Find f.f ''(𝜃) = sin(𝜃) + cos(𝜃),    f(0) = 4,    f '(0) = 4

The given equation is a second order differential equation. To solve it, we need to find two integrals.

Step 1: Integrate the equation once The integral of sin(𝜃) is -cos(𝜃) and the integral of cos(𝜃) is sin(𝜃). So, the first integral of the equation f''(𝜃) = sin(𝜃) + cos(𝜃) is f'(𝜃) = -cos(𝜃) + sin(𝜃) + C1, where C1 is the constant of integration.

Step 2: Integrate the equation again The integral of -cos(𝜃) is -sin(𝜃) and the integral of sin(𝜃) is -cos(𝜃). So, the second integral of the equation is f(𝜃) = -sin(𝜃) - cos(𝜃) + C1*𝜃 + C2, where C2 is the second constant of integration.

Step 3: Use the initial conditions to find the constants of integration We know that f(0) = 4 and f'(0) = 4. We can use these conditions to find the values of C1 and C2.

For f(0) = 4, we substitute 𝜃 = 0 into the equation f(𝜃) = -sin(𝜃) - cos(𝜃) + C1𝜃 + C2 to get 4 = -sin(0) - cos(0) + C10 + C2. Simplifying this gives C2 = 5.

For f'(0) = 4, we substitute 𝜃 = 0 into the equation f'(𝜃) = -cos(𝜃) + sin(𝜃) + C1 to get 4 = -cos(0) + sin(0) + C1. Simplifying this gives C1 = 5.

So, the solution to the differential equation is f(𝜃) = -sin(𝜃) - cos(𝜃) + 5𝜃 + 5.