# sol2 - MATHEMATICS 3161 Fall 2009(2009.9 2009.12 Assignment...

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Unformatted text preview: MATHEMATICS 3161: Fall 2009 (2009.9 - 2009.12) Assignment # 2 (Due Date: Oct. 5) 1. Find all singular points of the given equation and determine whether each one is regular or irregular. a) (1 − x 2 ) 2 y ′′ + x (1 − x ) y ′ + (1 + x ) y = 0 b) ( x 2 − x − 2) y ′′ + ( x + 1) y ′ + 2 y = 0, c) x 2 y ′′ + 2( e x − 1) y ′ + ( e − x cos x ) y = 0 Solution : a) p ( x ) = x (1 − x )(1 + x ) 2 , q ( x ) = 1 (1 + x )(1 − x ) 2 both are rational functions. x = ± 1 are singular points. lim x → 1 ( x − 1) p ( x ) = − 1 4 , lim x → 1 ( x − 1) 2 q ( x ) = 1 2 , lim x →− 1 ( x + 1) p ( x ) = ∞ , therefore, x = 1 is a regular singular point, x = − 1 is an irregular singular point. b) p ( x ) = 1 x − 2 , q ( x ) = 2 ( x + 1)( x − 2) both are rational functions. x = − 1 , 2 are singular points. lim x →− 1 ( x + 1) p ( x ) = 0 , lim x →− 1 ( x + 1) 2 q ( x ) = 0 , lim x → 2 ( x − 2) p ( x ) = 1 , lim x → 2 ( x − 2) 2 q ( x ) = 0 , therefore, x = − 1 , 2 are regular singular points. c) p ( x ) = 2( e x − 1) x 2 , q ( x ) = e − x cos x x 2 , they are not rational functions. x = 0 is singular point. xp ( x ) = 2( e x − 1) x = 2 ∞ summationdisplay n =1 x n − 1 n ! = 2 ∞ summationdisplay n =0 x n ( n + 1)! is convergent for all x , ( R = ∞ ), x 2 q ( x ) = e − x cos x is convergent for all x since both e − x and cos x are convergent for all x , so x = 0 is a regular singular point. 2. Determine the general solution of the given differential equation that is valid in any interval not including the singular point. a) x 2 y ′′ − 5 xy ′ + 9 y = 0 b) ( x − 2) 2 y ′′ + 5( x − 2) y ′ + 8 y = 0 c) x 2 y ′′ − 4 xy ′ + 4 y = 0 Solution : There are all Euler equations. a) the characteristic equation is r ( r − 1) − 5 r + 9 = r 2 − 6 r + 9 = ( r − 3) 2 = 0 then r = 3 (equal). So the general solution is y = c 1 x 3 + c 2 x 3 ln | x | , ( x negationslash = 0). b) the characteristic equation is r ( r − 1) + 5 r + 8 = r 2 + 4 r + 8 = 0 r = − 4 ± √ 16 − 32 2 = − 2 ± 2 i (complex) , x = 2 Then the general solution is y = | x − 2 | − 2 ( c 1 cos 2 ln | x − 2 | + c 2 sin ln2 | x − 2 | ) , ( x negationslash = 2)....
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## This note was uploaded on 03/18/2010 for the course MATH 316 taught by Professor Schoutz during the Spring '10 term at UBC.

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sol2 - MATHEMATICS 3161 Fall 2009(2009.9 2009.12 Assignment...

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