into $f(x)$: $f(0) = -\sin(0) - \cos(0) + 6 \cdot 0 + C_2 = 0 - 1 + 0 + C_2 = 5$ $C_2 = 6$ So, $f(x) = -\sin x - \cos x + 6x + 6$

Final Answer

$f(x) = -\sin x - \cos x + 6x + 6$

Would you like more details on any step, or have any further questions?

Additional Related Questions:

1. How would the answer change if $f(0)$ were a different value?
2. What if $f'(x)$ had additional terms beyond trigonometric functions?
3. Can we use alternative methods, like Laplace transforms, for such problems?
4. How can we verify our solution by differentiating $f(x)$ back?
5. What if $f(x)$ were required to satisfy an initial condition at a different point?

Tip:

Remember to check initial conditions carefully, as they determine the constants in integration solutions.