Thus y x a 0 1 1 x x 1 4 2 x 2 3 1 6 4 2 x 3 5 3 1 8

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Thus y ( x ) = a 0 1 + 1 x x 1 4 · 2 x 2 + 3 · 1 6 · 4 · 2 x 3 5 · 3 · 1 8 · 6 · 4 · 2 x 4 + 7 · 5 · 3 · 1 10 · 8 · 6 · 4 · 2 x 5 − · · · = a 0 1 + x . The radius of convergence of this binomial series is R = 1. 10.10.9: We use series methods to solve ( x 1) dy dx + 2 y = 0. Assume that y ( x ) = n =0 a n x n , so that y ( x ) = n =1 na n x n 1 = n =0 ( n + 1) a n +1 x n . Substitution in the given differential equation yields n =1 na n x n n =0 ( n + 1) a n +1 x n + n =0 2 a n x n = 0 . When n = 0, we have a 1 + 2 a 0 = 0, so that a 1 = 2 a 0 . If n 1, then na n ( n + 1) a n +1 + 2 a n = 0 : ( n + 1) a n +1 = ( n + 2) a n , 1248
and hence a n +1 = n + 2 n + 1 a n if n 0 . Therefore a 2 = 3 2 a 1 = 3 a 0 , a 3 = 4 3 a 2 = 4 a 0 , a 4 = 5 4 a 3 = 5 a 0 , . . . ; in general, a n = ( n + 1) a 0 if n 1. Therefore y ( x ) = a 0 n =0 ( n + 1) x n . Now y ( x ) = F ( x ) where F ( x ) = a 0 n =0 x n +1 = a 0 x 1 x . Consequently, y ( x ) = F ( x ) = a 0 (1 x + x ) (1 x ) 2 = a 0 (1 x ) 2 . The radius of convergence of the series for y ( x ) is R = 1. 10.10.10: We use series methods to solve 2( x 1) dy dx = 3 y . Assume that y ( x ) = n =0 a n x n , so that y ( x ) = n =1 na n x n 1 = n =0 ( n + 1) a n +1 x n . Substitution in the given differential equation yields n =1 2 na n x n n =0 2( n + 1) a n +1 x n = n =0 3 a n x n . It is convenient, and has no effect, if we change the range of the index in the first sum from 1 n < + to 0 n < + . Thus we find that, if n 0, then 2 na n 2( n + 1) a n +1 = 3 a n ; 2( n + 1) a n +1 = (2 n 3) a n ; a n +1 = 2 n 3 2 n + 2 a n . Therefore a 1 = 3 2 a 0 , a 2 = 1 4 a 1 = 1 · 3 4 · 2 a 0 , 1249
a 3 = 1 6 a 2 = 1 · 1 · 3 6 · 4 · 2 a 0 , a 4 = 3 8 a 3 = 3 · 1 · 1 · 3 8 · 6 · 4 · 2 a 0 , and so on. Thus y ( x ) = a 0 1 3 2 x + 1 · 3 4 · 2 x 2 + 1 · 1 · 3 6 · 4 · 2 x 3 + 3 · 1 · 1 · 3 8 · 6 · 4 · 2 x 4 + 5 · 3 · 1 · 1 · 3 10 · 8 · 6 · 4 · 2 x 5 + · · · = a 0 (1 x ) 3 / 2 . The radius of convergence of this binomial series is R = 1. 10.10.11: We use series methods to solve the differential equation y = y . Assume the existence of a solution of the form y ( x ) = n =0 a n x n . Then y ( x ) = n =1 na n x n 1 = n =0 ( n + 1) a n +1 x n and y ( x ) = n =2 n ( n 1) a n x n 2 = n =0 ( n + 2)( n + 1) a n +2 x n . Then substitution in the given differential equation yields a n +2 = a n ( n + 2)( n + 2) for n 0 . Therefore a 2 = a 0 2 · 1 , a 3 = a 1 3 · 2 , a 4 = a 0 4! , a 5 = a 1 5! , and so on. Hence y ( x ) = a 0 1 + x 2 2! + x 4 4! + x 6 6! + · · · + a 1 x + x 3 3! + x 5 5! + x 7 7! + · · · = a 0 cosh x + a 1 sinh x. The radius of convergence of all series here is R = + . The solution may also be expressed in the form y ( x ) = c 1 e x + c 2 e x . 10.10.12: We use series methods to solve the differential equation y = 4 y . Assume the existence of a solution of the form y ( x ) = n =0 a n x n . 1250
Then y ( x ) = n =1 na n x n 1 = n =0 ( n + 1) a n +1 x n and y ( x ) = n =2 n ( n 1) a n x n 2 = n =0 ( n + 2)( n + 1) a n +2 x n . Much as in the solution of Problem 11, substitution in the given differential equation leads to a n +2 = 4 a n ( n + 2)( n + 1) for n 0 .

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