sol2 - MATHEMATICS 3161: Fall 2009 (2009.9 - 2009.12)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

Page1 / 6

sol2 - MATHEMATICS 3161: Fall 2009 (2009.9 - 2009.12)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online