Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis
Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis
5th Edition
ISBN: 9780321816252
Author: C. Henry Edwards, David E. Penney, David Calvis
Publisher: PEARSON
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Chapter 4.3, Problem 1P

(a)

Program Plan Intro

Program Description: Purpose of the problem is to obtain the approximate values of x(0.2) and y(0.2) by Euler’s method.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The differential problem is x=x+2y,x(0)=0 and y=2x+y,y(0)=2 .

The exact solution of the differential equation is x(t)=e3tet and y(t)=e3t+et .

The two step size is 0.1 .

Calculation:

The differential problem can be represented as,

  f(x,y)=x+2y,g(x,y)=2x+y

The Euler’s formula for xn+1 is shown below.

  xn+1=xn+h(xn+2yn) …… (1)

The Euler’s formula for yn+1 is shown below.

  yn+1=yn+h(2xn+yn) …… (2)

Substitute 0 for n in equation (1)

  x0+1=x0+h(x0+2y0) …… (3)

Substitute 0 for n in equation (2)

  y0+1=y0+h(2x0+y0) …… (4)

Substitute 0 for x0 , 0.1 for h and 2 for y0 in equation (3).

  x0+1=0+0.1(0+2(2))=0.4

Substitute 0 for x0 , 0.1 for h and 2 for y0 in equation (4).

  y0+1=2+0.1(2(0)+2)=2+0.2=2.2

Substitute 1 for n in equation (1)

  x2=x1+h(x1+2y1) …… (5)

Substitute 1 for n in equation (2)

  y2=y1+h(2x1+y1) …… (6)

Substitute 0.4 for x1 , 0.1 for h and 2.2 for y1 in equation (5).

  x2=0.4+0.1(0.4+2( 2.2))=0.4+0.48=0.88

Substitute 0.4 for x1 , 0.1 for h and 2.2 for y1 in equation (6).

  y2=2.2+0.1(2( 0.4)+2.2)=2.2+0.3=2.5

Therefore, the value of x2 and y2 is 0.88 and 2.5

Conclusion:

Thus, the approximate values of x(0.2) and y(0.2) by Euler’s method is 0.8800 and 2.5000 .

(b)

Program Plan Intro

Program Description: Purpose of the problem is to find the approximate values of x(0.2) and y(0.2) by improved Euler’s method.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The differential problem is x=x+2y,x(0)=0 and y=2x+y,y(0)=2 .

The exact solution of the differential equation is x(t)=e3tet and y(t)=e3t+et .

The two step size is 0.2 .

Calculation:

The improved Euler’s formula for predicators un+1 is shown below.

  un+1=xn+h(xn+2yn) …… (7)

The improved Euler’s formula for predicators vn+1 is shown below.

  vn+1=yn+h(2xn+yn) …… (8)

Substitute 0 for n in equation (7)

  u0+1=x0+h(x0+2y0) …… (9)

Substitute 0 for n in equation (8)

  v0+1=y0+h(2x0+y0) …… (10)

Substitute 0 for x0 , 0.2 for h and 2 for y0 in equation (9).

  u0+1=0+0.2(0+2(2))=0.8

Substitute 0 for x0 , 0.2 for h and 2 for y0 in equation (10).

  v0+1=2+0.2(2(0)+2)=2+0.2=2.4

The improved Euler’s formula for correctors is shown below.

  xn+1=xn+h2((xn+2yn)+(u n+1+2v n+1)) …… (11)

The improved Euler’s formula for correctors x1 is shown below.

  yn+1=yn+h2((2xn+yn)+(2u n+1+v n+1)) …… (12)

Substitute 0 for n in equation (11)

  x1=x0+h2((x0+2y0)+(u1+2v1)) …… (13)

Substitute 0 for n in equation (12)

  y1=y0+h2((2x0+y0)+(2u1+v1)) …… (14)

Substitute 0 for x0 , 0.2 for h , 0.8 for u1 , 2.4 for v1 and 2 for y0 in equation (13).

  x1=0+0.22(( 0+2( 2 ))+( 0.8+2( 2.4 )))=0.1(9.6)=0.96

Substitute 0 for x0 , 0.2 for h , 0.8 for u1 , 2.4 for v1 and 2 for y0 in equation (14).

  y1=2+0.22(( 2( 0 )+2)+( 2( 0.8 )+2.4))=2+0.2+0.4=2.6

Therefore, the value of x1 and y1 is 0.96 and 2.6

Conclusion:

Thus, the approximate values of x(0.2) and y(0.2) by improved Euler’s method is 0.9600 and 2.6000 .

(c)

Program Plan Intro

Program Description: Purpose of the problem is to find the approximate values of x(0.2) and y(0.2) by Runge-Kutta method and also compare the result with all the obtained results.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The differential problem is x=x+2y,x(0)=0 and y=2x+y,y(0)=2 .

The exact solution of the differential equation is x(t)=e3tet and y(t)=e3t+et .

The two step size is 0.2 .

Calculation:

The Runge-Kutta iteration formulas are shown below.

  xn+1=xn+h6(F1+2F2+2F3+F4) …… (15)

The Runge-Kutta iteration formulas are shown below

  yn+1=yn+h6(G1+2G2+2G3+G4) …… (16)

The value of F1,F2,F3 and F4 is shown below.

  F1=x0+2y0F2=(x0+h2F1)+2(y0+h2G1)F3=(x0+h2F2)+2(y0+h2G2)F4=(x0+hF3)+2(y0+hG3)

The value of G1,G2,G3 and G4 is shown below.

  G1=2x0+y0G2=2(x0+h2F1)+(y0+h2G1)G3=2(x0+h2F2)+(y0+h2G2)G4=2(x0+hF3)+(y0+hG3)

Substitute 0 for x0 ,and 2 for y0 in F1=x0+2y0 as,

  F1=0+4=4

Substitute 0 for x0 , and 2 for y0 in G1=2x0+y0 as,

  G1=2(0)+2=2

Substitute 0 for x0 , 0.2 for h , 4 for F1 , 2 for G1 and 2 for y0 in F2=(x0+h2F1)+2(y0+h2G1) as,

  F2=(0+ 0.22(4))+2(2+ 0.22(2))=0.4+4.4=4.8

Substitute 0 for x0 , 0.2 for h , 4 for F1 , 2 for G1 and 2 for y0 in G2=2(x0+h2F1)+(y0+h2G1) as,

  G2=2(0+ 0.22(4))+(2+ 0.22(2))=0.8+2.2=3

Substitute 0 for x0 , 0.2 for h , 4.8 for F2 , 3 for G2 and 2 for y0 in F3=(x0+h2F2)+2(y0+h2G2) as,

  F3=(0+ 0.22( 4.8))+2(2+ 0.22(3))=0.48+4.6=5.08

Substitute 0 for x0 , 0.2 for h , 4.8 for F2 , 3 for G2 and 2 for y0 in G3=2(x0+h2F2)+(y0+h2G2) as,

  G3=2(0+ 0.22( 4.8))+(2+ 0.22(3))=0.96+2.3=3.26

Substitute 0 for x0 , 0.2 for h , 5.08 for F3 , 3.26 for G3 and 2 for y0 in equation F4=(x0+hF3)+2(y0+hG3) as,

  F4=(0+0.2( 5.08))+2(2+0.2( 3.26))=1.016+5.304=6.32

Substitute 0 for x0 , 0.2 for h , 5.08 for F3 , 3.26 for G3 and 2 for y0 in G4=2(x0+hF3)+(y0+hG3) as,

  G4=2(0+0.2( 5.08))+(2+0.2( 3.26))=2.032+2.652=4.684

Substitute 5.08 for F3 , 0 for x0 , 0.2 for h , 3 for G2 , 4 for F1 , and 6.32 for F4 in equation (15).

  xn+1=0+0.26(4+2( 4.8)+2( 5.08)+6.32)=0.26(30.08)=1.0027

Substitute 2 for G1 , 2 for y0 , 0.2 for h , 4.8 for F2 , 3.26 for G3 and 4.684 for G4 in equation (16).

  yn+1=2++0.26(2+2(3)+2( 3.26)+4.684)=2+0.26(19.204)=2+0.6401=2.64101

The exact solution x(0.2) can be obtained as,

  x(0.2)=e3( 0.2)e0.2=1.0034

The exact solution y(0.2) can be obtained as,

  y(0.2)=e3( 0.2)+e0.2=2.6408

It can be observed that the Runge-Kutta method is more closes to the exact solution of the differential equation.

Conclusion:

Thus, the approximate values of x(0.2) and y(0.2) by Runge-Kutta method is 1.0027 and 2.64101 .

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Chapter 4 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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