# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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than two.

Determine approximate values of the solution x = ?(t), y = ^(t) of the initial value problem

X = x - 4y, ? = -x + y, (8)

x (0) = 1, y(0) = 0, (9)

at the point t = 0.2. Use the Euler method with h = 0.1 and the Runge-Kutta method with h = 0.2. Compare the results with the values of the exact solution:

e-t + e3t e-t - e3t

$(t) = —2— ’ ^(t) = —4— • (10)

Let us first use the Euler method. For this problem fn = xn - 4yn and gn = —xn +

yn ;hence

= 1 - (4)(0) = 1, g0 = -1 + 0 = -1.

Then, from the Euler formulas (4) and (5) we obtain

x! = 1 + (0.1)(1) = 1.1, yx = 0 + (0.1)(-1) = -0.1.

At the next step

fx = 1.1 - (4)(-0.1) = 1.5, gj = -1.1 + (-0.1) = -1.2.

Consequently,

x2 = 1.1 + (0.1)(1.5) = 1.25, y2 = -0.1 + (0.1)(-1.2) = -0.22.

8.6 Systems of First Order Equations

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The values of the exact solution, correct to eight digits, are 0(0.2) = 1.3204248 and 0(0.2) = —0.25084701. Thus the values calculated from the Euler method are in error by about 0.0704 and 0.0308, respectively, corresponding to percentage errors of about 5.3% and 12.3%.

Now let us use the Runge-Kutta method to approximate 0(0.2) and 0(0.2). With h = 0.2 we obtain the following values from Eqs. (7):

k //(1.15, — 0.12)\ _ / 1.63 \

k03 = ^(1-15, —0.12),/ = V—1.27,/ ’

k / /(1.326, —0.254)\ / 2.342\

k04 = \g(1326, —0.254) j = y—1.580y .

Then, substituting these values in Eq. (6), we obtain

/1V 02 / 9.602\ / 1.3200667 \

X1 = \0/ + 6 \—7.52 j = ^—0.25066667/ .

These values of x1 and j1 are in error by about 0.000358 and 0.000180, respectively, with percentage errors much less than one-tenth of 1%.

This example again illustrates the great gains in accuracy that are obtainable by

using a more accurate approximation method, such as the Runge-Kutta method. In

the calculations we have just outlined, the Runge-Kutta method requires only twice as many function evaluations as the Euler method, but the error in the Runge-Kutta method is about 200 times less than the Euler method.

PROBLEMS In each of Problems 1 through 6 determine approximate values of the solution x = 0(t), ; y = 0(t) of the given initial value problem at t = 0.2, 0.4, 0.6, 0.8, and 1.0. Compare the results obtained by different methods and different step sizes.

(a) Use the Euler method with h = 0.1.

(b) Use the Runge-Kutta method with h = 0.2.

(c) Use the Runge-Kutta method with h = 0.1.

? 1. X = x + y + t, y = 4x — 2y; x(0) = 1, y(0) = 0

? 2. X = 2x + ty, y = xy; x(0) = 1, y(0) = 1

? 3. X =-tx — y — 1, / = x; x(0) = 1, y(0) = 1

? 4. x = x — y + xy, y = 3x — 2y — xy; x(0) = 0, y(0) = 1

? 5. 0 = x(1 — 0.5x — 0.5y), y = y(—0.25 + 0.5x); x(0) = 4, y(0) = 1

? 6. x = exp(—x + y) — cos x, y = sin(x — 3y); x(0) = 1, y(0) = 2

? 7. Consider the example problem X = x — 4y, y = — x + y with the initial conditions

x(0) = 1 and y(0) = 0. Use the Runge-Kutta method to solve this problem on the

interval 0 < t < 1. Start with h = 0.2 and then repeat the calculation with step sizes

h = 0.1, 0.05, ..., each half as long as in the preceding case. Continue the process until

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Chapter 8. Numerical Methods

the first five digits of the solution at t = 1 are unchanged for successive step sizes. Determine whether these digits are accurate by comparing them with the exact solution given in Eqs. (10) in the text.

? 8. Consider the initial value problem

x" + t2 y + 3x = t, x (0) = 1, y (0) = 2.

Convert this problem to a system of two first order equations and determine approximate values of the solution at t = 0.5 and t = 1.0 using the Runge-Kutta method with h = 0.1.

? 9. Consider the initial value problem X = f (t, x, y) and ? = g(t, x, y) with x(t0) = x0

and y(t0) = y0. The generalization of the Adams-Moulton predictor-corrector method of Section 8.4 is

xn+1 = xn + 24h(55 fn - 59 fn-1 + 37 4-2 - 9 fn_3),

yn+1 = yn + 24h(55gn - 59gn-1 + 37gn-2 - 9gn-3)

and

xn+1 = xn + 24h(9 fn+1 + 19 fn - 5 fn-1 + fn-2'), yn+1 = yn + 24h(9gn+1 + 19gn - 5gn-1 + gn-2).

Determine an approximate value of the solution at t = 0.4 for the example initial value

problem x = x - 4 y, y1 = - x + y with x (0) = 1, y(0) = 0. Take h = 0.1. Correct the

predicted value once. For the values of x1,..., y3 use the values of the exact solution rounded to six digits: x1 = 1.12883, x2 = 1.32042, x3 = 1.60021, y1 = —0.110527, y2 = -0.250847, and y3 = -0.429696.

REFERENCES There are many books of varying degrees of sophistication dealing with numerical analysis in general and the numerical solution of ordinary differential equations in particular. Among these are:

Ascher, Uri M., and Petzold, Linda R., Computer Methods for Ordinary Differential Equations and Differential-Algebraic Equations (Philadelphia: Society for Industrial and Applied Mathematics, 1998).

Gear, C. William, Numerical Initial Value Problems in Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice Hall, 1971).

Henrici, Peter, Discrete Variable Methods in Ordinary Differential Equations (New York: Wiley, 1962).

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