Problem 89 - Thomas Calculus in SI Units

Question

Solve the initial value problems. $$\begin{aligned} &y^{(4)}=-\sin t+\cos t\\\ &y^{\prime \prime \prime}(0)=7, \quad y^{\prime \prime}(0)=y^{\prime}(0)=-1, \quad y(0)=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( y = -\sin t + \cos t + t^3 - 1 \).

Step 1: Find the General Solution of the Differential Equation

The given equation is \( y^{(4)} = -\sin t + \cos t \). We need to integrate this equation step-by-step to reduce the derivative order and find the general solution. First, integrate \( y^{(4)} = -\sin t + \cos t \) to get \( y^{(3)} \):\[ y^{(3)} = \int (-\sin t + \cos t) \, dt = \cos t + \sin t + C_1 \]. Here, \( C_1 \) is a constant of integration.

Step 2: Integrate to Find Second Derivative

Now, integrate \( y^{(3)} = \cos t + \sin t + C_1 \) to find \( y^{(2)} \): \[ y^{(2)} = \int (\cos t + \sin t + C_1) \, dt = \sin t - \cos t + C_1 t + C_2 \]. Again, \( C_2 \) is a constant of integration.

Step 3: Integrate to Find First Derivative

Integrate \( y^{(2)} = \sin t - \cos t + C_1 t + C_2 \) to get \( y' \): \[ y' = \int (\sin t - \cos t + C_1 t + C_2) \, dt = -\cos t - \sin t + \frac{1}{2}C_1 t^2 + C_2 t + C_3 \]. Here, \( C_3 \) is another constant of integration.

Step 4: Integrate to Find the Function y

Integrate \( y' = -\cos t - \sin t + \frac{1}{2}C_1 t^2 + C_2 t + C_3 \) to find \( y \): \[ y = \int (-\cos t - \sin t + \frac{1}{2}C_1 t^2 + C_2 t + C_3) \, dt = -\sin t + \cos t + \frac{1}{6}C_1 t^3 + \frac{1}{2}C_2 t^2 + C_3 t + C_4 \]. Here, \( C_4 \) is the last constant of integration.

Step 5: Use Initial Conditions to Determine Constants

Now use the initial conditions to solve for the constants \( C_1, C_2, C_3, C_4 \). - From \( y^{(3)}(0) = 7 \), substituting into \( y^{(3)} = \cos t + \sin t + C_1 \), gives: \( \cos 0 + \sin 0 + C_1 = 7 \) Thus, \( 1 + 0 + C_1 = 7 \) which implies \( C_1 = 6 \). - Use \( y''(0) = -1 \), substituting into \( y'' = \sin t - \cos t + 6t + C_2 \), gives: \( \sin 0 - \cos 0 + C_2 = -1 \) Thus, \( 0 - 1 + C_2 = -1 \) implies \( C_2 = 0 \). - From \( y'(0) = -1 \), substituting into \( y' = -\cos t - \sin t + 3t^2 + C_3 \), gives: \( -\cos 0 - \sin 0 + C_3 = -1 \) \( -1 + C_3 = -1 \) implies \( C_3 = 0 \). - Finally, from \( y(0) = 0 \), substituting into \( y = -\sin t + \cos t + \frac{1}{2}t^3 + C_4 \) gives: \( -\sin 0 + \cos 0 + C_4 = 0 \) \( 1 + C_4 = 0 \) implies \( C_4 = -1 \).

Step 6: Write the Particular Solution

Substitute the constants back into the expression for \( y \). The solution is:\[ y = -\sin t + \cos t + \frac{1}{6}(6) t^3 + \frac{1}{2}(0) t^2 + 0 t - 1 \]Simplify:\[ y = -\sin t + \cos t + t^3 - 1 \]. This is the particular solution to the given initial value problem.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. This concept is pivotal in understanding how things change over time or space. In our exercise, we deal with a fourth-order differential equation, indicated by the notation \( y^{(4)} \), which signifies the fourth derivative of \( y \) with respect to some variable, in this case, \( t \). Differential equations can describe a wide variety of phenomena, such as physical systems, population dynamics, and economic behaviors.

• Higher-order differential equations involve derivatives of great complexity.
• They require methods like integration to solve and understand the behavior of the system mathematically.
• The goal is to reduce the differential equation to a simpler form through integration, ultimately finding an expression for \( y \).

Understanding the type and order of a differential equation is essential as this determines the strategy to solve it effectively.

Integration

Integration is the reverse process of differentiation. It's like piecing together a puzzle—taking derivative pieces to form the original puzzle. In our problem, we integrate the equation sequentially, reducing order by one each time, moving from the fourth derivative down to the function \( y \) itself.

• Step-by-step Reduction: Start with the highest-order derivative and integrate iteratively until the original function is found.
• Function Recovery: Each integration 'undoes' differentiation, slowly reconstructing the original function that describes the system.

Each integration step introduces a constant of integration, reflecting the undetermined nature of the indefinite integral. Only initial conditions can fix these constants to obtain a unique solution.

Constants of Integration

The constants of integration arise from the process of integration and represent the unknowns in our problem. Each time we integrate, a constant is added because integration does not undo differentiation exactly; instead, it recovers a family of functions.

• For the fourth integration, we have four constants \( C_1, C_2, C_3, \text{and} C_4 \).
• Constants account for all possible solutions; each solves the differential equation but represents different initial conditions.

The initial conditions provided are crucial in calculating these constants to derive a particular solution. Without initial conditions, these constants cannot be determined, and the solution remains general.

Particular Solution

A particular solution is a single, unique solution to a differential equation that satisfies both the equation and any initial conditions specified. In our context, the particular solution was found by substituting the initial conditions into the general solution.

• The provided initial conditions, such as \( y(0) = 0 \), determine the values of the constants of integration.
• The particular solution fits the specific behavior described by the initial conditions, differentiating it from the general solution.

Plugging these constants back into the function gives us the particular solution \( y = -\sin t + \cos t + t^3 - 1 \). This means at \( t = 0 \), all these conditions are satisfied, ensuring the solution perfectly describes this particular initial value problem.