Math Problem Statement

Solve the fourth-order initial value problem with the equation y^(4)(t) = -sin(t) + cos(t) and given initial conditions y(0) = 0, y'(0) = 0, y''(0) = 0, y'''(0) = 1.

Solution

The problem shown in the image involves solving a fourth-order initial value problem with the differential equation:

y(4)(t)=sin(t)+cos(t)y^{(4)}(t) = -\sin(t) + \cos(t)

Initial Conditions:

y(0)=0,y(0)=0,y(0)=0,y(3)(0)=1y(0) = 0, \quad y'(0) = 0, \quad y''(0) = 0, \quad y^{(3)}(0) = 1

The task requires integrating y(4)(t)y^{(4)}(t) four times to find y(t)y(t), while solving for constants C1,C2,C3,C4C_1, C_2, C_3, C_4 using the initial conditions.

Step-by-step Solution:

  1. First Integration (for y(3)(t)y^{(3)}(t)):

    y(3)(t)=(sin(t)+cos(t))dt=cos(t)+sin(t)+C1y^{(3)}(t) = \int \left( -\sin(t) + \cos(t) \right) \, dt = \cos(t) + \sin(t) + C_1

    Apply the initial condition y(3)(0)=1y^{(3)}(0) = 1: 1=cos(0)+sin(0)+C1    C1=01 = \cos(0) + \sin(0) + C_1 \implies C_1 = 0 So: y(3)(t)=cos(t)+sin(t)y^{(3)}(t) = \cos(t) + \sin(t)

  2. Second Integration (for y(t)y''(t)):

    y(t)=(cos(t)+sin(t))dt=sin(t)cos(t)+C2y''(t) = \int \left( \cos(t) + \sin(t) \right) \, dt = \sin(t) - \cos(t) + C_2

    Apply the initial condition y(0)=0y''(0) = 0: 0=sin(0)cos(0)+C2    C2=10 = \sin(0) - \cos(0) + C_2 \implies C_2 = 1 So: y(t)=sin(t)cos(t)+1y''(t) = \sin(t) - \cos(t) + 1

  3. Third Integration (for y(t)y'(t)):

    y(t)=(sin(t)cos(t)+1)dt=cos(t)sin(t)+t+C3y'(t) = \int \left( \sin(t) - \cos(t) + 1 \right) \, dt = -\cos(t) - \sin(t) + t + C_3

    Apply the initial condition y(0)=0y'(0) = 0: 0=cos(0)sin(0)+0+C3    C3=10 = -\cos(0) - \sin(0) + 0 + C_3 \implies C_3 = 1 So: y(t)=cos(t)sin(t)+t+1y'(t) = -\cos(t) - \sin(t) + t + 1

  4. Fourth Integration (for y(t)y(t)):

    y(t)=(cos(t)sin(t)+t+1)dt=sin(t)+cos(t)+t22+t+C4y(t) = \int \left( -\cos(t) - \sin(t) + t + 1 \right) \, dt = -\sin(t) + \cos(t) + \frac{t^2}{2} + t + C_4

    Apply the initial condition y(0)=0y(0) = 0: 0=sin(0)+cos(0)+022+0+C4    C4=10 = -\sin(0) + \cos(0) + \frac{0^2}{2} + 0 + C_4 \implies C_4 = -1 So: y(t)=sin(t)+cos(t)+t22+t1y(t) = -\sin(t) + \cos(t) + \frac{t^2}{2} + t - 1

Final Answer:

y(t)=sin(t)+cos(t)+t22+t1y(t) = -\sin(t) + \cos(t) + \frac{t^2}{2} + t - 1

Would you like further details on any step, or explanations for verification? Also, here are some follow-up questions for deeper understanding:

  1. What is the significance of initial conditions in solving differential equations?
  2. Can you derive the solution using Laplace transforms instead?
  3. How would you solve a second-order ODE with similar trigonometric forcing terms?
  4. What method can be used to check the solution's correctness?
  5. How does the number of initial conditions relate to the order of a differential equation?

Tip: When integrating, don't forget to add the constant of integration, which is crucial for solving based on initial conditions!

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Initial Value Problems
Integration

Formulas

y^(4)(t) = -sin(t) + cos(t)
y^(3)(t) = ∫(-sin(t) + cos(t)) dt
y''(t) = ∫(cos(t) + sin(t)) dt
y'(t) = ∫(sin(t) - cos(t)) dt
y(t) = ∫(-cos(t) - sin(t) + t + 1) dt

Theorems

Fundamental Theorem of Calculus
Initial Value Theorem for Differential Equations

Suitable Grade Level

College Level (Calculus II/III, Differential Equations)