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Chapter 5: Problem 2

Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values. a. \(\quad y^{\prime}=e^{t-y}, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.5\); actual solution \(y(t)=\ln \left(e^{t}+e-1\right)\). b. \(\quad y^{\prime}=\frac{1+t}{1+y}, \quad 1 \leq t \leq 2, \quad y(1)=2\), with \(h=0.5\); actual solution \(y(t)=\sqrt{t^{2}+2 t+6}-1\). c. \(y^{\prime}=-y+t y^{1 / 2}, \quad 2 \leq t \leq 3, \quad y(2)=2\), with \(h=0.25 ;\) actual solution \(y(t)=\) \(\left(t-2+\sqrt{2} e e^{-t / 2}\right)^{2}\) d. \(\quad y^{\prime}=t^{-2}(\sin 2 t-2 t y), \quad 1 \leq t \leq 2, \quad y(1)=2\), with \(h=0.25 ;\) actual solution \(y(t)=\) \(\frac{1}{2} t^{-2}(4+\cos 2-\cos 2 t)\)

Short Answer

Expert verified
This exercise aims at utilizing the Modified Euler method to approximate the solution to four different differential equations. You can complete this task by appropriately using the iterative process defined by the Euler method, taking into account the step size \( h \), the current \( t \) value and the given differential equation for each problem.

Step by step solution

01

Solve differential equation a using Modified Euler method

Let's start with differential equation a. The Modified Euler method, also known as Heun's method, can be summarized as: \( k_1 = h f(t_k,y_k) \), \( k_2 = h f ( t_{k+1}, y_k + k_1 ) \), \( y_{k+1} = y_k + 1/2 (k_1 + k_2) \). For \( y^{\prime}=e^{t-y}, y(0)=1, h=0.5 \), starting at \( t=0 \), you can iterate using this method to approximate the solution at \( t=1 \). Do compare the approximation with the actual solution \( \ln \left(e^{t}+e-1\right) \) at \( t=1 \).
02

Solve differential equation b using Modified Euler method

Next, apply the same Modified Euler method to differential equation b: \( y^{\prime}=\frac{1+t}{1+y}, y(1)=2, h=0.5 \). Starting at \( t=1 \), you can iterate using this method to approximate the solution at \( t=2 \). Compare this approximation with the actual solution \( \sqrt{t^{2}+2 t+6}-1 \) at \( t=2 \).
03

Solve differential equation c using Modified Euler method

For the third differential equation: \( y^{\prime}=-y+t y^{1 / 2}, y(2)=2, h=0.25 \). Starting at \( t=2 \), apply the Modified Euler method again to approximate the solution at \( t=3 \). Compare the approximation with the actual solution \( (t-2+\sqrt{2} e e^{-t / 2})^{2} \) at \( t=3 \).
04

Solve last differential equation using Modified Euler method

At last, for the differential equation: \( y^{\prime}=t^{-2}(\sin 2 t-2 t y), y(1)=2, h=0.25 \). Starting at \( t=1 \), apply the Modified Euler method once more to approximate the solution for \( t=2 \). Do compare the approximation with the actual solution \( \frac{1}{2} t^{-2}(4+\cos 2-\cos 2 t) \) at \( t=2 \).

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Key Concepts

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

Initial-Value Problems
An initial-value problem (IVP) is a specific type of differential equation along with a designated point where the solution value is given. This problem is fundamental in the world of differential equations as it represents a situation where the state of a system is known at a specific time, and we're looking to predict its future behavior.

For instance, in the exercise provided, each scenario gives us an ordinary differential equation (ODE) together with the initial condition, such as \(y(0)=1\). These initial conditions help determine the unique solution to an ODE and set the stage for the numerical methods that approximate the solution over a range of values.

IVPs appear in various scientific and engineering disciplines, commonly representing physical processes like motion dynamics, chemical reactions, or population growth, where knowing the starting point is crucial for forecasting evolution over time.
Numerical Approximation
Numerical approximation refers to the process of finding an approximate, rather than exact, solution to a problem. In the context of IVPs for ODEs, exact solutions are often challenging, if not impossible, to find. Hence, numerical methods, such as the Modified Euler method, are utilized.

This method is a refinement of the simple Euler approach. It improves the approximation by considering not only the beginning of the interval (as Euler's method does) but also the end, providing a midpoint for correction. The step-by-step solution illustrates how this method yields an approximate value to the solution of the differential equation at a certain point, utilizing the step size \(h\).

When choosing \(h\), a trade-off exists between accuracy and computational effort. Smaller values of \(h\) lead to better approximations but require more computations, while larger values can yield faster but less precise results. The art of numerical analysis is finding that sweet spot where the balance of precision and effort aligns with the specific needs of the problem at hand.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving derivatives of a function and the function itself. They serve as a mathematical model to explain how certain quantities evolve over time, given their rate of change and initial conditions.

ODEs can be simple or complex, linear or nonlinear, and they can vary greatly in the difficulty of finding their solutions. The fundamental purpose of the ODE in an IVP, similar to those given in the exercise examples, is to detail the laws of motion or change for a quantity, whether that's a physical object's velocity or a population's growth rate.

The solutions to these equations provide a formula that can reveal the state of the system at any given time. However, exact solutions are known only for a subset of ODEs with specific forms. That's why methods like the Modified Euler method are invaluable, as they let us approximate the behavior of systems even when we can't find exact analytical solutions.

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Most popular questions from this chapter

Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values. a. \(\quad y^{\prime}=t e^{3 t}-2 y, \quad 0 \leq t \leq 1, \quad y(0)=0\), with \(h=0.5 ;\) actual solution \(y(t)=\frac{1}{5} t e^{3 t}-\frac{1}{25} e^{3 t}+\) \(\frac{1}{25} e^{-2 t}\) b. \(\quad y^{\prime}=1+(t-y)^{2}, \quad 2 \leq t \leq 3, \quad y(2)=1\), with \(h=0.5 ;\) actual solution \(y(t)=t+\frac{1}{1-t}\). c. \(\quad y^{\prime}=1+y / t, \quad 1 \leq t \leq 2, \quad y(1)=2\), with \(h=0.25 ;\) actual solution \(y(t)=t \ln t+2 t\). d. \(\quad y^{\prime}=\cos 2 t+\sin 3 t, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.25 ;\) actual solution \(y(t)=\) \(\frac{1}{2} \sin 2 t-\frac{1}{3} \cos 3 t+\frac{4}{3}\).

Use Euler's method to approximate the solutions for each of the following initial-value problems. a. \(\quad y^{\prime}=e^{t-y}, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.5\) b. \(\quad y^{\prime}=\frac{1+t}{1+y}, \quad 1 \leq t \leq 2, \quad y(1)=2\), with \(h=0.5\) c. \(\quad y^{\prime}=-y+t y^{1 / 2}, \quad 2 \leq t \leq 3, \quad y(2)=2\), with \(h=0.25\) d. \(\quad y^{\prime}=t^{-2}(\sin 2 t-2 t y), \quad 1 \leq t \leq 2, \quad y(1)=2\), with \(h=0.25\)

Given the initial-value problem $$ y^{\prime}=\frac{1}{t^{2}}-\frac{y}{t}-y^{2}, \quad 1 \leq t \leq 2, \quad y(1)=-1 $$ with exact solution \(y(t)=-1 / t\) : a. Use Euler's method with \(h=0.05\) to approximate the solution, and compare it with the actual values of \(y\). b. Use the answers generated in part (a) and linear interpolation to approximate the following values of \(y\), and compare them to the actual values. i. \(y(1.052)\) ii. \(y(1.555)\) iii. \(\quad y(1.978)\) c. Compute the value of \(h\) necessary for \(\left|y\left(t_{i}\right)-w_{i}\right| \leq 0.05\) using Eq. (5.10).

Use the Adams Variable Step-Size Predictor-Corrector Algorithm with tolerance \(T O L=10^{-6}\), \(\operatorname{hmax}=0.5\), and \(\operatorname{hmin}=0.02\) to approximate the solutions to the given initial-value problems. Compare the results to the actual values. a. \(\quad y^{\prime}=y / t-(y / t)^{2}, \quad 1 \leq t \leq 4, \quad y(1)=1 ;\) actual solution \(y(t)=t /(1+\ln t)\). b. \(\quad y^{\prime}=1+y / t+(y / t)^{2}, \quad 1 \leq t \leq 3, \quad y(1)=0\); actual solution \(y(t)=t \tan (\ln t)\). c. \(\quad y^{\prime}=-(y+1)(y+3), \quad 0 \leq t \leq 3, \quad y(0)=-2 ;\) actual solution \(y(t)=-3+2\left(1+e^{-2 t}\right)^{-1}\). d. \(\quad y^{\prime}=\left(t+2 t^{3}\right) y^{3}-t y, \quad 0 \leq t \leq 2, \quad y(0)=\frac{1}{3} ;\) actual solution \(y(t)=\left(3+2 t^{2}+6 e^{t^{2}}\right)^{-1 / 2}\).

Consider the problem \(y^{\prime}=0,0 \leq t \leq 10, y(0)=0\), which has the solution \(y \equiv 0\). If the difference method of Exercise 4 is applied to the problem, then $$ \begin{gathered} w_{i+1}=4 w_{i}-3 w_{i-1}, \quad \text { for } i=1,2, \ldots, N-1 \\ w_{0}=0, \quad \text { and } \quad w_{1}=\alpha_{1} \end{gathered} $$ Suppose \(w_{1}=\alpha_{1}=\varepsilon\), where \(\varepsilon\) is a small rounding error. Compute \(w_{i}\) exactly for \(i=2,3, \ldots, 6\) to find how the error \(\varepsilon\) is propagated.
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