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Unformatted text preview: , 5 , 10 , . . . , ( 2 . 1503 , . 4854) , n = 1 , 6 , 11 , . . . , ( . 6114 , . 1854) , n = 2 , 7 , 12 , . . . , ( 2 . 5136 , . 1854) , n = 3 , 8 , 13 , . . . , ( . 9747 , . 4854) , n = 4 , 9 , 14 , . . . . Consequently, we can deduce that there is a subharmonic solution of period 10 . 3. With A = F = 1, = 0, = 1, b = . 1, and = 0 (because tan = ( 2 1) /b = 0) the solution (5) to equation (4) becomes x ( t ) = e . 05 t sin 3 . 99 2 t + 10 sin t. Thus v ( t ) = x ( t ) = e . 05 t . 05 sin 3 . 99 2 t + 3 . 99 2 cos 3 . 99 2 t + 10 cos t 325...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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