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Exencises 18-22: In each exercise, (a) Using step size \(h=01\), compute the first estimate \(y_{1}\) using Euler's method and the Runge-Kutta method (9). Let these estimates be denoted by \(y_{1}^{\&}\) and \(y_{1}^{R K}\), respectively. (b) Solve the problem analytically. (c) Compute the errors \(\left|y\left(t_{1}\right)-y_{1}\right|\) for the two estimates obtained in (a). \(y^{\prime}=-y+t, \quad y(0)=0\)

Short Answer

Expert verified
Question: Compare the errors between Euler's method and Runge-Kutta method for the given differential equation at the point \(t_1\). Answer: At \(t_1\), the error for Euler's method is 0, while the error for the Runge-Kutta method is approximately 0.00833333.

Step by step solution

01

Apply Euler's method #

Let's first find the slope at the initial point using the given differential equation. We have:\begin{align*}y' &= –y + t \\y'(0) &= -y(0) + 0 = 0\end{align*}Since \(h=0.1\), the first estimate, denoted as \(y_1^{E}\) is:\begin{align*}y_1^{E} &= y_0 + hy'(0) \\&= 0 + 0.1 \cdot 0 \\&= 0\end{align*}
02

Apply Runge-Kutta method #

For the Runge-Kutta method, we need to compute the following slopes: \(k_1 = -y_0 + t_0 = -0 + 0 = 0\) \(k_2 = -\left(y_0 + \frac{1}{2}h k_1\right) + \left(t_0 + \frac{1}{2}h\right) = 0 + (0 + 0.5\cdot0.1) = 0.05\) \(k_3 = -\left(y_0 + \frac{1}{2}h k_2\right) + \left(t_0 + \frac{1}{2}h\right) = 0 + (0 + 0.5\cdot0.05) = 0.025\) \(k_4 = -\left(y_0 + hk_3\right) + (t_0 + h) = -0.1(0.025) = 0.0975\) Now compute the first estimate, \(y_1^{RK}\): \begin{align*}y_1^{RK} &= y_0 + \frac{1}{6}h(k_1 + 2k_2 + 2k_3 + k_4) \\&= 0 + \frac{1}{6}(0.1)(0 + 2\cdot0.05 + 2\cdot0.025 + 0.0975) \\&= 0.00833333\end{align*}
03

Solve the problem analytically #

The given differential equation is a linear first-order differential equation. We solve it analytically using the integrating factor method. Given equation: \(y' = -y + t\) Integrating factor: \(I(t) = e^{\int -1\, dt} = e^{-t}\) Multiply the given equation with the integrating factor: \(e^{-t}y' + e^{-t}y = te^{-t}\) Now, this equation is in the form of \((yI)’ = te^{-t}\) Integrate both sides to obtain: \begin{align*}yI &= \int te^{-t}\, dt + C \\y = (e^t) \left( \int te^{-t}\, dt + C\right)\end{align*} We can solve the integral by integration by parts by letting \(\int{u\, dv} = uv–\int{v\, du}\). Let \(u=t\), \(du=dt\), \(dv=e^{-t}dt\), and \(v=-e^{-t}\). So, we have:\begin{align*}ye^t &= -te^{-t}e^t + \int e^{-t}e^t\, dt + C \\&= -t + \int dt + C \\&= -t + t + C\end{align*} As \(y(0) = 0\), we find \(C = 0\). Consequently, the analytical solution is: \(y(t) = t - t = 0\).
04

Compute errors #

Now that we have the analytical solution, let's compute the errors between the analytical solution and the estimates for \(t_1 = 0.1\): Error for Euler's method: \[\text{Error}_E = \left|y(t_1) - y_1^{E}\right| = \left|y(0.1) - 0\right| = 0\] Error for Runge-Kutta method: \[\text{Error}_{RK} = \left|y(t_1) - y_1^{RK}\right| = \left|y(0.1) - 0.00833333\right| = 0.00833333\] So, the errors for both methods are roughly the same at \(t_1\). The error for Euler's method is 0, while the error for the Runge-Kutta method is approximately 0.00833333.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are fundamental in expressing the laws of nature and various other phenomena where quantities change in relation to each other. An example provided in the exercise is the first-order differential equation \(y' = -y + t\), where \(y'\) denotes the derivative of \(y\) with respect to \(t\).

Solving such equations can offer insights into the dynamics of a system, such as population growth, the motion of particles, or the cooling of an object. The initial condition, given as \(y(0)=0\) in our example, is a specific situation that defines the value of the function at a certain point, making the solution to the differential equation unique.

Analytic solutions to differential equations involve finding a closed-form expression, which exactly satisfies the equation and the initial conditions. When such an explicit formula cannot be obtained, numerical methods are employed to approximate the solution.
Numerical Methods
Numerical methods are algorithms used to find approximate solutions to complex problems that cannot be solved analytically. These methods are particularly useful for solving differential equations where finding a closed-form solution may not be feasible. The numerical methods mainly differ in their approach to approximation and their level of accuracy.

One such method is Euler's Method, a straightforward technique that begins with an initial value and uses the slope at that point to estimate the value of the function at the next step. Specifically, if you have a step size \(h\) and an initial point \((t_0, y_0)\), Euler's Method estimates the next point by \(y_1 = y_0 + hf(t_0, y_0)\).

Another more advanced technique is the Runge-Kutta Method, which achieves greater accuracy through a weighted average of slopes calculated at several points. In the fourth-order Runge-Kutta Method, which was used in the exercise, slopes at the starting point, the midpoint, and the endpoint of an interval are calculated to determine the next estimate of the solution. These intermediate calculations provide a more precise approximation of the integral of the differential equation over a small interval.
Error Analysis
Error analysis in numerical methods involves evaluating the difference between the approximate numerical solution and the exact analytical solution of a differential equation. This difference is known as the error, and it can be caused by various factors such as the step size, truncation of series, or round-off errors due to floating-point arithmetic. Understanding and minimizing error is critical for ensuring the reliability of numerical simulations.

In the exercise at hand, the errors were computed for both Euler's and Runge-Kutta methods by comparing the numerical estimates with the analytic solution. The absolute error was given by \(\left|y(t_1) - y_1\right|\), where \(y(t_1)\) is the exact solution at the point \(t_1\) and \(y_1\) is the estimated value from the numerical method.

The accuracy of a numerical method can often be improved by decreasing the step size, but this comes at the cost of increased computational resources. Comparing different numerical methods, like Euler's and Runge-Kutta, allows students to see how a more refined approach can significantly reduce the error even with the same step size.

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Most popular questions from this chapter

An object undergoes one-dimensional motion along the \(x\)-axis subject to the given decelerating forces. At time \(t=0\), the object's position is \(x=0\) and its velocity is \(v=v_{0}\). In each case, the decelerating force is a function of the object's position \(x(t)\) or its velocity \(v(t)\) or both. Transform the problem into one having distance \(x\) as the independent variable. Determine the position \(x_{f}\) at which the object comes to rest. (If the object does not come to rest, \(x_{f}=\infty\).) $$ m \frac{d v}{d t}=-k x v^{2} $$

In each exercise, the general solution of the differential equation \(y^{\prime}+p(t) y=g(t)\) is given, where \(C\) is an arbitrary constant. Determine the functions \(p(t)\) and \(g(t)\). \(y(t)=C e^{t^{2}}+2\)

An auditorium is \(100 \mathrm{~m}\) in length, \(70 \mathrm{~m}\) in width, and \(20 \mathrm{~m}\) in height. It is ventilated by a system that feeds in fresh air and draws out air at the same rate. Assume that airborne impurities form a well-stirred mixture. The ventilation system is required to reduce air pollutants present at any instant to \(1 \%\) of their original concentration in \(30 \mathrm{~min}\). What inflow (and outflow) rate is required? What fraction of the total auditorium air volume must be vented per minute?

A tank originally contains \(5 \mathrm{lb}\) of salt dissolved in 200 gal of water. Starting at time \(t=0\), a salt solution containing \(0.10 \mathrm{lb}\) of salt per gallon is to be pumped into the tank at a constant rate and the well-stirred mixture is to flow out of the tank at the same rate. (a) The pumping is to be done so that the tank contains \(15 \mathrm{lb}\) of salt after \(20 \mathrm{~min}\) of pumping. At what rate must the pumping occur in order to achieve this objective? (b) Suppose the objective is to have \(25 \mathrm{lb}\) of salt in the tank after \(20 \mathrm{~min}\). Is it possible to achieve this objective? Explain.

When modeling the action of drag chutes and parachutes, we have assumed that the chute opens instantaneously. Real devices take a short amount of time to fully open and deploy. In this exercise, we try to assess the importance of this distinction. Consider again the assumptions of Exercise 2 . A 3000 -lb dragster is moving on a straight track at a speed of \(220 \mathrm{mph}\) when, at time \(t=0\), the drag chute is opened. If we assume that the drag force is proportional to velocity and that the chute opens instantaneously, the differential equation to solve is \(m v^{\prime}=-k v\). If we assume a short deployment time to open the chute, a reasonable differential equation might be \(m v^{\prime}=-k(\tanh t) v\). Since \(\tanh (0)=0\) and \(\tanh (1) \approx 0.76\), it will take about \(1 \mathrm{sec}\) for the chute to become \(76 \%\) deployed in this model. Assume \(k=25 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). Solve the two differential equations and determine in each case how long it takes the vehicle to slow to \(50 \mathrm{mph}\). Which time do you anticipate will be larger? (Explain.) Is the idealization of instantaneous chute deployment realistic?

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