2.2 Problem 6 Solved

# 2.2 Problem 6 Solved - r r ( ) = , where = " ! r ( ) =...

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u tt = c 2 u rr + n ! 1 r u r " # \$ % u r , t ( ) = ! r ( ) f t " # r ( ) ( ) u t = r ( ) \$ f t " r ( ) ( ) u tt = r ( ) \$\$ f t " r ( ) ( ) u r = \$ r ( ) f t " r ( ) ( ) " \$ r ( ) r ( ) \$ f t " r ( ) ( ) u rr = \$\$ r ( ) f t " r ( ) ( ) " 2 \$ r ( ) \$ r ( ) \$ f t " r ( ) ( ) " \$\$ r ( ) r ( ) \$ f t " r ( ) ( ) + \$ r ( ) ( ) 2 r ( ) \$\$ f t " r ( ) ( ) Letting t ! " r ( ) = w we have: r ( ) "" f w ( ) # c 2 "" r ( ) f w ( ) # 2 " \$ r ( ) " r ( ) " f w ( ) # "" r ( ) r ( ) " f w ( ) + " r ( ) ( ) 2 r ( ) "" f w ( ) + n # 1 r " r ( ) f w ( ) # " r ( ) r ( ) " f w ( ) ( ) % ( ) * * * = 0 Collecting terms we have: !! f w ( ) r ( ) # c 2 ! r ( ) ( ) 2 r ( ) % ( + ! f w ( ) 2 c 2 ! r ( ) ! r ( ) + c 2 !! r ( ) r ( ) + c 2 n # 1 r ! r ( ) r ( ) % ) ( * + f w ( ) # c 2 !! r ( ) # c 2 n # 1 r ! r ( ) % ) ( * = 0 Setting the coefficients of !! f , ! f , f equal to zero we have: i. r ( ) " c 2 # r ( ) ( ) 2 r ( ) = 0 ii. 2 c 2 ! r ( ) ! r ( ) + c 2 !! r ( ) r ( ) + c 2 n \$ 1 r ! r ( ) r ( ) = 0 iii. ! c 2 "" r ( ) ! c 2 n ! 1 r " r ( ) = 0 From the first equation we have: ! r ( ) = 1 c # r ( ) = r c From the second equation we have: ! r ( ) + n # 1 2 r r ( ) = 0 \$ r ( ) = c 1 r 1 # n 2 From the third equation we have: ! r ( ) + n # 1

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Unformatted text preview: r r ( ) = , where = " ! r ( ) = c 2 r 1 " n # r ( ) = c 2 r 2 " n 2 " n + c 3 Setting r ( ) from equation 2 & 3 equal we have: c 1 r 1 ! n 2 = c 2 r 2 ! n 2 ! n + c 3 Setting the exponents on the r term equal we have: 2 ! n = 1 ! n 2 " n = 3 n = 1 also works, since: c 1 r 1 ! 1 2 = c 1 = " r 2 ! 1 2 ! 1 + c 1...
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## This note was uploaded on 04/08/2008 for the course MATH 3028 taught by Professor Desilva during the Spring '08 term at Columbia.

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2.2 Problem 6 Solved - r r ( ) = , where = " ! r ( ) =...

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