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lecture_25 (dragged) 2

lecture_25 (dragged) 2 - n ³ j =0 a n − j d j dt j y =...

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MA 36600 LECTURE NOTES: MONDAY, MARCH 23 3 Case #1: Real and Unequal Roots. Say for the moment that we have the characteristic polynomial Z ( r )= p ° k =1 ± r r k ² for some distinct real numbers r k . We will write the corresponding operator L [ y ] in the following form: L [ y ]= n ³ j =0 a n j d j dt j y = ´ p ° k =1 µ d dt r k · y where n = p . Clearly a fundamental set of solutions is { y 1 ,...,y p } ,where y k ( t )= e r k t . Hence the general solution is y ( t )= n ³ k =1 C k y k ( t )= p ³ k =1 C k e r k t . Case #2: Complex and Unequal Roots. Say for the moment that we have the characteristic polynomial Z ( r )= q ° k =1 ¸ ± r [ λ k + k ] ²± r [ λ k k ] ² ¹ for some distinct complex numbers r k = λ k ± k . We will write the corresponding operator L [ y ]inthe following form: L [ y ]=
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Unformatted text preview: n ³ j =0 a n − j d j dt j y = ´ q ° k =1 µ d dt − º λ k + i μ k » ¶µ d dt − º λ k − i μ k ² ¶ · y where n = 2 q . Recall that Euler’s Formula states that e r k t = e λ k t cos μ k t ± i e λ k t sin μ k t where r k = λ k ± i μ k . Clearly a fundamental set of solutions is { y 11 , y 21 , . . . , y 1 q , y 2 q } , where y 1 k ( t ) = e λ k t cos μ k t and y 2 k ( t ) = e λ k t sin μ k t . Hence the general solution is y ( t ) = n ³ k =1 C k y k ( t ) = q ³ k =1 ± A k e λ k t cos μ k t + B k e λ k t sin μ k t ² ....
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