DOC PREVIEW
TAMU PETE 301 - Numerical Methods for Engineers Ch. 26 Solutions

This preview shows page 1-2-3-4-5-6 out of 19 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 19 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

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 = 1105. (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 22 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)+e2(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%3.904508t4 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) + e2(2.5)]1 = 3.208397 Predictor Modifier: y20 = 3.208397 + 4/5(3.9025343.913253) = 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%3.198639t5 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


View Full Document

TAMU PETE 301 - Numerical Methods for Engineers Ch. 26 Solutions

Documents in this Course
Load more
Download Numerical Methods for Engineers Ch. 26 Solutions
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Numerical Methods for Engineers Ch. 26 Solutions and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Numerical Methods for Engineers Ch. 26 Solutions 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?