lecture_26

# lecture_26 - MA 36600 LECTURE NOTES MONDAY MARCH 30 Higher...

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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 X j =0 a n - j d j dt j y = d dt - r k 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 X k =1 C k y k ( t ) = " s k X m =1 C m t m - 1 # e r k t . To see why, define 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 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 u m ( t ) = du m dt - r k u m = 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 s k u s k = d dt - r k s k - 1 u s k - 1 = · · · = d dt - r k u 1 = 0 . Hence { u 1 , . . . , u n } is a fundamental set of solutions. Recap. Consider the n th order nonhomogeneous linear differential equation P 0 ( t ) d n y dt n + P 1 ( t ) d n - 1 y dt n - 1 + · · · + P n - 1 ( t ) dy dt + P n ( t ) y = G ( t ) . We have seen that if we can find ( 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 X j =0 P n - j ( t ) y ( j ) i = 0 for i = 1 , 2 , . . . , n .

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