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466 DIFFERENTIAL EQUATIONS Table 49.8 x y y ′ 1. 0 2 2 2. 0.1 2.205 2.105 3. 0.2 2.421025 2.221025 4. 0.3 2.649232625 2.349232625 5. 0.4 2.89090205 2.49090205 6. 0.5 3.147446765 For line 3, x 1 = 0.2 y P1 = y 0 + h(y ′ ) 0 = 2.205 + (0.1)(2.105) = 2.4155 y C1 = y 0 + 1 2 h[(y′ ) 0 + f (x 1 , y P1 )] = 2.205 + 2 1 (0.1)[2.105 + (2.4155 − 0.2)] = 2.421025 (y ′ ) 0 = y C1 − x 1 = 2.421025 − 0.2 = 2.221025 For line 4, x 1 = 0.3 y P1 = y 0 + h(y ′ ) 0 = 2.421025 + (0.1)(2.221025) = 2.6431275 y C1 = y 0 + 1 2 h[(y′ ) 0 + f (x 1 , y P1 )] = 2.421025 + 1 2 (0.1)[2.221025 = 2.649232625 + (2.6431275 − 0.3)] (y ′ ) 0 = y C1 − x 1 = 2.649232625 − 0.3 = 2.349232625 For line 5, x 1 = 0.4 y P1 = y 0 + h(y ′ ) 0 = 2.649232625 + (0.1)(2.349232625) = 2.884155887 y C1 = y 0 + 1 2 h[(y′ ) 0 + f (x 1 , y P1 )] = 2.649232625 + 1 2 (0.1)[2.349232625 = 2.89090205 (y ′ ) 0 = y C1 − x 1 = 2.89090205 − 0.4 = 2.49090205 For line 6, x 1 = 0.5 y P1 = y 0 + h(y ′ ) 0 + (2.884155887 − 0.4)] = 2.89090205 + (0.1)(2.49090205) = 3.139992255 y C1 = y 0 + 1 2 h[(y′ ) 0 + f (x 1 , y P1 )] = 2.89090205 + 1 2 (0.1)[2.49090205 = 3.147446765 + (3.139992255 − 0.5)] Problem 4 is the same example as Problem 3 and Table 49.9 shows a comparison of the results, i.e. it compares the results of Tables 49.3 and 49.8. dy = y − x may be solved analytically by the integrating factor method of Chapter 48 with the solution dx y = x + 1 + e x . Substituting values of x of 0, 0.1, 0.2, ...give the exact values shown in Table 49.9. The percentage error for each method for each value of x is shown in Table 49.10. For example when x = 0.3, % error with Euler method ( ) actual − estimated = × 100% actual ( ) 2.649858808 − 2.631 = × 100% 2.649858808 = 0.712% Table 49.9 Euler method Euler-Cauchy method Exact value x y y y= x + 1 + e x 1. 0 2 2 2 2. 0.1 2.2 2.205 2.205170918 3. 0.2 2.41 2.421025 2.421402758 4. 0.3 2.631 2.649232625 2.649858808 5. 0.4 2.8641 2.89090205 2.891824698 6. 0.5 3.11051 3.147446765 3.148721271
NUMERICAL METHODS FOR FIRST ORDER DIFFERENTIAL EQUATIONS 467 % error with Euler-Cauchy method ( ) 2.649858808 − 2.649232625 = × 100% 2.649858808 = 0.0236% This calculation and the others listed in Table 49.10 show the Euler-Cauchy method to be more accurate than the Euler method. Table 49.10 x Error in Error in Euler method Euler-Cauchy method 0 0 0 0.1 0.234% 0.00775% 0.2 0.472% 0.0156% 0.3 0.712% 0.0236% 0.4 0.959% 0.0319% 0.5 1.214% 0.0405% Problem 5. Obtain a numerical solution of the differential equation dy = 3(1 + x) − y dx in the range 1.0(0.2)2.0, using the Euler-Cauchy method, given the initial conditions that x = 1 when y = 4. This is the same as Problem 1 on page 461, and a comparison of values may be made. dy dx = y′ = 3(1 + x) − y i.e. y ′ = 3 + 3x − y x 0 = 1.0, y 0 = 4 and h = 0.2 (y ′ ) 0 = 3 + 3x 0 − y 0 = 3 + 3(1.0) − 4 = 2 x 1 = 1.2 and from equation (3), y P1 = y 0 + h(y ′ ) 0 = 4 + 0.2(2) = 4.4 y C1 = y 0 + 1 2 h[(y′ ) 0 + f (x 1 , y P1 )] = y 0 + 1 2 h[(y′ ) 0 + (3 + 3x 1 − y P1 )] = 4 + 2 1 (0.2)[2 + (3 + 3(1.2) − 4.4)] = 4.42 (y ′ ) 1 = 3 + 3x 1 − y P1 = 3 + 3(1.2) − 4.42 = 2.18 Thus the first two lines of Table 49.11 have been completed. Table 49.11 x 0 y 0 y 0 ′ 1. 1.0 4 2 2. 1.2 4.42 2.18 3. 1.4 4.8724 2.3276 4. 1.6 5.351368 2.448632 5. 1.8 5.85212176 2.54787824 6. 2.0 6.370739847 For line 3, x 1 = 1.4 y P1 = y 0 + h(y ′ ) 0 = 4.42 + 0.2(2.18) = 4.856 y C1 = y 0 + 1 2 h[(y′ ) 0 + (3 + 3x 1 − y P1 )] = 4.42 + 2 1 (0.2)[2.18 + (3 + 3(1.4) − 4.856)] = 4.8724 (y ′ ) 1 = 3 + 3x 1 − y P1 = 3 + 3(1.4) − 4.8724 = 2.3276 For line 4, x 1 = 1.6 y P1 = y 0 + h(y ′ ) 0 = 4.8724 + 0.2(2.3276) = 5.33792 y C1 = y 0 + 2 1 h[(y′ ) 0 + (3 + 3x 1 − y P1 )] = 4.8724 + 2 1 (0.2)[2.3276 + (3 + 3(1.6) − 5.33792)] = 5.351368 (y ′ ) 1 = 3 + 3x 1 − y P1 = 3 + 3(1.6) − 5.351368 = 2.448632 For line 5, x 1 = 1.8 y P1 = y 0 + h(y ′ ) 0 = 5.351368 + 0.2(2.448632) = 5.8410944 y C1 = y 0 + 1 2 h[(y′ ) 0 + (3 + 3x 1 − y P1 )] = 5.351368 + 1 2 (0.2)[2.448632 = 5.85212176 (y ′ ) 1 = 3 + 3x 1 − y P1 = 3 + 3(1.8) − 5.85212176 = 2.54787824 + (3 + 3(1.8) − 5.8410944)] I
Page 1 and 2: Differential equations 46 Solution
Page 3 and 4: SOLUTION OF FIRST ORDER DIFFERENTIA
Page 5 and 6: α is the temperature coefficient o
Page 7 and 8: SOLUTION OF FIRST ORDER DIFFERENTIA
Page 9 and 10: Differential equations 47 Homogeneo
Page 11 and 12: HOMOGENEOUS FIRST ORDER DIFFERENTIA
Page 13 and 14: Differential equations 48 Linear fi
Page 15 and 16: Hence from equation (4): ye x = xe
Page 17 and 18: LINEAR FIRST ORDER DIFFERENTIAL EQU
Page 19 and 20: NUMERICAL METHODS FOR FIRST ORDER D
Page 23: NUMERICAL METHODS FOR FIRST ORDER D
Page 33 and 34: Differential equations 50 Second or
Page 35 and 36: SECOND ORDER DIFFERENTIAL EQUATIONS
Page 37 and 38: When thus t = 0, dV dt = 3ω, 3ω =
Page 39 and 40: Differential equations 51 Second or
Page 49 and 50: Differential equations 52 Power ser
Page 51 and 52: POWER SERIES METHODS OF SOLVING ORD
Page 71 and 72: AN INTRODUCTION TO PARTIAL DIFFEREN
Page 73 and 74: AN INTRODUCTION TO PARTIAL DIFFEREN
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AN INTRODUCTION TO PARTIAL DIFFEREN
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Differential equations Assignment 1
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