239_pdfsam_math 54 differential equation solutions odd

239_pdfsam_math 54 differential equation solutions odd - y...

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Exercises 4.8 b =10 b =10 b = 10. r = 10 ± 100 64 2 = 5 ± 3 . Thus, r = 2, 8, and a general solution is given by y = C 1 e 2 t + C 2 e 8 t .I n i t i a l conditions yield y (0) = ( C 1 e 2 t + C 2 e 8 t ) ± ± t =0 = C 1 + C 2 =1 , y 0 (0) = ( 2 C 1 e 2 t 8 C 2 e 8 t ) ± ± t =0 = 2 C 1 8 C 2 =0 C 1 =4 / 3 , C 2 = 1 / 3 , and, therefore, y ( t )=(4 / 3) e 2 t (1 / 3) e 8 t is the solution to the initial value problem. The graphs of the solutions are depicted in Figures B.19–B.22 in the answers in the text. 5. The auxiliary equation associated with given di±erential equation is r 2 +10 r + k =0,andits roots are r = 5 ± 25 k k =20 k =20 k =20 . Inth iscase , r = 5 ± 25 20 = 5 ± 5. Thus, a general solution is given by y = C 1 e ( 5+ 5) t + C 2 e ( 5 5) t . The initial conditions yield y (0) = h C 1 e ( 5+ 5) t + C 2 e ( 5 5) t i ± ± ± t =0 = C 1 + C 2 =1 , y 0 (0) = h ( 5+ 5) C 1 e ( 5+ 5) t +( 5 5) C 2 e ( 5 5) t i ± ± ± t =0 =( 5+ 5) C 1 +( 5 5) C 2 =0 C 1 = ( 1+ 5 ) / 2 , C 2 = ( 1
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Unformatted text preview: y ( t ) = [ ( 1 + 5 ) / 2] e ( 5+ 5) t + [ ( 1 5 ) / 2] e ( 5 5) t is the solution to the initial value problem. k = 25 k = 25 = 25. Then r = 5 25 25 = 5. Thus, r = 5 is a double root of the characteristic equation. So, a general solution has the form y = ( C 1 t + C ) e 5 t . For constants C 1 and C 2 , using the initial conditions, we obtain the system y (0) = ( C 1 t + C ) e 5 t t =0 = C = 1 , y (0) = ( 5 C 1 t 5 C + C 1 ) e 5 t t =0 = C 1 5 C = 0 C = 1 , C 1 = 5 , and so y ( t ) = (5 t + 1) e 5 t . 235...
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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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