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Chapter 3: Q 3.6-8E (page 130)

Use the improved Euler’s method subroutine with step size h= 0.2 to approximate the solution to the initial value problemy'=1x(y2+y),y(1)=1 at the points x= 1.2, 1.4, 1.6, and 1.8. (Thus, input N= 4.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 6, page 28).

Short Answer

Expert verified

xn

yn

1.2

1.48

1.4

2.24788

1.6

3.6518

1.8

6.88733

Step by step solution

01

Find the equation of approximation value

Here, y'=1x(y2+y),y(1)=0 for 1⩽x⩽1.8

For h=0.2, x=1, y=1, N=4

F=f(x,y)=1x(y2+y)G=f(x+h,y+hF)=1x+0.2y+0.2x(y2+y)2+y+0.2x(y2+y)

02

Solve for x1 and y1

Apply initial points xo=1,yo=1,h=0.2

F(1,1)=2G(1,1)=2.8

x1=1+0.2=1.2y1=1+0.22(2+2.8)=1.48

03

Evaluate the value of  x2 and y2

F(1.2,1.48)=3.05867G(1.2,1.48)=4.61934

x2=1.2+0.2=1.4y2=1.48+0.1(3.05867+4.61934)=2.24788

04

Determine the value of  x3 and y3

F(1.4,2.24788)=5.21489G(1.4,2.2488)=8.82538

x3=1.4+0.2=1.6y3=2.24788+0.1(5.21489+8.82538)=3.6518

05

Determine the value of  x4 and y4

F(1.6,3.6518)=10.6172G(1.6,3.6518)=21.7381

x4=1.6+0.2=1.8y4=3.6518+0.1(10.6172+21.7381)=6.88733

Hence the solution is

xn

yn

1.2

1.48

1.4

2.24788

1.6

3.6518

1.8

6.88733

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Most popular questions from this chapter

A cold beer initially at 35°F warms up to 40°F in 3 min while sitting in a room of temperature 70°F. How warm will the beer be if left out for 20 min?

Local versus Global Error. In deriving formula (4) for Euler’s method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With

\({\bf{g(t) = f(t,f(t))}}\), this approximation can be written as \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt}} \approx {\bf{hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \)where \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .

  1. Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error\(\left( {\bf{O}} \right){{\bf{h}}^{\bf{2}}}\); that is, for some constant M, \(\left| {\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} } \right| \le {\bf{M}}{{\bf{h}}^{\bf{2}}}\).This is called the local truncation error of the scheme. [Hint: Write \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} {\bf{ = }}\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {\left[ {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right]{\bf{dt}}} \). Next, using the mean value theorem, show that\(\left| {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right| \le {\bf{B}}\left| {{\bf{t - }}{{\bf{x}}_{\bf{n}}}} \right|\) . Then integrate to obtain the error bound\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]
  2. In applying Euler’s method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after n steps is (O)h. This is the global error, which is the same as the[ss1] [m2] convergence rate of Euler’s method.


Determine the recursive formulas for the Taylor method of order 2 for the initial value problemy'=cos(x+y),y(0)=π.

An object of mass 60 kg starts from rest at the top of a 45º inclined plane. Assume that the coefficient of kinetic friction is 0.05 (see Problem 18). If the force due to air resistance is proportional to the velocity of the object, say, -3, find the equation of motion of the object. How long will it take the object to reach the bottom of the inclined plane if the incline is 10 m long?

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