Moler Ch7 solutions

Moler Ch7 solutions - Solution Manual for Numerical...

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Unformatted text preview: Solution Manual for: Numerical Computing with MATLAB by Cleve B. Moler John L. Weatherwax * July 25, 2007 Chapter 7 (Ordinary Differential Equations) Problem 7.1 Defining the vector y as y = v u ˙ v ˙ u (1) Then we have for its time derivative the following dy dt = ˙ v ˙ u ¨ v ¨ u = y 3 y 4- y 2 1+ t 2 + cos( r )- y 1 1+ t 2- sin( r ) (2) Where r given in terms of components of y is r = q y 2 4 + y 2 3 (3) Given the initial conditions in terms of u , v , ˙ u , and ˙ v we have y = 1 (4) * [email protected] 1 Problem 7.2 See the matlab code in the file prob 7 2.m . Problem 7.3 The algorithm BS23 can be represented by the following sequence of steps s 1 = f ( t n , y n ) (5) s 2 = f ( t n + h 2 , y n + h 2 s 1 ) (6) s 3 = f ( t n + 3 4 h, y n + 3 4 s 2 ) (7) y n +1 = y n + h 9 (2 s 1 + 3 s 2 + 4 s 3 ) (8) Part (a): The specific ODE’s (and their exact solutions) to consider are given by dy dt = 1 ⇒ y = t + C (9) dy dt = t ⇒ y = t 2 2 + C (10) dy dt = t 2 ⇒ y = t 3 3 + C (11) dy dt = t 3 ⇒ y = t 4 4 + C (12) For the first system, dy/dt = 1, so f ( t, y ) = 1 then the BS23 procedure gives the following s 1 = 1 (13) s 2 = 1 (14) s 3 = 1 (15) y n +1 = y n + h 9 (2 + 3 + 4) = y n + h (16) Does this equal the true solution at t n + h ? We evaluate the above to obtain y ( t n + h ) = ( t n + h ) + C = t n + C + h = y n + h (17) and the answer is yes. For the second function f ( t, y ) = t we see that the BS23 procedure is s 1 = t n (18) s 2 = t n + h 2 (19) s 3 = t n + 3 4 h (20) y n +1 = y n + h 2 (2 t n + 3( t n + h 2 ) + 4( t n + 3 4 h )) (21) Simplifying this last equation we have = y n + h 9 (9 t n + ( 3 2 + 3) h ) (22) = y n + ht n + h 9 ( 9 2 ) h (23) = y n + ht n + h 2 2 (24) Again we ask, does this equal the true solution at t n +1 ? Lets check by evaluating y ( t n +1 ) = y ( t n + h ) = 1 2 ( t n + h ) 2 + C = t 2 n 2 + C + ht n + h 2 2 = y ( t n ) + ht n + h 2 2 (25) and we see that the answer is yes. For the second function f ( t, y ) = t 2 , then the BS23 gives s 1 = t 2 n (26) s 2 = ( t n + h 2 ) 2 (27) s 3 = ( t n + 3 4 h ) 2 (28) y n +1 = y n + h 9 (2 t 2 n + 3( t n + h 2 ) 2 + 4( t n + 3 4 h ) 2 ) (29) Now simplifying the last equation we have y n +1 = y n + h 9 (2 t 2 n + (3 + 6) ht n + ( 3 4 + 9 4 ) h 2 ) (30) = y n + h 9 (9 t 2 n + 9 2 ht n + 12 4 h 2 ) (31) = y n + ht 2 n + h 2 t n + h 3 1 3 (32) Does this equal the exact solution at t n + h ? To decide we compute y ( t n + h ) = ( t n...
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This note was uploaded on 05/18/2010 for the course MAE 107 taught by Professor Goddard during the Spring '10 term at San Diego.

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Moler Ch7 solutions - Solution Manual for Numerical...

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