253_pdfsam_math 54 differential equation solutions odd

253_pdfsam_math 54 differential equation solutions odd -...

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Review Problems 17. To solve the auxiliary equation, r 3 +10 r 11 = 0, we note that r 1 = 1 is a root. Dividing the polynomial r 3 +10 r 11 by r 1weget r 3 +10 r 11 = ( r 1)( r 2 + r + 11) , and so the other two roots are r 2 , 3 = 1 ± p 1 4(1)(11) 2 = 1 2 ± 43 2 i. A general solution is then given by y ( t )= c 1 e t + e t/ 2 c 2 cos 43 t 2 + c 3 sin 43 t 2  . 19. By inspection, we fnd that r = 3 as a root oF the auxiliary equation, 4 r 3 +8 r 2 11 r +3 = 0. Using, say, the long division, we get 4 r 3 +8 r 2 11 r +3=( r + 3)(4 r 2 4 r +1)=( r + 3)(2 r 1) 2 . Thus, in addition, r =1 / 2 is a double root oF the auxiliary equation. A general solution then
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Unformatted text preview: has the Form y ( t ) = c 1 e 3 t + c 2 e t/ 2 + c 3 te t/ 2 . 21. irst, we solve the corresponding homogeneous equation, y 3 y + 7 y = 0 . Since the roots oF the auxiliary equation, r 2 3 r + 7 = 0, are r = 3 9 28 2 = 3 19 i 2 , a general solution to the homogeneous equation is y h ( t ) = c 1 cos 19 t 2 + c 2 sin 19 t 2 e 3 t/ 2 . 249...
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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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