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# lecture_24 (dragged) 1 - 2 MA 36600 LECTURE NOTES FRIDAY...

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2 MA 36600 LECTURE NOTES: FRIDAY, MARCH 13 We write this using the product of matrices: y 0 Y ( t 0 ) y (1) 0 Y (1) ( t 0 ) y (2) 0 Y (2) ( t 0 ) . . . y ( n 1) 0 Y ( n 1) ( t 0 ) = y 1 ( t 0 ) y 2 ( t 0 ) · · · y n ( t 0 ) y (1) 1 ( t 0 ) y (1) 2 ( t 0 ) · · · y (1) n ( t 0 ) y (2) 1 ( t 0 ) y (2) 2 ( t 0 ) · · · y (2) n ( t 0 ) . . . . . . . . . . . . y ( n 1) 1 ( t 0 ) y ( n 1) 2 ( t 0 ) · · · y ( n 1) n ( t 0 ) c 1 c 2 c 3 . . . c n . We can solve for the c i because the n × n matrix is nonsingular by (2). Hence we have the solution as claimed. Linear Independence. Consider a collection of functions { f 1 , f 2 , . . . , f n } . We say that this is a linearly independent set if the only solution to the equation k 1 f 1 ( t ) + k 2 f 2 ( t ) + · · · + k n f n ( t ) = 0 for all t is k 1 = k 2 = · · · = k n = 0. We say that this is a linearly dependent set otherwise. We explain the relationship with the Wronskian
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