lecture_26

lecture_26 - MA 36600 LECTURE NOTES: MONDAY, MARCH 30...

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Unformatted text preview: 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 ( 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 ) ....
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.

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lecture_26 - MA 36600 LECTURE NOTES: MONDAY, MARCH 30...

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