Question

In each exercise, for the given \(t_{0}\), (a) Obtain the fifth degree Taylor polynomial approximation of the solution, $$ P_{5}(t)=y\left(t_{0}\right)+y^{\prime}\left(t_{0}\right)\left(t-t_{0}\right)+\frac{y^{\prime \prime}\left(t_{0}\right)}{2 !}\left(t-t_{0}\right)^{2}+\cdots+\frac{y^{(5)}\left(t_{0}\right)}{5 !}\left(t-t_{0}\right)^{5} . $$ (b) If the exact solution is given, calculate the error at \(t=t_{0}+0.1\). \(y^{\prime \prime \prime}-y^{\prime}=0, \quad y(0)=1, \quad y^{\prime}(0)=2, \quad y^{\prime \prime}(0)=0 ; \quad t_{0}=0 .\)

Short answer

Answer: The fifth degree Taylor polynomial approximation for the given function is \(P_5(t) = 1 + 2t\), and the error at \(t = t_0 + 0.1\) is 0.01.

Step-by-step solution

Find the fourth and fifth order derivatives of y.

Given the third order derivative of y, we can find the fourth and fifth-order derivatives by integrating the equation given. $$y''' - y' = 0$$ Integrate with respect to t: $$y'' - y = C_1$$ Integrate again with respect to t and obtaining y'(t) and y(t): $$y' = y + C_1 t$$ $$y = C_2 t^2 + C_3 t + C_4$$ Let's use the initial conditions \(y(0)=1, y'(0)=2, y''(0)=0\) to find the constants \(C_1, C_2, C_3, C_4\).

Find the constants using initial conditions.

Use the initial conditions and solve for the constants: $$y(0) = C_2(0)^2 + C_3(0) + C_4 = 1$$ $$y'(0) = C_1(0) + 0 = 2$$ $$y''(0) = 0 - 0 = 0$$ From these equations, we get \(C_1=2, C_2=0, C_3=0, C_4=1\). Therefore, the functions for y'(t) and y(t) are as follows: $$y'(t) = 2t$$ $$y(t) = t^2 + 1$$ Now that we found the function for y(t) and its derivatives up to the third order, we can find the fourth and fifth order derivatives.

Calculate the fourth and fifth order derivatives at \(t_0\).

Differentiate \(y'(t)\) with respect to t to find \(y'''(t)\), and differentiate \(y'''(t)\) with respect to t to find \(y^{(4)}(t)\): $$y'''(t) = 2$$ $$y^{(4)}(t) = 0$$ Differentiate \(y^{(4)}(t)\) with respect to t to find \(y^{(5)}(t)\): $$y^{(5)}(t) = 0$$ The higher order derivatives \(y^{(4)}(t_0)\) and \(y^{(5)}(t_0)\) at \(t_0=0\) are: $$y^{(4)}(0) = 0$$ $$y^{(5)}(0) = 0$$

Construct the fifth-degree Taylor polynomial.

Now we can construct the fifth degree Taylor polynomial approximation of the solution: $$P_5(t) = y(0) + y'(0)(t - 0) + \frac{y''(0)}{2!}(t - 0)^2 + \frac{y^{(3)}(0)}{3!}(t - 0)^3 + \frac{y^{(4)}(0)}{4!}(t - 0)^4 + \frac{y^{(5)}(0)}{5!}(t - 0)^5$$ $$P_5(t) = 1 + 2t$$

Calculate the error at \(t=t_0+0.1\).

The exact solution \(y(t) = t^2 + 1\). Let's calculate the error at \(t=t_0+0.1\) for both the exact solution and the Taylor polynomial approximation: $$|Error| = |y(t_0 + 0.1) - P_5(t_0 + 0.1)|$$ $$|Error| = |(0.1)^2 + 1 - (1 + 2(0.1))|$$ $$|Error| = |0.01|$$ The error at \(t=t_0+0.1\) is 0.01.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives.
The function typically represents a real-world quantity, and its derivatives establish how that quantity changes over time.
For example, in this exercise, the differential equation is given by \(y''' - y' = 0\).
This describes the relationship between the third derivative of a function \(y(t)\) and its first derivative. Here's how differential equations are used:

• Describe dynamic systems such as motion, heat, and waves.
• Model growth and decay, like population dynamics or radioactive decay.
• Help predict future behavior of a system based on current or past trends.

To solve a differential equation, we usually integrate it. For instance, integrating \(y''' - y' = 0\) with respect to \(t\) gives the functions \(y'(t)\) and \(y(t)\) in the exercise.

Initial Value Problem

An initial value problem (IVP) involves finding a function or set of functions that satisfy a differential equation and meet specified conditions, known as initial conditions.
In this context, we look for a solution where specific values are assigned to the function and its derivatives at a particular point.
In the problem we're examining, we have several initial conditions:

• \(y(0) = 1\) - The initial value of the function at \(t_0\).
• \(y'(0) = 2\) - The initial rate of change at \(t_0\).
• \(y''(0) = 0\) - The initial acceleration at \(t_0\).

These conditions are crucial as they help determine the constants in the general solution obtained from integrating the differential equation.
Through solving the initial value problem, we precisely pin down \(C_1, C_2, C_3,\) and \(C_4\) so that the solution fits the specific situation described by the initial conditions.

Error Calculation

When approximating complex solutions using methods like Taylor polynomials, calculating the error gives us insight into how close our approximation is to the exact solution.
This error is determined by comparing the approximate solution with the actual function at a given point.For this problem:

• The exact function is \(y(t) = t^2 + 1\).
• The Taylor polynomial approximation up to the fifth degree is \(P_5(t) = 1 + 2t\).

To compute the error, substitute \(t = t_0 + 0.1\) (which is 0.1 here) into both the exact solution and the approximation:

• Calculate the exact value: \(y(0.1) = (0.1)^2 + 1 = 1.01\).
• Calculate the polynomial value: \(P_5(0.1) = 1 + 2 \times 0.1 = 1.2\).

The absolute error is then \(|1.01 - 1.2| = 0.01\).
This tells us the approximation diverges by 0.01 units from the exact solution at \(t = 0.1\). This kind of calculation is essential to understand the accuracy of approximations in mathematical and real-world applications.

Higher Order Derivatives

Higher order derivatives are derivatives of a function that go beyond the first derivative.
These derivatives give us deeper insight into the behavior of a function.In our example, we examined derivatives up to the fifth order. Here's a quick breakdown:

• The first derivative, \(y'(t)\), represents the rate of change or the slope of the function.
• The second derivative, \(y''(t)\), indicates the function's curvature or acceleration.
• Higher derivatives like \(y'''(t)\), \(y^{(4)}(t)\), and \(y^{(5)}(t)\) reveal subtle changes in the function's behavior over time.

In this exercise, you find these derivatives by repeatedly differentiating: after resolving \(y'(t) = 2t\), continue to find \(y'''(t) = 2\) and beyond.
These derivatives contribute to forming the accurate Taylor polynomial, which approximates the function.
Understanding higher order derivatives equips students to model and analyze real-world systems more effectively.