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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A computer will be required for the problem. In this Problem, an initial value problem and its exact solution are given. In each of these four problems, use the Runge–Kutta method with step sizes h = 0.1 and h = 0.05 to approximate to five decimal places the values x(1) and y(1). Compare the approximations with the actual values.

x'' + x = sin t , x(0) = 0;

x(t) = 1/2 (sin t - t cos t)

Expert Solution

Step 1

Given

$x'' + x = \sin t, x (0) = 0$

and $x (t) = \frac{1}{2} (\sin t - t \cos t)$

and h=0.1

Step 2

Then

$x'' + x = \sin t x'' = - x + \sin t$

Consider $x' = y \Rightarrow x'' = y'$

$y' = - x + \sin t f (t_{n}, x_{n}, y_{n}) = - x_{n} + \sin (t_{n}) . . . . (1)$

Step 3

By using the Runge–Kutta method with h = 0.1

First iteration

$k_{1} = h y_{0} = 0.1 \times 0 = 0 k_{2} = h (y_{0} + \frac{k_{1}}{2}) = 0.1 (0 + \frac{0}{2}) = 0 k_{3} = h (y_{0} + \frac{k_{2}}{2}) = 0.1 (0 + \frac{0}{2}) = 0 k_{4} = h (y_{0} + k_{3}) = 0.1 (0 + 0) = 0 x_{1} = x_{0} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0 + \frac{1}{6} (0) = 0 a n d k_{1} = h f (t_{0}, x_{0}, y_{0}) = 0.1 (- x_{0} + \sin (t_{0})) = 0.1 (- 0 + \sin (0)) = 0 k_{2} = h f (t_{0} + \frac{h}{2}, x_{0} + \frac{k_{1}}{2}, y_{0} + \frac{k_{1}}{2}) = h f (0 + \frac{0.1}{2}, 0 + \frac{0}{2}, 0 + \frac{0}{2}) = 0.1 (- 0 + \sin (\frac{0.1}{2})) = 0.00499 k_{3} = h f (t_{0} + \frac{h}{2}, x_{0} + \frac{k_{2}}{2}, y_{0} + \frac{k_{2}}{2}) = h f (0 + \frac{0.1}{2}, 0 + \frac{0.00499}{2}, 0 + \frac{0.00499}{2}) = 0.1 (- \frac{0.00499}{2} + \sin (\frac{0.1}{2})) = 0.00474 k_{4} = h f (t_{0} + h, x_{0} + k_{3}, y_{0} + k_{3}) = h f (0 + 0.1, 0 + 0.00474, 0 + 0.00474) = 0.1 (- 0.00474 + \sin (0.1)) = 0.00950 y_{1} = y_{0} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0 + \frac{1}{6} (0 + 2 \times 0.00499 + 2 \times 0.00474 + 0.00950) = 0.00482 a n d t_{1} = 0.1$

Second iteration

$k_{1} = h y_{1} = 0.1 \times 0.00482 = 0.000482 k_{2} = h (y_{1} + \frac{k_{1}}{2}) = 0.1 (0.00482 + \frac{0.000482}{2}) = 0.0005061 k_{3} = h (y_{1} + \frac{k_{2}}{2}) = 0.1 (0.00482 + \frac{0.0005061}{2}) = 0.000507305 k_{4} = h (y_{1} + k_{3}) = 0.1 (0.00482 + 0.000507305) = 0.0005327305 x_{2} = x_{1} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0 + \frac{1}{6} (0.000482 + 2 (0.0005061) + 2 (0.000507305) + 0.0005327305) = = 0.00051$

and

$k_{1} = h f (t_{1}, x_{1}, y_{1}) = 0.1 (- x_{1} + \sin (t_{1})) = 0.1 (- 0 + \sin (0.1)) = 0.00998 k_{2} = h f (t_{1} + \frac{h}{2}, x_{1} + \frac{k_{1}}{2}, y_{1} + \frac{k_{1}}{2}) = h f (0.1 + \frac{0.1}{2}, 0 + \frac{0.00998}{2}, 0.00482 + \frac{0.00998}{2}) = 0.1 (- \frac{0.00998}{2} + \sin (0.1 + \frac{0.1}{2})) = 0.01444 k_{3} = h f (t_{1} + \frac{h}{2}, x_{1} + \frac{k_{2}}{2}, y_{1} + \frac{k_{2}}{2}) = h f (0.1 + \frac{0.1}{2}, 0 + \frac{0.01444}{2}, 0.00482 + \frac{0.01444}{2}) = 0.1 (- \frac{0.01444}{2} + \sin (0.1 + \frac{0.1}{2})) = 0.01422 k_{4} = h f (t_{1} + h, x_{1} + k_{3}, y_{1} + k_{3}) = h f (0.1 + 0.1, 0 + 0.01422, 0.00482 + 0.01422) = 0.1 (- 0.01422 + \sin (0.2)) = 0.01844 y_{2} = y_{1} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0.00482 + \frac{1}{6} (0.00998 + 2 \times 0.01444 + 2 \times 0.01422 + 0.01844) = 0.01911$

and $t_{2} = 0.2$

Step 4

Third Iteration

$k_{1} = h y_{2} = 0.1 \times 0.01911 = 0.001911 k_{2} = 0.1 (0.01911 + \frac{0.001911}{2}) = 0.00201 k_{3} = 0.1 (0.01911 + \frac{0.00200655}{2}) = 0.00201 k_{4} = 0.1 (0.01911 + 0.0020113275) = 0.00211 x_{3} = 0.00051 + \frac{1}{6} (0.001911 + 2 (0.00201) + 2 (0.00201) + 0.00211) = 0.00252$

and

$k_{1} = h f (t_{2}, x_{2}, y_{2}) = 0.1 (- x_{2} + \sin (t_{2})) = 0.1 (- 0.00051 + \sin (0.2)) = 0.01981 k_{2} = h f (0.2 + \frac{0.1}{2}, 0.00051 + \frac{0.01981}{2}, 0.01911 + \frac{0.01981}{2}) = 0.1 (- (0.00051 + \frac{0.01981}{2}) + \sin (0.2 + \frac{0.1}{2})) = 0.02369 k_{3} = h f (0.2 + \frac{0.1}{2}, 0.00051 + \frac{0.02369}{2}, 0.01911 + \frac{0.02369}{2}) = 0.1 (- (0.00051 + \frac{0.02369}{2}) + \sin (0.2 + \frac{0.1}{2})) = 0.02350 k_{4} = h f (0.2 + 0.1, 0.00051 + 0.02350, 0.01911 + 0.02350) = 0.1 (- (0.00051 + 0.02350) + \sin (0.2 + 0.1)) = 0.02715 y_{3} = y_{2} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0.01911 + \frac{1}{6} (0.01981 + 2 \times 0.02369 + 2 \times 0.02350 + 0.02715) = 0.04266$

and $t_{3} = 0.3$

Fourth iteration

$k_{1} = h y_{3} = 0.1 \times 0.04266 = 0.004266 k_{2} = 0.1 (0.04266 + \frac{0.004266}{2}) = 0.00448 k_{3} = 0.1 (0.04266 + \frac{0.00448}{2}) = 0.00449 k_{4} = 0.1 (0.04266 + 0.00449) = 0.004715 x_{4} = 0.00252 + \frac{1}{6} (0.004266 + 2 (0.00448) + 2 (0.00449) + 0.004715) = 0.00701$

and

$k_{1} = h f (t_{3}, x_{3}, y_{3}) = 0.1 (- x_{3} + \sin (t_{3})) = 0.1 (- 0.00252 + \sin (0.3)) = 0.02930 k_{2} = h f (0.3 + \frac{0.1}{2}, 0.00252 + \frac{0.02930}{2}, 0.04266 + \frac{0.02930}{2}) = 0.1 (- (0.00252 + \frac{0.02930}{2}) + \sin (0.3 + \frac{0.1}{2})) = 0.03257 k_{3} = h f (0.3 + \frac{0.1}{2}, 0.00252 + \frac{0.03257}{2}, 0.04266 + \frac{0.03257}{2}) = 0.1 (- (0.00252 + \frac{0.03257}{2}) + \sin (0.3 + \frac{0.1}{2})) = 0.03241 k_{4} = h f (0.3 + 0.1, 0.00252 + 0.03241, 0.04266 + 0.03241) = 0.1 (- (0.00252 + 0.03241) + \sin (0.3 + 0.1)) = 0.03544 y_{4} = y_{3} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0.04266 + \frac{1}{6} (0.02930 + 2 \times 0.03257 + 2 \times 0.03241 + 0.03544) = 0.07511$

and $t_{4} = 0.4$

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