# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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e~* + e3t e~* e3t

τ() = 2, ^(t) = 4. (10)

Let us first use the Euler method. For this problem fn = xn 4yn and gn = xn + Σο ; hence

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

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

x1 = 1 + (0.1)(1) = 1.1, Σ1 = 0 + (0.1)(1) = -0.1.

At the next step

f1 = 1.1 (4)(0.1) = 1.5, g1 = -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 τ(0.2) = 1.3204248 and ¦τ-(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 τ(0.2) and ty(0.2). With h = 0.2 we obtain the following values from Eqs. (7):

, / f(1.15, 0.12)\ _ / 1.63 \

k°3 = V.g(115, 0.12)σ = \1.27) ;

k ( f (1.326, 0.254)\ ( 2.342\

^o4 = yg(1.326, 0.254) j = y1.580y .

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

/Λ 02 f 9.602\ / 1.3200667 \

x! = \0/ + 6 V7.52/ = \0.25066667J ^

These values of x1 and y1 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 = τ(΄),

³ y = ft(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, Σ = 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, y = x; x(0) = 1, y(0) = 1

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

> 5. x = 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 X + 3x = t, x (0) = 1, Σ (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 f = 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 Λ - 59 fn-1 + 37 fn-2 - 9 fn-3)

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

and

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

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