lecture_26 (dragged)

lecture_26 (dragged) - − 1 P n − 1 t dy dt P n t...

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MA 36600 LECTURE NOTES: MONDAY, MARCH 30 Higher Order Linear Equations Case #3: Repeated Roots. Say for the moment that we have Z ( r )= ° r r k ± s k for some complex number r k . We will write the corresponding operator L [ y ] in the following form: L [ y ]= n ² j =0 a n j d j dt j y = ³´ d dt r k µ s k y where n = s k . We will show that a fundamental set of solutions consists of { y 1 ,...,y s k } where y m ( t )= t m 1 e r k t . Hence, the general solution is y ( t )= n ² k =1 C k y k ( t )= · s k ² m =1 C m t m 1 ¸ e r k t . To see why, deFne the function u m ( t )= t m 1 ( m 1)! e r k t = y m ( t ) ( m 1)! for m =1 , 2 , 3 ,...,s k . We know that u 1 ( t )= e r k t = ´ d dt r k µ u 1 ( t )= du 1 dt r k u 1 = r k e r k t r k e r k t =0 . In fact, inductively, we have ´ d dt r k µ u m ( t )= du m dt r k u m = ³ t m 2 ( m 2)! e r k t + r k t m 1 ( m 1)! e r k t r k u m = u m 1 ( t ); so that whem m = s k we have L [ u s k ]= ´ d dt r k µ s k u s k = ´ d dt r k µ s k 1 u s k 1 = ··· = ´ d dt r k µ u 1 =0 . Hence { u 1 ,...,u n } is a fundamental set of solutions. Recap. Consider the n th order nonhomogeneous linear di±erential equation P 0 ( t ) d n y dt n + P 1 ( t ) d n 1 y dt n
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Unformatted text preview: − 1 + · · · + P n − 1 ( t ) dy dt + P n ( t ) y = G ( t ) . We have seen that if we can Fnd ( n + 1) functions y 1 = y 1 ( t ) , ..., y n ( t ) and Y = Y ( t ) such that (1) Each y i = y i ( t ) is a solution to the homogeneous equation n ² j =0 P n − j ( t ) y ( j ) i = 0 for i = 1 , 2 , ..., n . (2) The n × n determinant is nonzero: W ° y 1 , y 2 , ..., y n ± ( t ) = ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ y 1 ( t ) y 2 ( t ) · · · y n ( t ) y (1) 1 ( t ) y (1) 2 ( t ) · · · y (1) n ( t ) . . . . . . . . . . . . y ( n − 1) 1 ( t ) y ( n − 1) 2 ( t ) · · · y ( n − 1) n ( t ) ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ° = 0 . 1...
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