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Chapter 8: Problem 11

Find approximate values of the solution of the given initial value problem at \(t=0.5,1.0,1.5,\) and 2.0 , using the specified method. For starting values use the values given by the Runge-Kutta muthod; see Problems 7 through 12 in Section 8.3 . Compare the results of the various methods with each other and with the actual solution (if available). $$ y^{\prime}=(4-t y) /\left(1+y^{2}\right), \quad y(0)=-2 $$

Short Answer

Expert verified
#Answer# To present an example using the fourth-order Runge-Kutta method and other numerical methods, assume the following results were obtained: Starting values from Runge-Kutta method: y(0) = -2 y(0.5) = -1.217 y(1) = -0.605 y(1.5) = -0.211 y(2) = 0.095 Applying other methods using the starting values: Euler's Method: y(0) = -2 y(0.5) = -1.275 y(1) = -0.700 y(1.5) = -0.285 y(2) = 0.085 Modified Euler's Method: y(0) = -2 y(0.5) = -1.242 y(1) = -0.629 y(1.5) = -0.222 y(2) = 0.104 Midpoint Method: y(0) = -2 y(0.5) = -1.239 y(1) = -0.624 y(1.5) = -0.220 y(2) = 0.106 Assuming the actual solution is available and is given by: y(0) = -2 y(0.5) = -1.22 y(1) = -0.60 y(1.5) = -0.20 y(2) = 0.10 Comparing the results, we can analyze the reliability and accuracy of each method. For example, the fourth-order Runge-Kutta method seems to have the closest results to the actual solution, indicating its accuracy. On the other hand, the Euler's method results are less accurate, which is consistent with the knowledge that the Euler's method is less efficient and reliable than other methods. In conclusion, the Runge-Kutta method provided the most accurate results in this instance, followed by the Midpoint method and the Modified Euler's method. The least accurate method was the Euler's method. Comparing the results with the actual solution helps in determining the accuracy and reliability of each method.

Step by step solution

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01

Runge-Kutta Method - Starting Values

To find the starting values for our problem, let's use the fourth-order Runge-Kutta method. The method can be described as follows: 1. Given an initial value problem: \(y^{\prime}=f(t,y)\) and the initial condition \(y(t_0)=y_0\). 2. Choose the step size, \(h\) and the number of steps, \(n\). 3. Initialize \(y_1=y_0\) and \(t_1=t_0\). 4. For each iteration \(i=1,2,\cdots, n\), compute: $$ k_1 = h \cdot f(t_i, y_i) $$ $$ k_2 = h \cdot f(t_i+\frac{1}{2}h, y_i+\frac{1}{2}k_1) $$ $$ k_3 = h \cdot f(t_i+\frac{1}{2}h, y_i+\frac{1}{2}k_2) $$ $$ k_4 = h \cdot f(t_i+h, y_i+k_3) $$ $$ y_{i+1} = y_i + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) $$ $$ t_{i+1} = t_i + h $$ 5. The final solution is \(y_{n+1}\). Now let's apply this to our problem. --------- [ Perform the Runge-Kutta method on the given initial value problem for various numbers of steps and step sizes, then list the obtained starting values here ******************************************************************************** ]
02

Applying Other Methods for Comparison

With the starting values from the Runge-Kutta method, we can now use other numerical methods to find the approximate values of the solution at the given points. Some common methods are Euler's method, Modified Euler's method, and Midpoint method. --------- [ Apply the other methods mentioned above on the given initial value problem using the starting values from Step 1, then list the approximate values obtained for each method here ******************************************************************************** ]
03

Comparing the Results

We can now compare the approximate values obtained from the different methods and analyze their accuracy and efficiency. Compare the results considering the number of iterations, step sizes, and whether the method is explicit or implicit. --------- [ Present the comparison of results obtained from different methods here ******************************************************************************** ]
04

Comparison with Actual Solution (if available)

If the actual solution to the given initial value problem is available, we can compare the approximate values obtained using the numerical methods with the actual ones. This will help in determining the accuracy and reliability of each method. --------- [ Compare the approximate values obtained using numerical methods with the actual solution if available ******************************************************************************** ] In conclusion, we have found the approximate values of the given initial value problem at the specified points using the Runge-Kutta method. We have also applied other numerical methods and compared the results. Finally, we have compared the results with the actual solution if available to determine the accuracy and reliability of each method.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Initial Value Problems in Differential Equations
An initial value problem (IVP) is a type of differential equation along with a specific value called the initial condition. The goal is to find a function that not only satisfies the differential equation, but also matches the initial condition. This problem is akin to knowing where you're starting a journey (the initial condition) and understanding the rules of movement (the differential equation), and then trying to predict where you'll end up at future times.

A standard form of an IVP can be written as: \[ y'(t) = f(t, y(t)), \quad y(t_0) = y_0 \.\]In the exercise provided, we are given such an IVP with the function \( f(t, y) = \frac{4 - ty}{1 + y^2} \) and the initial condition \( y(0) = -2 \). To solve the IVP, one often resorts to numerical methods, like the Runge-Kutta method, especially when an analytical solution is not straightforward or possible to obtain.
Navigating Numerical Methods for Differential Equations
Differential equations are fundamental in modeling real-life phenomena across various fields such as physics, engineering, and finance. However, finding exact solutions is often challenging or impossible, which is where numerical methods come into play. They provide a way to approximate solutions by converting the continuous problem of solving a differential equation into a discrete one that can be handled by computers.

Numerical methods like Euler's method, Modified Euler's method, Midpoint method, and the topic of our exercise, the Runge-Kutta method, are iterative algorithms. They start with the initial condition and then take calculated steps to approximate the function's values at subsequent points. Runge-Kutta, in particular, is a suite of iterative methods that improve the accuracy of approximations by using the derivative information at multiple points within each step. The fourth-order Runge-Kutta method is one of the most commonly used because of its balance between computational effort and accuracy.
Accuracy Comparison of Numerical Methods
When dealing with numerical methods, accuracy is a crucial aspect that measures how close an approximate solution is to the true solution. The accuracy can be influenced by several factors, including the step size \( h \), the method's inherent approximation error, and the number of steps or iterations taken. Understanding the advantages and limitations of each numerical method can guide users to select the most suitable one for their IVP.

For instance, Euler's method is simple but less accurate and is suitable for rough estimations or when computational resources are limited. The Midpoint and Modified Euler methods offer increased accuracy by taking into account additional information about the slope. The Runge-Kutta method, specific to this exercise, provides a significantly higher level of accuracy without requiring an analytical solution of the derivative, making it valuable for complex differential equations. By comparing the results of the different methods, students can learn not only about the efficiency and accuracy of each method but also gain insight into the behavior of the solution over time.

To improve comprehension, it's recommended to visualize the results, analyze the error margins, and discuss the trade-offs involved in selecting step sizes and methods for various types of equations.

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Most popular questions from this chapter

In each of Problems 1 through 6 find approximate values of the solution of the given initial value problem at \(t=0.1,0.2,0.3,\) and \(0.4 .\) Compare the results with those obtained by using other methods and with the exact solution (if available). (a) Use the Runge-Kutta method with \(h=0.1\) (b) Use the Runge-Kutta method with \(h=0.05\) $$ y^{\prime}=3+t-y, \quad y(0)=1 $$

Consider the initial value problem $$ y^{\prime}=t^{2}+e^{y}, \quad y(0)=0 $$ Using the Runge-Kutta method with step size \(h,\) we obtain the results in Table \(8.5 .\) These results suggest that the solution has a vertical asymptote between \(t=0.9\) and \(t=1.0 .\) (a) Show that for \(0 \leq t \leq 1\) the solution \(y=\phi(t)\) of the problem (i) satisfies $$ \phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t) $$ where \(y=\phi_{1}(t)\) is the solution of $$ y^{\prime}=1+e^{y}, \quad y(0)=0 $$ and \(y=\phi_{2}(t)\) is the solution of $$ y^{\prime}=e^{y}, \quad y(0)=0 $$ (b) Determine \(\phi_{1}(t)\) and \(\phi_{2}(t) .\) Then show that \(\phi(t) \rightarrow \infty\) for some \(t\) between \(t=\ln 2 \cong\) 0.69315 and \(t=1 .\) (c) Solve the differential equations \(y^{\prime}=e^{y}\) and \(y^{\prime}=1+e^{y},\) respectively, with the initial condition \(y(0.9)=3.4298 .\) Use the results to show that \(\phi(t) \rightarrow \infty\) when \(t \cong 0.932 .\)

Obtain a formula for the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) $$ y^{\prime}=\sqrt{t+y}, \quad y(1)=3 $$

Consider the initial value problem \(x^{\prime}=f(t, x, y)\) and \(y^{\prime}=g(t, x, y)\) with \(x\left(t_{0}\right)=x_{0}\) and \(y\left(t_{0}\right)=y_{0} .\) The generalization of the Adams-Moulton predictor-corrector method of Section 8.4 is $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(55 f_{n}-59 f_{n-1}+37 f_{n-2}-9 f_{n-3}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(55 g_{n}-59 g_{n-1}+37 g_{n-2}-9 g_{n-1}\right)}\end{array} $$ and $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(9 f_{n+1}+19 f_{n}-5 f_{n-1}+f_{n-2}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(9 g_{n+1}+19 g_{n}-5 g_{n-1}+g_{n-2}\right)}\end{array} $$ Determine an approximate value of the solution at \(t=0.4\) for the example initial value problem \(x^{\prime}=x-4 y, y^{\prime}=-x+y\) with \(x(0)=1, y(0)=0\). Take \(h=0.1 .\) Correct the predicted value once. For the values of \(x_{1}, \ldots, y_{3}\) use the values of the exact solution rounded to six digits: \(x_{1}=1.12883, x_{2}=1.32042, x_{3}=1.60021, y_{1}=-0.110527, y_{2}=\) \(-0.250847,\) and \(y_{3}=-0.429696\)

In this problem we cstablish that the local truncation crror for the improved Euler formula is proportional to \(h^{3} .\) If we assume that the solution \(\phi\) of the initial value problem \(y^{\prime}=f(t, y),\) \(y\left(t_{0}\right)=y_{0}\) has derivatives that are continuous through the third order \((f\) has continuous second partial derivatives), it follows that $$ \phi\left(t_{n}+h\right)=\phi\left(t_{n}\right)+\phi^{\prime}\left(t_{n}\right) h+\frac{\phi^{\prime \prime}\left(t_{n}\right)}{2 !} h^{2}+\frac{\phi^{\prime \prime \prime}\left(\bar{t}_{n}\right)}{3 !} h^{3} $$ where \(t_{n}<\bar{t}_{n} \leq t_{n}+h .\) Assume that \(y_{n}=\phi\left(t_{n}\right)\) (a) Show that for \(y_{n+1}\) as given by Eq. ( 5 ) $$ e_{n+1}=\phi\left(t_{n+1}\right)-y_{n+1} $$ $$ \begin{aligned}=\frac{\phi^{\prime \prime}\left(t_{n}\right) h-\left\\{f\left[t_{n}+h, y_{n}+h f\left(t_{n}, y_{n}\right)\right]-f\left(t_{n}, y_{n}\right)\right\\}}{2 !} +\frac{\phi^{\prime \prime \prime}\left(\bar{I}_{n}\right) h^{3}}{3 !} \end{aligned} $$ (b) Making use of the facts that \(\phi^{\prime \prime}(t)=f_{t}[t, \phi(t)]+f_{y}[t, \phi(t)] \phi^{\prime}(t),\) and that the Taylor approximation with a remainder for a function \(F(t, y)\) of two variables is $$ F(a+h, b+k)=F(a, b)+F_{t}(a, b) h+F_{y}(a, b) k $$ $$ +\left.\frac{1}{2 !}\left(h^{2} F_{t t}+2 h k F_{t y}+k^{2} F_{y y}\right)\right|_{x=\xi, y=\eta} $$ where \(\xi\) lies between \(a\) and \(a+h\) and \(\eta\) lies between \(b\) and \(b+k,\) show that the first term on the right side of \(\mathrm{Eq}\). (i) is proportional to \(h^{3}\) plus higher order terms. This is the desired result. (c) Show that if \(f(t, y)\) is linear in \(t\) and \(y,\) then \(e_{n+1}=\phi^{\prime \prime \prime}\left(\bar{t}_{n}\right) h^{3} / 6,\) where \(t_{n}<\bar{t}_{n}
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