62_pdfsam_math 54 differential equation solutions odd

# 62_pdfsam_math 54 differential equation solutions odd - j...

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Chapter 2 for j odd dT j dt = T j dT j T j = dt ln | T j | = t + c j T j ( t )= C j e t ; where C j 6 = 0 are constants. From (2.12) we have: 0= T 0 (9) = ( 1 C 0 e t ) ± ± ± t =9 =1 C 0 e 9 C 0 = e 9 . Also from (2.13), for even values of j (say, j =2 k )weget ( 1 C 2 k e t ) ± ± ± t =9+(2 k +1) = C 2 k +1 e t ± ± ± t =9+(2 k +1) 1 C 2 k e (10+2 k ) = C 2 k +1 e (10+2 k ) C 2 k +1 = e 10+2 k C 2 k . Similarly from (2.13) for odd values of j (say, j =2 k +1)weget C 2 k +1 e t ± ± ± t =9+(2 k +2) = ( 1 C 2 k +2 e t ) ± ± ± t =9+(2 k +2) C 2 k +1 e (11+2 k ) =1 C 2 k +2 e (11+2 k ) C 2 k +2 = e 11+2 k C 2 k +1 . In general we see that for any integer
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Unformatted text preview: j (even or odd) the following formula holds: C j = e 9+ j C j 1 . Using this recurrence formula we successively compute C 1 = e 10 C = e 10 e 9 = e 9 ( e 1) C 2 = e 11 C 1 = e 11 e 10 + e 9 = e 9 ( e 2 e + 1) . . . C j = e 9 j k =0 ( 1) j k e k . 58...
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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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