360_pdfsam_math 54 differential equation solutions odd

360_pdfsam_math 54 differential equation solutions odd - ....

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Chapter 6 27. Solving the auxiliary equation, r 4 +2 r 3 3 r 2 r +(1 / 2) = 0, using computer software yields the roots r 1 =1 . 119967680 ,r 2 =0 . 2963247800 ,r 3 = 0 . 5202201098 ,r 4 = 2 . 896072350 . Thus, all the roots are real and distinct. A general solution to the given equation is, therefore, y ( x )= c 1 e r 1 x + c 2 e r 2 x + c 3 e r 3 x + c 4 e r 4 x c 1 e 1 . 120 x + c 2 e 0 . 296 x + c 3 e 0 . 520 x + c 4 e 2 . 896 x . 29. The auxiliary equation in this problem is r 4 +2 r 3 +4 r 2 +3 r +2=0. Let g ( r )= r 4 +2 r 3 +4 r 2 +3 r +2 g 0 ( r )=4 r 3 +6 r 2 +8 r +3 . Then the Newton’s recursion formula (2) in Appendix A of the text becomes r n +1 = r n r 4 n +2 r 3 n +4 r 2 n +3 r n +2 4 r 3 n +6 r 2 n +8 r n +3 . With initial guess r 0 =1+ i , this formula yields r 1 =(1+ i ) (1 + i ) 4 +2(1+ i ) 3 +4(1+ i ) 2 +3(1+ i )+2 4(1 + i ) 3 +6(1+ i ) 2 +8(1+ i )+3 0 . 481715 + 0 . 837327 i, r 2 = r 1 r 4 1 +2 r 3 1 +4 r 2 1 +3 r 1 +2 4 r 3 1 +6 r 2 1 +8 r 1 +3
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Unformatted text preview: . 052833 + 0 . 763496 i , r 3 = r 2 r 4 2 + 2 r 3 2 + 4 r 2 2 + 3 r 2 + 2 4 r 3 2 + 6 r 2 2 + 8 r 2 + 3 . 284333 + 0 . 789859 i , . . . r 7 = r 6 r 4 6 + 2 r 3 6 + 4 r 2 6 + 3 r 6 + 2 4 r 3 6 + 6 r 2 6 + 8 r 6 + 3 . 500000 + 0 . 866025 i , r 8 = r 7 r 4 7 + 2 r 3 7 + 4 r 2 7 + 3 r 7 + 2 4 r 3 7 + 6 r 2 7 + 8 r 7 + 3 . 500000 + 0 . 866025 i . Therefore, Frst two roots of the auxiliary equation are r . 5 + 0 . 866 i and r = . 5 + 0 . 866 i = . 5 . 866 i . 356...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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