https://file.aiquickdraw.com/mathbot/file/1728480799244wfs5hsdz.jpg

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

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

### Initial Conditions:
\[
y(0) = 0, \quad y'(0) = 0, \quad y''(0) = 0, \quad y^{(3)}(0) = 1
\]

The task requires integrating \( y^{(4)}(t) \) four times to find \( y(t) \), while solving for constants \( C_1, C_2, C_3, C_4 \) using the initial conditions.

### Step-by-step Solution:

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

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

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

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

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

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

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

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

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

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

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

   Apply the initial condition \( y(0) = 0 \):
   \[
   0 = -\sin(0) + \cos(0) + \frac{0^2}{2} + 0 + C_4 \implies C_4 = -1
   \]
   So:
   \[
   y(t) = -\sin(t) + \cos(t) + \frac{t^2}{2} + t - 1
   \]

### Final Answer:
\[
y(t) = -\sin(t) + \cos(t) + \frac{t^2}{2} + t - 1
\]

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!

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.

Learn how to solve a fourth-order differential equation with trigonometric forcing terms using initial conditions. This example walks through integration and applying initial values step by step.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation