Final answer:
In order to solve the given initial value problem f'''(x) = cos(x), with f(0) = 1, f'(0) = 2, and f''(0) = 3, we integrate the equation three times and substitute the initial conditions to find the function f(x) = -sin(x) + x^2/2 + 2x + 1.
Step-by-step explanation:
In order to address this initial value problem, our objective is to identify the function f(x) satisfying the differential equation f'''(x) = cos(x), along with the initial conditions f(0) = 1, f'(0) = 2, and f''(0) = 3.
Commencing with the differential equation, we perform successive integrations to derive expressions for f''(x), f'(x), and f(x).
These integrals yield f''(x) = sin(x) + C1, f'(x) = -cos(x) + C1x + C2, and f(x) = -sin(x) + C1x^2/2 + C2x + C3.
Substituting the given initial conditions into these expressions produces a system of three equations. Solving this system yields the values for the constants C1, C2, and C3.
Subsequently, these constants are inserted back into the expression for f(x), resulting in the final solution: f(x) = -sin(x) + x^2/2 + 2x + 1.
This systematic approach ensures a comprehensive solution to the specified initial value problem.