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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 ( 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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