527solns3

# 527solns3 - 642:527 SOLUTIONS ASSIGNMENT 3 FALL 2007...

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642:527 SOLUTIONS: ASSIGNMENT 3 FALL 2007 4.3.6 (e) This is an Euler equation and one can look for its solution in the form x r without need for the full series. Plugging this into the equation x 2 y ′′ + xy y = 0 leads to x r ( r ( r 1) + r 1) = 0. Hence x r will be a solution if r is a root of r ( r 1) + r 1 = r 2 1 = 0. There are two roots r = 1 and r = 1, so one finds two independent solutions y 1 ( x ) = x and y 2 ( x ) = x 1 . (f) The equation is x 2 y ′′ x 2 y 2 y = 0. Plugging the Frobenius series y = n =0 a n x r + n into the equation leads to summationdisplay n =0 [ γ ( r + n ) a n ( n 1 + r ) a n 1 ] x n + r = 0 , where γ ( r ) = r ( r 1) 2 = ( r 2)( r + 1). The indicial equation is γ ( r ) = 0 and it has two roots, r 1 = 2 and r 1 = 1. For root r 1 = 2 there is a solution of the form x 2 0 a n x n obtained by solving the recursion relations: a n = n + 1 n ( n + 3) a n 1 . Solving this and setting a 0 = 1 one finds the solution y 1 ( x ) = x 2 summationdisplay n =0 6( n + 1) ( n + 3)! x n . The overall factor of 6 could be replaced by 1. The roots of the indicial equation differ by an integer. We look for the second solution in the form y 2 ( x ) = C ln( x ) y 1 ( x ) + x 1 n =0 d n x n . Substituting this into the equation summationdisplay n =0 [ γ ( n 1) d n ( n 2) d n 1 + C (2 n 3) a n 3 Ca n 4 ] x n 1 . As usual, any coefficients with negative indices are equal to 0. Since γ ( n 1) = n ( n 3), it must follow that n ( n 3) d n ( n 2) d n 1 + C (2 n 3) a n 3 Ca n 4 = 0 , n 1 . (3.1) Setting n = 1, n = 2, n = 3 respectively 2 d 1 + d 0 = 0 2 d 2 = 0 d 2 + C 3 a 0 = 0 The last two equations combined imply d 2 = 0 and C = 0. Thus, the logarithm term will not appear in the second solution. For n > 3, the factor n ( n 3) > 0 and the recursion (3.1) may be written (with C = 0) simply as d n = n 2 n ( n 3) d n 1 . (3.2) 1

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642:527 SOLUTIONS: ASSIGNMENT 3 FALL 2007 For n = 4, this is d 4 = (2 / 4) d 3 and d 3 may be chosen arbitrarily. To get a simple solution take d 3 = 0; (letting d 3 negationslash = 0 will produce a term d 3 y 1 ( x ) in the solution and we already have y 1 ( x )). Then iterating (3.2) it follows that d n = 0 for all n 3. Only d 0 and d 1 may be non-zero and 2 d 1 + d 0 = 0. Thus, if we set d 0 = 1, then d 1 = (1 / 2) and y 2 ( x ) = x 1 + 1 2 . (u) The equation is xy ′′ + y y = 0 . There is a regular singular point at x 0 = 0. Substituting the Frobenius form n =0 a n x r + n leads to summationdisplay n =0 [ γ ( r + n ) a n a n 1 ] x n + r 1 , (3.3) where γ ( r ) = r 2 . The indicial equation is therefore r 2 = 0 and it has only one root at r = 0. We first find the solution y 1 ( x ), which has no logarithm by solving the recursion relations γ ( r + n ) a n a n 1 = 0 with r = 0, i.e., γ ( n ) a n = n 2 a n = a n 1 . This leads to a 1 = a 0 , a 2 = a 1 2 2 = a 0 2 2 , a 2 = a 2 3 2 = a 0 2 2 3 2 , . . . , a n = a 0 ( n !) 2 . . . .
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