Homework1_Solutions

Homework1_Solutions - MA 107 Modeling And Analysis of...

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MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao 1. Prove that Item (1) and (2) above are true by direct evaluation. Consider only one repeated characteristic root in (1), i.e. p=2. Item (1): to prove that 11 12 ( ) ... n rt r t hn yt k e k t e ke =+ + + , where r i are roots of the characteristic equation: 1 0 () . . . 0 nn n As s a s as a + + += , is the solution of 1 0 ... 0 n Dy a D y aD y ay ++ + + = , we simply plug the solutions into the differential equation and verify the equality: 1 1 1 1 1 1 1 0 1 0 ,... ... ( ... ) 0 n n n n De re D e r e ra r a r a e == + + = + + = 1 1 1 1 1 1 2 1 1 1 1 1 1 0 1 0 1 1 1 ( 1 ) ()( 1 ) ( 2 ) ,... ... ( ... ) ( ( 1) ... ) n Dte rte D te r rt e r De r n r te r a r a t e n r an r a e −− + = + + + = + + + + + + Note that the ( ) in the first term is zero since r1 is a characteristic root. To prove that second term is also zero we must somehow make use of the fact that r1 is a repeated characteristic root, which means that 1 0 1 2 1 1 1 1 . . . ( ) () ( ... 2( ) ( ) ( ) ( ... 0 n n n sr s s r Ps dA s dP s n s a n s a srP s sr ds ds dA s nr a n r a ds = + + +=− + + = + + + = Thus the ( ) in the second term is also zero.
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Homework1_Solutions - MA 107 Modeling And Analysis of...

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