This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 ∑ ∞ 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 = 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 − 2 d 2 = 0 − d 2 + C 3 a = 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) 1 642:527 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 and d 1 may be nonzero and − 2 d 1 + d = 0. Thus, if we set d = 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. 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....
View
Full
Document
This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.
 Fall '08
 Staff
 Math

Click to edit the document details