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Solve the following problems using Euler's method with stepsizes of \(h=0.2,0.1\), 0.05. Compute the error and relative error using the true answer \(Y(x) .\) For selected values of \(x\), observe the ratio by which the error decreases when \(h\) is halved. \(\begin{array}{ll}\text { (a) } \quad & Y^{\prime}(x)=[\cos (Y(x))]^{2}, \quad 0 \leq x \leq 10, \quad Y(0)=0 ; \\\ Y(x)=\tan ^{-1}(x) & & \\ \text { (b) } \quad Y^{\prime}(x)=\frac{1}{1+x^{2}}-2[Y(x)]^{2}, \quad 0 \leq x \leq 10, \quad Y(0)=0 \\ Y(x)=\frac{x}{1+x^{2}} & \\ \text { (c) } \quad Y^{\prime}(x)=\frac{1}{4} Y(x)\left[1-\frac{1}{20} Y(x)\right], \quad 0 \leq x \leq 20, \quad Y(0)=1 ; \\ Y(x)= & \frac{20}{1+19 e^{-x / 4}}\end{array}\) (d) \(Y^{\prime}(x)=-[Y(x)]^{2}, \quad 1 \leq x \leq 10, \quad Y(1)=1 ; \quad Y(x)=\frac{1}{x}\) (e) \(\quad Y^{\prime}(x)=x e^{-x}-Y(x), \quad 0 \leq x \leq 10, \quad Y(0)=1 ;\) \(y(x)=\left(1=\frac{1}{2} x^{2}\right) e^{-x}\) (f) \(\quad Y^{\prime}(x)=\frac{x^{3}}{Y(x)}, \quad 0 \leq x \leq 10, \quad Y(0)=1 ; \quad Y(x)=\sqrt{\frac{1}{2} x^{4}+1}\) (g) \(\quad Y^{\prime}(x)=\left(3 x^{2}-1\right) Y(x)^{2}, \quad 0 \leq x \leq 10, \quad Y(0)=1\); \(y(x)=\sqrt{3\left(x^{3}-x\right)+1}\)

Short Answer

Expert verified
Apply Euler's method iteratively for each step size, compute errors and relative errors, analyze error reduction when halving step size.

Step by step solution

01

Title - Understanding Euler's Method

Euler's method is a numerical technique for solving ordinary differential equations (ODEs) by approximating solutions at discrete points. It is given by the formula: \[ Y_{n+1} = Y_n + h f(x_n, Y_n) \] where \( h \) is the step size, \( Y_n \) is the current value, and \( f(x_n, Y_n) \) is the value of the derivative at the point \( (x_n, Y_n) \).
02

Title - Setting Up the Problem

For each problem, first determine the initial conditions and the corresponding differential equation. Then we select step sizes of \(h = 0.2, 0.1,\) and \(0.05\).
03

Title - Computing With Euler's Method

For each step size, compute the approximated values using Euler’s method by applying the formula iteratively from the initial condition to the desired value of \(x\). Keep track of the values at each step.
04

Title - Calculating Errors and Relative Errors

Calculate the true value \( Y(x) \) using the provided true solution. Calculate the error as : \[ \text{Error} = |Y_\text{true} - Y_\text{approx}| \] and the relative error as :\[ \text{Relative Error} = \frac{|Y_\text{true} - Y_\text{approx}|}{|Y_\text{true}|} \]
05

Title - Examining Error Reduction

For selected values of \(x\), compute the errors for different step sizes. Observe how the error changes as the step size \(h\) is halved. Use the ratio \( \frac{\text{Error}_h}{\text{Error}_{h/2}} \) to analyze the decrease in error.
06

Title - Example Application (Part (a))

For Part (a): Given \( Y'(x) = [\cos(Y(x))]^2, Y(0)=0, Y(x) = \tan^{-1}(x) \), follow the previous steps. Start with \( x = 0, Y(0) = 0\) and use Euler’s formula with each step size \(h\). Compare the approximation with \( Y(x) = \tan^{-1}(x) \).
07

Title - Repeat for Remaining Problems

Repeat the steps for each problem (b) through (g) using the corresponding differential equations and true solutions provided.

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Key Concepts

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

Numerical Techniques
Numerical techniques are methods used to solve mathematical problems that are difficult or impossible to address analytically. These techniques often involve approximations and are essential in fields like engineering, physics, and economics. Euler’s method is one such numerical technique.
Euler's method specifically addresses the need to solve ordinary differential equations (ODEs) or problems involving rates of change. These techniques are especially useful when the exact analytical solution is not feasible or straightforward to derive.
Ordinary Differential Equations
Ordinary differential equations (ODEs) involve functions and their derivatives. They are used to model various real-world phenomena, such as population growth, heat transfer, and motion of objects. An ODE relates an unknown function, typically dependent on one variable such as time, to its rates of change.
For instance, consider an ODE of the form \(Y'(x) = f(x, Y(x))\), where \(Y(x)\) is the unknown function and \(f(x, Y(x))\) is a given function. The goal is to determine \(Y(x)\) for a given initial condition such as \(Y(0) = Y_0\). Euler's method helps approximate this function by considering small increments (or steps) and iterating over the range of interest.
Step Size
The step size, denoted by \(h\), is a critical component in numerical methods like Euler's method. It represents the increment in the independent variable (such as time or distance) for each step of the computation. Choosing an appropriate step size is crucial for balancing accuracy and computational effort.
If the step size is too large, the method may miss important changes in the function's behavior, leading to significant errors. Conversely, a very small step size improves accuracy but increases the number of computations needed, impacting performance.
In our given problems, step sizes of \(h = 0.2, 0.1, \text{ and } 0.05\) are used to observe how the error changes with different increments.
Error Calculation
When using numerical methods, it's essential to quantify how approximate results differ from true values. This difference is known as the error. Two common forms of error calculations are absolute error and relative error.
The absolute error is given by:
\[\text{Error} = |Y_{\text{true}} - Y_{\text{approx}}|\]
The relative error, which helps in understanding the error magnitude in relation to the true value, is calculated as:
\[\text{Relative Error} = \frac{|Y_{\text{true}} - Y_{\text{approx}}|}{|Y_{\text{true}}|}\]
By comparing errors at different step sizes, one can analyze the effectiveness and accuracy of the numerical method. Observing how the error decreases when the step size is halved provides insights into the method's convergence properties.
Iterative Methods
Euler's method is an example of an iterative method, where the solution is approximated through a series of repeated steps. Starting from an initial condition, the method calculates successive values using a specified formula. For Euler’s method, the iterative formula is:
\[ Y_{n+1} = Y_n + h f(x_n, Y_n)\]
Here, \(h\) is the step size, \(Y_n\) is the current value, and \(f(x_n, Y_n)\) is the derivative at the current point. The next value, \(Y_{n+1}\), is then calculated, and the process repeats for the range of the independent variable.
Iterative methods are powerful as they enable solving complex problems by breaking them down into smaller, manageable steps. The accuracy of the solution depends on the step size and the number of iterations performed.

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Most popular questions from this chapter

Solve the equation $$ Y^{\prime}(x)=\lambda Y(x)+\frac{1}{1+x^{2}}-\lambda \tan ^{-1}(x), \quad Y(0)=0 $$ \(Y(x)=\tan ^{-1}(x)\) is the true solution. Use Euler's method, the backward Euler method, and the trapezoidal method. Let \(\lambda=-1,-10,-50\), and \(h=0.5,0.1\), \(0.001\). Discuss the results.

Consider the motion of a particle of mass in falling vertically under the earth's gravitational field, and suppose the downward motion is opposed by a frictional force \(p(v)\) dependent on the velocity \(v(t)\) of the particle. Then the velocity satisfies the equation $$ m v^{\prime}(t)=-m g+p(v), \quad t \geq 0, \quad v(0) \text { given } $$ Let \(m=1 \mathrm{~kg}, g=9.8 \mathrm{~m} / \mathrm{sec}^{2}\), and \(v(0)=0\). Solve the differential equation for \(0 \leq t \leq 20\) and for the following choices of \(p(v)\) : (a) \(p(v)=-0.1 v\), which is positive for a falling body. (b) \(\quad p(v)=0.1 v^{2}\). Find answers to at least three digits of accuracy. Graph the functions \(v(t) .\) Compare the solutions.

In this exercise, we consider a method with third-order truncation errors that is not convergent or stable. (a) If \(Y(x)\) is three times continuously differentiable, show that $$ \begin{aligned} Y\left(x_{n+1}\right)=& 3 Y\left(x_{n}\right)-2 Y\left(x_{n-1}\right)+\frac{h}{2}\left[Y^{\prime}\left(x_{n}\right)-3 Y^{\prime}\left(x_{n-1}\right)\right] \\ &+\frac{7}{12} h^{3} Y^{\prime \prime \prime}\left(x_{n}\right)+O\left(h^{4}\right) \end{aligned} $$ Thus, a numerical method for solving the differential equation $$ Y^{\prime}(x)=f(x, Y(x)) $$ is $$ y_{n+1}=3 y_{n}-2 y_{n-1}+\frac{h}{2}\left[f\left(x_{n}, y_{n}\right)-3 f\left(x_{n-1}, y_{n-1}\right)\right], \quad n \geq 1 $$ This is a numerical method whose truncation error is \(O\left(h^{3}\right) .\) It is an example of a multistep method (cf. Section 8.6). To use the method, we need a value for \(y_{1}\), called an artificial initial value, in addition to the initial value \(y_{0}=Y_{0}\). Hint: To prove (8.65), use Taylor expansions about the point \(x_{n}\). (b) Now apply the method to solve the very simple initial value problem $$ Y^{\prime}(x) \equiv 0, \quad Y(0)=1 $$ whose solution is \(Y(x) \equiv 1\). Show that if the initial values are chosen to be \(y_{0}=1, y_{1}=1+h\), then the numerical solution is \(y_{n}=1-h+h 2^{n}\). Note that \(\left|y_{1}-Y(h)\right|=h \rightarrow 0\) as \(h \rightarrow 0\). Let \(x_{n}=1\). Show that \(\left|Y(1)-y_{n}\right| \rightarrow\) \(\infty\) as \(h \rightarrow 0\). Thus, the method is not convergent. (c) A slight variant of the arguments of (b) can be used to show the instability of the method. Show that with the initial values \(y_{0}=y_{1}=1\), the numerical solution is \(y_{n}=1\) for all \(n\), whereas if the initial values are perturbed to \(y_{\epsilon, 0}=\) 1, \(y_{\epsilon, 1}=1+\epsilon\), then the numerical solution becomes \(y_{\epsilon, n}=1-\epsilon+\epsilon 2^{n}\). Show that at any fixed node point \(x_{n}=\bar{x}>0,\left|y_{\epsilon, n}-y_{n}\right| \rightarrow \infty\) as \(h \rightarrow 0\). Hence, the method is unstable.

Let $$ \begin{gathered} A=\left[\begin{array}{ll} 1 & -2 \\ 2 & -1 \end{array}\right], \quad \boldsymbol{Y}=\left[\begin{array}{l} Y_{1} \\ Y_{2} \end{array}\right] \\ \boldsymbol{G}(x)=\left[\begin{array}{l} -2 e^{-x}+2 \\ -2 e^{-x}+1 \end{array}\right], \quad \boldsymbol{Y}_{0}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \end{gathered} $$ Write out the two equations that make up the system $$ \boldsymbol{Y}^{\prime}=A \boldsymbol{Y}+\boldsymbol{G}(x), \quad \boldsymbol{Y}\left(x_{0}\right)=\boldsymbol{Y}_{0} $$ The true solution is \(\boldsymbol{Y}=\left[e^{-x}, 1\right]^{T}\).

Convert the following system of second-order equations to a larger system of firstorder equations. It arises from studying the gravitational attraction of one mass by another. $$ x^{\prime \prime}(t)=\frac{-c x(t)}{r(t)^{3}}, \quad y^{\prime \prime}(t)=\frac{-c y(t)}{r(t)^{3}}, \quad z^{\prime \prime}(t)=\frac{-c z(t)}{r(t)^{3}} $$ with \(c\) positive constant and \(r(t)=\left[x(t)^{2}+y(t)^{2}+z(t)^{2}\right]^{1 / 2}, t=\) time.

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