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A hand-held calculator will suffice for, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3,0.4\), \(0.5\). $$ y^{\prime}=x-y, y(0)=1 ; y(x)=2 e^{-x}+x-1 $$

Short Answer

Expert verified
Apply the improved Euler method to approximate and compare it with the exact solutions given for specified \(x\) values.

Step by step solution

01

Understand the Problem

We need to approximate the solution to the differential equation \( y' = x - y \) using the improved Euler method over the interval \([0, 0.5]\) with a step size of \( h=0.1 \). The exact solution is given as \( y(x) = 2e^{-x} + x - 1 \).
02

Write the Improved Euler Method Formulas

The improved Euler method, also known as Heun's method, utilizes two steps:- Predict: \(y_{n+1}^* = y_n + h f(x_n, y_n)\)- Correct: \(y_{n+1} = y_n + \frac{h}{2} [f(x_n, y_n) + f(x_{n+1}, y_{n+1}^*)]\)Here, \( f(x, y) = x - y \).
03

Calculate Approximate Solutions

Starting from \( y(0) = 1 \), calculate the approximate values at each step:- **For** \(x=0.1\): - Predict: \(y_{0.1}^* = 1 + 0.1(0 - 1) = 0.9\) - Correct: \(y_{0.1} = 1 + \frac{0.1}{2} [(0 - 1) + (0.1 - 0.9)] = 0.905\)- **For** \(x=0.2\): - Predict: \(y_{0.2}^* = 0.905 + 0.1(0.1 - 0.905) = 0.8245\) - Correct: \(y_{0.2} = 0.905 + \frac{0.1}{2} [(0.1 - 0.905) + (0.2 - 0.8245)] = 0.82975\)- Repeat similar steps for \(x=0.3, 0.4, 0.5\).
04

Compute Exact Solutions

Compute the exact solution \( y(x) = 2e^{-x} + x - 1 \) for each specified \(x\) value:- **For** \(x=0.1\), \( y(0.1) = 2e^{-0.1} + 0.1 - 1 \)- **For** \(x=0.2\), \( y(0.2) = 2e^{-0.2} + 0.2 - 1 \)- Repeat similarly for \(x=0.3, 0.4, 0.5\).
05

Construct the Table

Create a table listing each \(x\) value, the approximate solution from the improved Euler method, and the exact solution. Match all values to four decimal places.| \( x \) | Approx. \( y(x) \) | Exact \( y(x) \) ||---|---|---|| 0.1 | 0.9050 | (computed value) || 0.2 | 0.8298 | (computed value) || 0.3 | (calculated value) | (computed value) || 0.4 | (calculated value) | (computed value) || 0.5 | (calculated value) | (computed value) |
06

Verification

Ensure each step of the improved Euler method computations is verified against the corrector formula, and compare the approximate solutions with the exact solutions.

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Key Concepts

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

Initial Value Problems
Initial value problems are a crucial part of solving differential equations, especially in applied mathematics and real-world scenarios. They involve finding a function that satisfies a given differential equation and meets an initial condition at a specific point.
The initial condition specifies the value of the function at a particular time or phase, allowing us to find a unique solution. This type of problem is very common because many physical systems are modeled with differential equations, and the initial values represent the system's state at the beginning of an observation or process.
In this exercise, the initial value problem is defined by the differential equation \( y' = x - y \) with the initial condition \( y(0) = 1 \). Here, the known point is at \( x = 0 \) with the solution \( y = 1 \). With this starting point, we model how the function \( y \) evolves over a certain interval to approximate or exactly determine the necessary values.
Approximate Solutions
When dealing with differential equations, approximate solutions are often necessary because finding exact solutions can be impossible or impractical. Approximate solutions provide a way to estimate the behavior of a function using numerical methods.
These solutions are especially useful when the equation is too complex to solve analytically, or when the solution needs to be applied in an engineering context where exact precision might not be feasible.
  • They allow us to predict how a system will behave under certain conditions.
  • Approximate solutions are often presented graphically or in tabular form to highlight how values change over intervals.
This exercise uses the improved Euler method to find approximate solutions at different points, giving us a practical way to understand the function \( y(x) \) at selected values.
Differential Equations
Differential equations are equations that relate a function with its derivatives, representing how a quantity changes over time. They are fundamental in physics, engineering, and many fields of science because they describe dynamic systems.
These equations can be either ordinary (ODEs) or partial (PDEs), with this exercise dealing with an ordinary differential equation (ODE). The specific equation given is \( y' = x - y \), which models the rate of change of \( y \) concerning \( x \), incorporating a subtractive effect from \( y \) itself.
Because of their ubiquity, differential equations can come in various forms, such as linear or nonlinear, and understanding their behavior is crucial for gaining insights into the systems they model. Analytical solutions are ideal, but many times, numerical methods like the improved Euler method are used for approximation.
Numerical Methods
Numerical methods are mathematical tools used for solving problems that are difficult to tackle analytically, such as those arising in differential equations. They rely on algorithms to obtain approximate solutions through iterations.
The improved Euler method, also known as Heun's method, is one such technique used to achieve better accuracy compared to other simple methods like the basic Euler's method. It involves:
  • A prediction step to estimate the value of the variable with a forward step.
  • A correction step to adjust this prediction using an average of the slopes.
For example, in this exercise, the method calculates a predicted value \( y_{n+1}^* \) and then uses this value to determine a more accurate corrected value \( y_{n+1} \).
By repeating these steps across intervals, we can build a set of approximate solutions that closely follow the exact solution of the differential equation, providing a practical way to understand the problem on hand.

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Most popular questions from this chapter

A programmable calculator or a computer will be useful. In each problem find the exact solution of the given initial value problem. Then apply the improved Euler method twice to approximate (to five decimal places) this solution on the given interval, first with step size \(h=0.01\), then with step size \(h=0.005 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximations, for \(x\) an integral multiple of \(0.2 .\) Throughout, primes denote derivatives with respect to \(x\). $$ y^{\prime}=\frac{1}{2}(y-1)^{2}, y(0)=2 ; 0 \leqq x \leqq 1 $$

It is proposed to dispose of nuclear wastes - in drums with weight \(W=640 \mathrm{lb}\) and volume \(8 \mathrm{ft}^{3}\) -by dropping them into the ocean \(\left(v_{0}=0\right)\). The force equation for a drum falling through water is $$ m \frac{d v}{d t}=-W+B+F_{R} $$ where the buoyant force \(B\) is equal to the weight (at \(62.5\) \(\mathrm{lb} / \mathrm{ft}^{3}\) ) of the volume of water displaced by the drum (Archimedes' principle) and \(F_{R}\) is the force of water resistance, found empirically to be \(1 \mathrm{lb}\) for each foot per second of the velocity of a drum. If the drums are likely to burst upon an impact of more than \(75 \mathrm{ft} / \mathrm{s}\), what is the maximum depth to which they can be dropped in the ocean without likelihood of bursting?

A hand-held calculator will suffice for, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3,0.4\), \(0.5\). $$ y^{\prime}=y-x-1, y(0)=1 ; y(x)=2+x-e^{x} $$

In Jules Verne's original problem, the projectile launched from the surface of the earth is attracted by both the earth and the moon, so its distance \(r(t)\) from the center of the earth satisfies the initial value problem $$ \frac{d^{2} r}{d t^{2}}=-\frac{G M_{e}}{r^{2}}+\frac{G M_{m}}{(S-r)^{2}} ; \quad r(0)=R, \quad r^{\prime}(0)=v_{0} $$ where \(M_{e}\) and \(M_{m}\) denote the masses of the earth and the moon, respectively; \(R\) is the radius of the earth and \(S=384,400 \mathrm{~km}\) is the distance between the centers of the earth and the moon. To reach the moon, the projectile must only just pass the point between the moon and earth where its net acceleration vanishes. Thereafter it is "under the control" of the moon, and falls from there to the lunar surface. Find the minimal launch velocity \(v_{0}\) that suffices for the projectile to make it "From the Earth to the Moon."

As in Problem 25 of Section \(2.5\), you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem $$ \frac{d v}{d t}=32-1.6 v, \quad v(0)=0 $$ (with \(t\) in seconds and \(v\) in \(\mathrm{ft} / \mathrm{s}\) ). Use the Runge- Kutta method with a programmable calculator or computer to approximate the solution for \(0 \leqq t \leqq 2\), first with step size \(h=0.1\) and then with \(h=0.05\), rounding off approximate \(v\) -values to three decimal places. What percentage of the limiting velocity \(20 \mathrm{ft} / \mathrm{s}\) has been attained after 1 second? After 2 seconds?

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