The Chemistry Maths Book, Second Edition

(Grace) #1

594 Chapter 20Numerical methods



  1. (i) Use Gauss elimination to find the value of λfor which the following equations have a


solution.


(ii)Solve the equations for this value of λ:


x 1 + y 1 + z 1 = 12


−x 1 + 12 y 1 − 13 z 1 = 132


3 x 1 + 5 z 1 = 1 λ


Section 20.8


32.Use Gauss–Jordan elimination to find the inverse of the matrix


33.Given the system of equations,


−x 1 − y 1 + 12 z 1 = 1 b


1

3 x 1 − y 1 + z 1 = 1 b


2

−x 1 + 13 y 1 + 14 z 1 = 1 b


3

use the result of Exercise 32 to express x,y, and zin terms of the arbitrary numbersb


1

,b


2

andb


3

.


Section 20.9


NOTE:The following exercises can be performed using a pocket calculator, but the


arithmetic is tedious. You are advised to use a spreadsheet or write your own computer


programs to perform the tasks.


34.Apply Euler’s method to the initial value problem


y′(x) 1 = 1 −y(x), y(0) 1 = 11


with step sizes (i)h 1 = 1 0.2, (ii)h 1 = 1 0.1, (iii)h 1 = 1 0.05to calculate approximate values of


y(x)forx 1 = 1 0.2, 0.4, 0.6, 0.8, 1.0. Compare these with the values obtained from the exact


solutiony 1 = 1 e


−x

.


For initial value problems in Exercises 35 to 37, (i)apply Euler’s method with step sizeh 1 = 1 0.1


to compute an approximate value ofy(1), (ii)confirm the given exact solution and compute


the error:


35.y′ 1 = 121 − 12 y, y(0) 1 = 1 0; y 1 = 11 −e


− 2 x


  1. y(0) 1 = 1 1;

  2. y(0) 1 = 1 2; y 1 = 1 x


2

1 + 12 x 1 + 121 − 1 2(x 1 + 1 1) ln(x 1 + 1 1)


38.Apply the second-order Runge–Kutta method to the initial value problem in Exercise 34


with step sizes (i)h 1 = 1 0.2, (ii)h 1 = 1 0.1.


39.Apply the fourth-order Runge–Kutta method to the initial value problem in Exercise 34


with step sizes (i)h 1 = 1 0.2, (ii)h 1 = 1 0.1.


Apply the fourth-order Runge–Kutta method to the initial value problems in Exercises 35, 36,


and 37 with step sizeh 1 = 1 0.1to compute (i)y(1)and (ii)the error:


40.As Exercise 35 41.As Exercise 36 42.As Exercise 37


y


yx


x


′=


+−






2

2


1


,


y


x


=


−+


1


11ln( )


y


y


x


′=






,


2

1

















112


311


134

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