1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 26 26.1 (a) h < 2/200,000 = 1105. (b) The implicit Euler can be written for this problem as 111200,000 199,999ixii iyyyeh which can be solved for 11199,9991 200,000ixiiyehyh The results of applying this formula for the first few steps are shown below. A plot of the entire solution is also displayed x y 0 0 0.1 0.904788 0.2 0.818731 0.3 0.740818 0.4 0.67032 0.5 0.606531 00.51012 26.2 The implicit Euler can be written for this problem as 111130(cos ) 3sinii ii iyytyth which can be solved for 11130cos 3sin130ii iiyththyh The results of applying this formula are tabulated and graphed below. x y 0 1 0.4 0.96308 0.8 0.783414 1.2 0.480781 1.6 0.102298 2 -0.29233 2.4 -0.640812 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2.8 -0.88811 3.2 -0.99521 3.6 -0.94518 4 -0.74593 -1.5-1-0.500.511.501234 26.3 (a) The explicit Euler can be written for this problem as 1, 1 1, 1, 2,1999 2999ii i ixxxxh 2, 1 2, 1, 2,2000 3000ii i ixxxxh Because the step-size is much too large for the stability requirements, the solution is unstable, t x1 x2 dx1/dt dx2/dt0 1 1 4998 -5000 0.05 250.9 -249 -245202 245200 0.1 -12009.2 12011 12014608 -1.2E+07 0.15 588721.2 -588720 -5.9E+08 5.89E+08 0.2 -2.9E+07 28847084 2.88E+10 -2.9E+10 (b) The implicit Euler can be written for this problem as 1, 1 1, 1, 1 2, 11999 2999ii i ixxx xh 2, 1 2, 1, 1 2, 12000 3000ii i ixxxxh or collecting terms 1, 1 2, 1 1,1, 1 2, 1 2,(1 1999 ) 29992000 (1 3000 )iiiiiihx hx xhx h x x or substituting h = 0.05 and expressing in matrix format 1, 1 1,2, 1 2,98.95 149.95100 151iiiixxxx Thus, to solve for the first time step, we substitute the initial conditions for the right-hand side and solve the 22 system of equations. The best way to do this is with LU decomposition since we will have to solve the system repeatedly. For the present case, because it’s easier to display, we will use the matrix inverse to obtain the solution. Thus, if the matrix is inverted, the solution for the first step amounts to the matrix multiplication,3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1, 12, 12.819795 2.800187 1 5.6199811.86741 1.84781 1 3.71522iixx For the second step (from x = 0.05 to 0.1), 1, 12, 12.819795 2.800187 5.619981 5.4438851.86741 1.84781 3.71522 3.62983iixx The remaining steps can be implemented in a similar fashion to give t x1 x2 0 1 1 0.05 5.619981 -3.71522 0.1 5.443885 -3.62983 0.15 5.186447 -3.45877 0.2 4.939508 -3.2941 The results are plotted below, along with a solution with the explicit Euler using a step of 0.0005. -5051000.10.2x1x2-5051000.10.2x1x2 26.4 The analytical solution is 0.4 210.625 0.625ttyee Therefore, the exact results are y(2.5) = 3.904508 and y(3) = 3.198639. First step: Predictor: y10 = 5.800007+[0.4(4.762673)+e2(2)]1 = 3.913253 Corrector: 2(2) 2(2.5)110.4(4.762673) 0.4(3.913253)4.762673 0.5 3.9013442eey The corrector can be iterated to yield j yi+1j a, % 1 3.901344 2 3.902534 0.0305 The true error can be computed as 3.904508 3.902534100% 0.0506%3.904508t4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Second step: Predictor: y20 = 4.762673 + [0.4(3.902534) + e2(2.5)]1 = 3.208397 Predictor Modifier: y20 = 3.208397 + 4/5(3.9025343.913253) = 3.199822 Corrector: 2(2.5) 2(3)120.4(3.902534) 0.4(3.199822)3.902534 0.5 3.1946032eey The corrector can be iterated to yield j yi+1j a, % 1 3.194603 2 3.195125 0.0163 The true error can be computed as 3.198639 3.195125100% 0.11%3.198639t 26.5 The analytical solution is 0.4 210.625 0.625ttyee Therefore, the exact results are y(2.5) = 3.904508 and y(3) = 3.198639. The Adams method can be implemented with the results predictor = 3.8918505 Corrector Iteration x y ea 2.5 3.905861533 3.59E-01 2.5 3.904810706 2.69E-02 2.5 3.904889518 2.02E-03 predictor = 3.194487763 Corrector Iteration x y ea 3 3.1994013 1.54E-01 3 3.199032785 1.15E-02 3 3.199060424 8.64E-04 Therefore, the true percent relative errors can be computed as 3.904508 3.9048895100% 0.0098%3.904508t 3.198639 3.19906100% 0.0132%3.198639t5 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or
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