Differential Equations Solutions 113

Differential Equations Solutions 113 - 123 CHALLENGE 20.13....

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123 CHALLENGE 20.13. y ( t )= u ( t ) v ( t ) w ( t ) , M ( t 100 010 000 , A ( t 7 60 4 20 111 , f ( t 4 t 0 24 . CHALLENGE 20.14. We calculate y ( t · u ( t ) p ( t ) ¸ , M = · I n u 0 00 ¸ , A = · CB B T 0 ¸ , f ( t 0 . Therefore, P 1 = I n u 0 0 0 A BI n u 0 B T 0 , N 1 = B T 0 0 0 0 , so rank( P 1 )=2( n u + n p ) 2 n p . Therefore we take n a =2 n p . We need a basis for the nullspace of P T 1 , and we can take, for example, Z T = · 0 I n p B T I n p ¸ . Now we calculate b N 1 = · B T 0 B T T B ¸ , so we can take, for example, T = · X ( B T B ) 1 B T CX ¸ , which has n d = n u n p columns. Then MT = · I n u 0 ¸· X ( B T B )
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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