238_pdfsam_math 54 differential equation solutions odd

# 238_pdfsam_math 54 differential equation solutions odd - 2...

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Chapter 4 b =0 b =0 b =0. r = ± 64 2 = ± 4 i. A general solution has the form y = C 1 cos 4 t + C 2 sin 4 t . Constants C 1 and C 2 can be found from the initial conditions. y (0) = ( C 1 cos 4 t + C 2 sin 4 t ) ± ± t =0 = C 1 =1 , y 0 (0) = ( 4 C 1 sin 4 t +4 C 2 cos 4 t ) ± ± t =0 =4 C 2 =0 C 1 =1 , C 2 =0 and so y ( t )=cos4 t . b =6 b =6 b =6. r = 6 ± 36 64 2 = 3 ± 7 i. A general solution has the form y =( C 1 cos 7 t + C 2 sin 7 t ) e 3 t .F o rcon s tan t s C 1 and C 2 ,wehavethesystem y (0) = ( C 1 cos 7 t + C 2 sin 7 t ) e 3 t ± ± t =0 = C 1 =1 , y 0 (0) = ² ( 7 C 2 3 C 1 )cos 7 t ( 7 C 1 +3 C 2 )sin 7 t ³ e 3 t ± ± t =0 = 7 C 2 3 C 1 =0 C 1 =1 , C 2 =3 / 7 , and so y ( t )= ´ cos 7 t + 3 7 sin 7 t µ e 3 t = 4 7 e 3 t sin 7 t + φ · , where φ = arctan( 7 / 3) 0 . 723 . b =8 b =8 b =8. r =
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Unformatted text preview: 2 = − 4 . Thus, r = − 4 is a double root of the characteristic equation. So, a general solution has the form y = ( C 1 t + C ) e − 4 t . For constants C 1 and C 2 , we obtain the system y (0) = ( C 1 t + C ) e − 4 t ± ± t =0 = C = 1 , y (0) = ( − 4 C 1 t − 4 C + C 1 ) e − 4 t ± ± t =0 = C 1 − 4 C = 0 ⇒ C = 1 , C 1 = 4 , and so y ( t ) = (4 t + 1) e − 4 t . 234...
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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 Berkeley.

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