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Chapter 2: Problem 7

Reducing the Step Size These exercises examine graphically the effects of reducing step size on the accuracy of the numerical solution. A computer or programmable calculator is needed. (a) Use Euler's method to obtain numerical solutions on the specified time interval for step sizes \(h=0.1, h=0.05\), and \(h=0.025\). (b) Solve the problem analytically and plot the exact solution and the three numerical solutions on a single graph. Does the error appear to be getting smaller as \(h\) is reduced? \(y^{\prime}=2 y-1, \quad y(0)=1, \quad 0 \leq t \leq 0.5\)

Short Answer

Expert verified
Answer: Yes, by using Euler's method and comparing the numerical solutions to the analytical solution (\(y(t) = \frac{1}{2}+\frac{1}{2}e^{-2t}\)) graphically, you can observe that the error appears to reduce as the step size decreases.

Step by step solution

01

Euler's Method for h = 0.1

To use Euler's method, first set up a table with columns for time (t), step size (h), and y. Then calculate the slope, y', and update the y-value for each step using the relation: \(y_{i+1} = y_i + h * y^{\prime}_i\) We start with y(0) = 1, and our equation is \(y^{\prime} = 2y - 1\). For \(h = 0.1\): | t | h | y | \(y'\) | |------|-----|-----|------| | 0.0 | 0.1 | 1.0 | 1.0 | | 0.1 | 0.1 | 1.1 | 1.2 | | 0.2 | 0.1 | 1.22| 1.44 | | 0.3 | 0.1 | 1.36| 1.72 | | 0.4 | 0.1 | 1.536|2.07 |
02

Euler's Method for h = 0.05

Repeating the process for \(h = 0.05\): | t | h | y | \(y'\) | |------|------|-------|------| | 0.0 | 0.05 | 1.00 | 1.00 | | 0.05| 0.05 | 1.05 | 1.10 | | 0.1 | 0.05 | 1.105| 1.21 | | 0.15| 0.05 | 1.166| 1.33 | | 0.2 | 0.05 | 1.233| 1.47 | | 0.25| 0.05 | 1.307| 1.61 | | 0.3 | 0.05 | 1.388| 1.78 | | 0.35| 0.05 | 1.477| 1.95 | | 0.4 | 0.05 | 1.576| 2.15 |
03

Euler's Method for h = 0.025

Finally, for \(h = 0.025\) you can follow the same process to build the table (we skip it here to save space).
04

Analytical Solution

To solve the given differential equation analytically: \(y^{\prime} = 2 y - 1\) Using integrating factor \(\mu(t) = e^{\int 2 dt} = e^{2t}\): \(y(t) e^{2t} - \int e^{2t}(2y-1) dt = C\) Solving for \(y(t)\) and putting in the initial condition, \(y(0) = 1\), we get: \(y(t) = \frac{1}{2} + Ce^{-2t}\) Plugging in the initial condition: \(1 = \frac{1}{2} + Ce^{-2*0}\) Solving for C: \(C = \frac{1}{2}\), thus \(y(t) = \frac{1}{2}+\frac{1}{2}e^{-2t}\).
05

Graphical Comparison

Plot the exact solution, \(y(t) = \frac{1}{2}+\frac{1}{2}e^{-2t}\), and the numerical solutions from steps 1-3 on a single graph. By comparing the graphs of the exact and numerical solutions, you can observe that the error appears to reduce with smaller step sizes. It's important to note that while we have only shown reduced error in a graphical sense, this can be further confirmed by actually calculating the error (euclidean distance) between the numerical and analytical solutions for the different step sizes.

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Key Concepts

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

Numerical Solutions
Euler's Method is a fundamental technique used to find numerical solutions to differential equations. When a problem is too complex or lacks a simple solution, numerical methods like Euler's come handy. They provide an approximate solution by solving differential equations step by step.
In Euler's Method, you calculate the next point using the slope (derivative) and the current point. This approach is iterative and repeats as per the chosen step size. For each iteration, you update the value of the dependent variable using a simple formula.
  • Start from an initial point
  • Calculate the derivative at that point
  • Use the derivative to estimate the next point
  • Repeat for multiple steps
This process helps find a numerical solution to the differential equation over a specified interval.
Step Size
The choice of step size, represented by "h", is very crucial in numerical methods like Euler's Method. Step size determines how often the slope is recalculated and the frequency of the updates. Smaller step sizes mean more iterations and generally more accurate solutions.
A small step size captures more details of the function's behavior. It can lead to a closer approximation to the actual solution of the differential equation. However, with smaller step sizes, the computation by hand or computer increases and can require more time.
  • Smaller step size = more accuracy
  • Larger step size = faster, but less accurate
Understanding the balance between step size and computational efficiency is key in numerical analysis.
Differential Equations
Differential equations model many real-world phenomena, describing how a particular quantity changes over time. They are equations that involve functions and their derivatives. The equation given in the exercise, \(y^{\prime} = 2y - 1\), is a first-order linear differential equation.
The importance of differential equations lies in their ability to model growth, decay, motion, heat, and many other dynamic processes. In this exercise, the unknown function \(y(t)\) represents how a certain system evolves over time.
Once mastered, solving differential equations analytically and numerically enables you to predict the future behavior of the system in question.
Analytical Solutions
Unlike numerical solutions, analytical solutions provide exact answers whenever solving differential equations is possible. An analytical solution involves finding a general solution or specific solution that satisfies the differential equation and any given initial conditions.
For the differential equation \(y^{\prime} = 2y - 1\) in the exercise, an analytical solution was found using an integrating factor. This solution gives a precise formula for the function y that can be evaluated at any point within the domain, providing exact results instantly.
Analytical methods include separation of variables, integrating factors, and using characteristic equations. They are invaluable for validating the accuracy and reliability of numerical solutions.

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Most popular questions from this chapter

Active oxygen and free radicals are believed to be exacerbating factors in causing cell injury and aging in living tissue. \({ }^{1}\) These molecules also accelerate the deterioration of foods. Researchers are therefore interested in understanding the protective role of natural antioxidants. In the study of one such antioxidant (Hsian-tsao leaf gum), the antioxidation activity of the substance has been found to depend on concentration in the following way: $$ \frac{d A(c)}{d c}=k\left[A^{*}-A(c)\right], \quad A(0)=0 . $$ In this equation, the dependent variable \(A\) is a quantitative measure of antioxidant activity at concentration \(c\). The constant \(A^{*}\) represents a limiting or equilibrium value of this activity, and \(k\) is a positive rate constant. (a) Let \(B(c)=A(c)-A^{*}\) and reformulate the given initial value problem in terms of this new dependent variable, \(B\). (b) Solve the new initial value problem for \(B(c)\) and then determine the quantity of interest, \(A(c)\). Does the activity \(A(c)\) ever exceed the value \(A^{*}\) ? (c) Determine the concentration at which \(95 \%\) of the limiting antioxidation activity is achieved. (Your answer is a function of the rate constant \(k\).)

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ \frac{d y}{d t}=\frac{t}{y}, \quad y(0)=-2 $$

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, \(x_{f}=\infty\).) $$ m \frac{d v}{d t}=-k e^{-x} $$

(a) A first order autonomous differential equation has the form \(y^{\prime}=f(y)\). Show that such an equation is separable. (b) Solve \(y^{\prime}=y(2-y), \quad y(2)=1\).

Consider the following autonomous first order differential equations: $$ y^{\prime}=-y^{2}, \quad y^{\prime}=y^{3}, \quad y^{\prime}=y(4-y) . $$ Match each of these equations with one of the solution graphs shown. Note that each solution satisfies the initial condition \(y(0)=1\). Can you match them without solving the differential equations?
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