Chem Differential Eq HW Solutions Fall 2011 77

Chem Differential Eq HW Solutions Fall 2011 77 - = 0 ....

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Section 4.8 Bessel Series Expansions 77 29. By Theorem 2 with p = 1, we have A j = 1 2 J 2 ( α 1 ,j ) 2 ± 2 0 J 1 ( α 2 ,j x/ 2) xdx = 2 α 2 1 ,j J 2 ( α 1 ,j ) 2 ± α 1 ,j 0 J 1 ( s ) sds (let α 1 ,j x/ 2= s ) . Since we cannot evaluate the defnite integral in a simpler Form, just leave it as it is and write the Bessel series expansion as 1= ² j =1 2 α 2 1 ,j J 2 ( α 1 ,j ) 2 ³± α 1 ,j 0 J 1 ( s ) sds ´ J 1 ( α 1 ,j x/ 2) For 0 <x< 2 . 33. p = 1 2 , y = c 1 J 1 2 ( λx )+ c 2 Y 1 2 ( λx ). ±or y to be bounded near 0, we must take c 2 = 0. ±or y ( π ) = 0, we must take λ = λ j = α 1 2 ,j π = j , j =1 , 2 ,... (see Exercises 21); and so y = y i = c 1 ,j J 1 ( α 1 2 ,j π x )= c 1 ,j µ 2 πx sin( jx ) (see Example 1, Section 4.7). One more formula. To complement the integral Formulas From this section, consider the Following interesting Formula. Let a , b , c , and p be positive real numbers with a ± = b . Then ± c 0 J p ( ax ) J p ( bx ) xdx = c b 2 - a 2 aJ p ( bc ) J p - 1 ( ac ) - bJ p ( ac ) J p - 1 ( bc ) · . To prove this Formula, we note that y 1 = J p ( ax ) satisfes x 2 y ±± 1 + xy ± 1 +( a 2 x 2 - p 2 ) y 1 =0 and y 2 = J p ( bx ) satisfes x 2 y ±± 2 + xy ± 2 +( b 2 x 2 - p 2 ) y 2
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Unformatted text preview: = 0 . Write these equations in the Form ( xy 1 ) + y 1 + a 2 x 2-p 2 x y 1 = 0 and ( xy 2 ) + y 2 + b 2 x 2-p 2 x y 1 = 0 . Multiply the frst by y 2 and the second by y-1, subtract, simpliFy, and get y 2 ( xy 1 ) -y 1 ( xy 2 ) = y 1 y 2 ( b 2-a 2 ) x. Note that y 2 ( xy 1 ) -y 1 ( xy 2 ) = d dx y 2 ( xy 1 )-y 1 ( xy 2 ) . So ( b 2-a 2 ) y 1 y 2 x = d dx y 2 ( xy 1 )-y 1 ( xy 2 ) , and, aFter integrating, ( b 2-a 2 ) c y 1 ( x ) y 2 ( x ) xdx = y 2 ( xy 1 )-y 1 ( xy 2 ) c = x y 2 y 1-y 1 y 2 c ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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