Tutorial9

# Tutorial9 - Four sequences of coecients starting with a0 a1...

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Four sequences of coefficients, starting with a 0 , a 1 , a 2 and a 3 decouple. In particular the sequences are a 0 , a 4 , a 8 , a 12 , . . . ; a 1 , a 5 , a 9 , a 13 , . . . ; a 2 , a 6 , a 10 , a 14 , . . . ; a 3 , a 7 , a 11 , a 15 , . . . . Since a 2 = a 3 = 0 , by recurrence relation the last two sequences vanish identically. Starting with a 0 , we have a 0 , a 4 = - k 2 3 · 4 a 0 , a 8 = k 4 3 · 4 · 7 · 8 a 0 , a 12 = - k 6 3 · 4 · 7 · 8 · 11 · 12 a 0 ; and starting with a 1 , a 1 , a 5 = - k 2 4 · 5 a 1 , a 9 = k 4 4 · 5 · 8 · 9 a 1 , a 13 = - k 6 4 · 5 · 8 · 9 · 12 · 13 a 1 . Finally the general solution can therefore written as y ( x ) = a 0 y 0 ( x ) + a 1 y 1 ( x ) , where y 0 ( x ) = 1 - k 2 x 4 3 · 4 + k 4 x 8 3 · 4 · 7 · 8 - k 6 x 12 3 · 4 · 7 · 8 · 11 · 12 + · · · = 1 + X m =0 ( - 1) m +1 ( k 2 x 4 ) m +1 3 · 4 · 7 · 8 · · · (4 m + 3)(4 m + 4) , y 1 ( x ) = x - k 2 x 5 4 · 5 + k 4 x 9 4 · 5 · 8 · 9 - k 6 x 13 4 · 5 · 8 · 9 · 12 · 13 + · · · = x " 1 + X m =0 ( - 1) m +1 ( k 2 x 4 ) m +1 4 · 5 · 8 · 9 · · · (4 m + 4)(4 m + 5) # . 2 Remark. By the relations y (0) = a 0 , y 0 (0) = a 1 , the solution of Example 1 2
can be determined uniquely by the initial conditions y (0) , y 0 (0) .

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• Spring '09
• T.Qian
• Characteristic polynomial, Complex number, Zagreb, Highways in Croatia, Recurrence relation, Regular singular point

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