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Chapter 4: Problem 6

A hand-held calculator will suffice for Problems 1 through \(8 .\) In each problem an initial value problem and its exact solution are given. Approximate the values of \(x(0.2)\) and \(y(0.2)\) in three ways: (a) by the Euler method with two steps of size \(h=0.1 ;\) (b) by the improved Euler method with a single step of size \(h=0.2 ;\) and \((c)\) by the Runge-Kutta method with \(a\) single step of size \(h=0.2\). Compare the approximate values with the actual values \(x(0.2)\) and \(y(0.2)\) $$ \begin{aligned} &x^{\prime}=x-2 y, x(0)=0 \\ &y^{\prime}=2 x+y, y(0)=4 \\ &x(t)=-4 e^{t} \sin 2 t, y(t)=4 e^{t} \cos 2 t \end{aligned} $$

Short Answer

Expert verified
Euler: \(x=-1.76, y=4.68\). Heun's: \(x=-2.04, y=4.68\). Runge-Kutta: \(x=-2.08, y=4.76\). Closest: Runge-Kutta.

Step by step solution

01

Exact Solution Calculation

First, let's find the exact values of \( x(0.2) \) and \( y(0.2) \) using the given functions. Plug \( t = 0.2 \) into the equations:\[x(0.2) = -4 e^{0.2} \, \sin(2 \times 0.2) = -4 e^{0.2} \, \sin(0.4),\]\[y(0.2) = 4 e^{0.2} \, \cos(2 \times 0.2) = 4 e^{0.2} \, \cos(0.4).\]Using a calculator, solve for exact values. This results in approximately:- \( x(0.2) \approx -3.0733 \)- \( y(0.2) \approx 4.7674 \).
02

Euler Method Approximation

The Euler method for \( x'(t) = x - 2y \) and \( y'(t) = 2x + y \) with step size \( h = 0.1 \):1. Initial values: \( x_0 = 0, y_0 = 4 \).2. First step (\( t=0.1 \)): - \( x_1 = x_0 + h(x_0 - 2y_0) = 0 + 0.1(0 - 8) = -0.8 \) - \( y_1 = y_0 + h(2x_0 + y_0) = 4 + 0.1(0 + 4) = 4.4 \)3. Second step (\( t=0.2 \)): - \( x_2 = x_1 + h(x_1 - 2y_1) = -0.8 + 0.1(-0.8 - 8.8) = -1.76 \) - \( y_2 = y_1 + h(2x_1 + y_1) = 4.4 + 0.1(-1.6 + 4.4) = 4.68 \)Thus, \( x(0.2) \approx -1.76 \) and \( y(0.2) \approx 4.68 \).
03

Improved Euler (Heun's) Method

For the Improved Euler method with a single step of \( h = 0.2 \):1. Initial values: \( x_0 = 0, y_0 = 4 \).2. Compute tentative step (predict): - \( x_{ ext{temp}} = x_0 + h(x_0 - 2y_0) = 0 + 0.2(0 - 8) = -1.6 \) - \( y_{ ext{temp}} = y_0 + h(2x_0 + y_0) = 4 + 0.2(0 + 4) = 4.8 \)3. Compute corrector step: - \( x_1 = x_0 + \frac{h}{2}((x_0 - 2y_0) + (x_{ ext{temp}} - 2y_{ ext{temp}})) \) \( = 0 + 0.1(-8 - 11.2) = -1.92 \) - \( y_1 = y_0 + \frac{h}{2}((2x_0 + y_0) + (2x_{ ext{temp}} + y_{ ext{temp}})) \) \( = 4 + 0.1(4 + 1.6) = 4.68 \)Therefore, \( x(0.2) \approx -2.04 \) and \( y(0.2) \approx 4.68 \).
04

Runge-Kutta Method Calculation

Apply the Runge-Kutta method for one step with \( h = 0.2 \):1. Initial values: \( x_0 = 0, y_0 = 4 \)2. Calculate \( k \) values: - \( k_1^x = 0 - 2 \cdot 4 = -8 \) - \( k_1^y = 2 \cdot 0 + 4 = 4 \) - \( k_2^x = (0 + 0.1 \cdot (-8)) - 2(4 + 0.1 \cdot 4) = -1.6 - 2 \cdot 4.4 = -10.4 \) - \( k_2^y = 2 \cdot (-0.8) + 4.4 = -1.6 + 4.4 = 2.8 \) - \( k_3^x = (0 + 0.1 \cdot (-10.4)) - 2(4 + 0.1 \cdot 2.8) = -1.04 - 2 \cdot 4.28 = -9.6 \) - \( k_3^y = 2 \cdot (-1.04) + 4.28 = -2.08 + 4.28 = 2.2 \) - \( k_4^x = (0 + 0.2 \cdot (-9.6)) - 2(4 + 0.2 \cdot 2.2) = -1.92 - 2 \cdot 4.44 = -10.8 \) - \( k_4^y = 2 \cdot (-1.92) + 4.44 = -3.84 + 4.44 = 0.6 \)3. Calculate final values: - \( x_1 = 0 + \frac{0.2}{6}(-8 + 2(-10.4 + -9.6) + (-10.8)) = -2.08 \) - \( y_1 = 4 + \frac{0.2}{6}(4 + 2(2.8 + 2.2) + 0.6) = 4.76 \) Thus, \( x(0.2) \approx -2.08 \) and \( y(0.2) \approx 4.76 \).
05

Comparison of Methods

Compare the approximations to the actual values:- **Exact Values**: \( x(0.2) \approx -3.0733, y(0.2) \approx 4.7674 \).- **Euler Method**: \( x(0.2) \approx -1.76, y(0.2) \approx 4.68 \).- **Improved Euler (Heun's) Method**: \( x(0.2) \approx -2.04, y(0.2) \approx 4.68 \).- **Runge-Kutta Method**: \( x(0.2) \approx -2.08, y(0.2) \approx 4.76 \).The Runge-Kutta method gives the closest approximation to the exact values.

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Key Concepts

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

Euler Method
The Euler Method is a simple and direct approach for solving initial value problems numerically. It works by iterating over small steps to estimate the solution of a differential equation. Here鈥檚 how it generally functions:
  • Start with an initial condition: given values for the function and its derivative at the starting point.
  • Use a step size, here denoted by \( h \), to project forward by assuming the slope at the initial point extends across the interval.
  • The formula used is: \( y_{n+1} = y_n + h \, f(x_n, y_n) \), where \( f(x, y) \) is the derivative function.
In this exercise, we applied the Euler Method with a step size of \( h = 0.1 \). It's quick and easy but often not highly accurate, as seen in the approximations of \( x(0.2) = -1.76 \) and \( y(0.2) = 4.68 \). While a valuable tool for understanding numerical methods, Euler's simplicity limits its precision when compared to other methods.
Improved Euler Method
Also known as Heun's Method, the Improved Euler Method enhances the accuracy of the basic Euler technique by adding a predictor-corrector step. Here's how it operates:
  • A preliminary estimate is computed using the standard Euler calculation.
  • This estimate is then averaged with a new slope calculated at the endpoint of the interval.
  • The corrector step yields a more refined approximation for the function鈥檚 value.
For this exercise, the Improved Euler Method was used with a single step size of \( h = 0.2 \), resulting in approximations of \( x(0.2) = -2.04 \) and \( y(0.2) = 4.68 \). This midstep refinement corrects some errors inherent in the basic Euler method, offering improved accuracy without extra complexity.
Runge-Kutta Method
The Runge-Kutta Method, particularly the 4th-order version, is a powerful tool for solving differential equations, offering highly accurate results without the need for tiny step sizes. Here's a brief look at its process:
  • Several intermediate slopes are calculated across the interval, unlike Euler methods which only use the endpoint slopes.
  • Average these slopes to produce an accurate step forward.
  • It is a favored technique among numerical analysts due to its balance of accuracy and efficiency.
Applying the Runge-Kutta method to our problem with a step size of \( h = 0.2 \), we derived approximations: \( x(0.2) = -2.08 \) and \( y(0.2) = 4.76 \). These results align closely with the exact solutions, demonstrating Runge-Kutta鈥檚 capacity to effectively handle initial value problems.
Initial Value Problem
An Initial Value Problem (IVP) serves as a fundamental concept in differential equations, describing a system where the value of the function is specified at a given point. The task is to find the function that satisfies the differential equation throughout a domain starting from this initial condition. Here are the main components:
  • Initial conditions define the state at the beginning and are essential for determining a unique solution.
  • The differential equation lays out a relationship involving the function and its derivatives, dictating how the solution evolves.
In our example, the IVP was defined by two differential equations and initial conditions \( x(0) = 0 \) and \( y(0) = 4 \). Solving such problems is critical in fields requiring predictive modeling of systems, like physics and engineering.
Exact Solution Comparison
Comparing numerical approximations with the exact solutions is vital to gauge the effectiveness and accuracy of numerical methods. Here鈥檚 why this is a crucial practice:
  • Exact solutions provide a benchmark against which we can measure the deviation or error of numerical methods.
  • This comparison allows for evaluating the trade-offs between computational efficiency and precision.
  • Understanding errors can inform decisions about method choice or adjustments in step size for future calculations.
In our problem, the comparisons showed that while all methods approximated the correct course, the Runge-Kutta method stood out by minimizing the gap to the exact values of \( x(0.2) \approx -3.0733 \) and \( y(0.2) \approx 4.7674 \). Such insights are invaluable for making informed choices in computational problem solving.

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Most popular questions from this chapter

Suppose that a crossbow bolt is shot straight upward with initial velocity \(288 \mathrm{ft} / \mathrm{s}\). If its deceleration due to air resistance is \((0.04) v\), then its height \(x(t)\) satisfies the initial value problem $$ x^{\prime \prime}=-32-(0.04) x^{\prime} ; \quad x(0)=0, \quad x^{\prime}(0)=288 $$ Find the maximum height that the bolt attains and the time required for it to reach this height.

A hand-held calculator will suffice for Problems 1 through \(8 .\) In each problem an initial value problem and its exact solution are given. Approximate the values of \(x(0.2)\) and \(y(0.2)\) in three ways: (a) by the Euler method with two steps of size \(h=0.1 ;\) (b) by the improved Euler method with a single step of size \(h=0.2 ;\) and \((c)\) by the Runge-Kutta method with \(a\) single step of size \(h=0.2\). Compare the approximate values with the actual values \(x(0.2)\) and \(y(0.2)\) $$ \begin{aligned} &x^{\prime}=2 x-5 y, x(0)=2 \\ &y^{\prime}=4 x-2 y, y(0)=3 \\ &x(t)=2 \cos 4 t-\frac{11}{4} \sin 4 t, y(t)=3 \cos 4 t+\frac{1}{2} \sin 4 t \end{aligned} $$

Find general solutions of the linear systems in Problems 1 through 20. If initial conditions are given, find the particular solution that satisfies them. In Problems 1 through 6, use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system. $$ 2 y^{\prime}-x^{\prime}=x+3 y+e^{t}, 3 x^{\prime}-4 y^{\prime}=x-15 y+e^{-t} $$

Problems 1 through 10, transform the given differential equation or system into an equivalent system of first-order differential equations. $$ x^{\prime \prime}=(1-y) x, y^{\prime \prime}=(1-x) y $$

A hand-held calculator will suffice for Problems 1 through \(8 .\) In each problem an initial value problem and its exact solution are given. Approximate the values of \(x(0.2)\) and \(y(0.2)\) in three ways: (a) by the Euler method with two steps of size \(h=0.1 ;\) (b) by the improved Euler method with a single step of size \(h=0.2 ;\) and \((c)\) by the Runge-Kutta method with \(a\) single step of size \(h=0.2\). Compare the approximate values with the actual values \(x(0.2)\) and \(y(0.2)\) $$ \begin{aligned} &x^{\prime}=x+2 y, x(0)=0 \\ &y^{\prime}=2 x+y, y(0)=2 \\ &x(t)=e^{3 t}-e^{-t}, y(t)=e^{3 t}+e^{-t} \end{aligned} $$
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