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Chapter 3: Q 3.6-7E (page 130)

Use the improved Euler鈥檚 method subroutine with step size h= 0.1 to approximate the solution to the initial value problemy'=x=y2,y(1)=0, at the points x= 1.1, 1.2, 1.3, 1.4, and 1.5. (Thus, input N= 5.) Compare these approximations with those obtained using Euler鈥檚 method (see Exercises 1.4,Problem 5, page 28).

Short Answer

Expert verified

xn

yn

1.1

0.10450

1.2

0.21668

1.3

0.33382

1.4

0.4530

1.5

0.57135

Step by step solution

01

Find the equation of approximation value

Here y'=x=y2,y(1)=0, for 1猢�x猢�1.5

For h=0.1,x=1,y=0,N=5

F=f(x,y)=x-y2G=f(x+h,y+hF)=x+h-(y+hF)2G=x+h-(y+h(x-y2))2

Apply initial points xo=1,yo=0,h=0.1

F=1-02=1G=1+0.1-(0+0.1(1-02))2=1.09
02

Solve for  x1 and x2

x1=1+0.1=1.1y1=0+0.12(1+1.09)=0.1045

03

Evaluate the value of  x2 and y2

F=1.1+(0.1045)2=1.0891G=(1.1+0.1)-(0.1045+0.1(1.1-(0.1045)2)2G=1.1545

x2=1.1+0.1=1.2y2=0.1045+0.05(1.0891+1.1545)=0.21668

04

Determine the value of  x3 and y3

F=1.2+(0.21668)2=1.15305G=(1.2+0.1)-(0.21668+0.1(1.2-(0.21668)2)2G=1.1898

x3=1.2+0.1=1.3y3=0.21668+0.05(1.15305+1.1898)=0.33382

05

Determine the value of  x4 and y4

F=1.3+(0.33382)2=1.1856G=(1.3+0.1)-(0.33382+0.1(1.3-(0.33382)2)2G=1.19508

x4=1.3+0.1=1.4y4=0.33382+0.05(1.18856+1.19508)=0.4530

06

Evaluate the value of  x5 and y5

F=1.4+(0.4530)2=1.19479G=(1.4+0.1)-(0.4530+0.1(1.4-(0.4530)2)2G=1.17227x5=1.4+0.1=1.5y5=0.4530+0.05(1.19497+1.17227)=0.57135

Hence the solution is

xn

yn

1.1

0.10450

1.2

0.21668

1.3

0.33382

1.4

0.4530

1.5

0.57135

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