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Chem Differential Eq HW Solutions Fall 2011 77

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

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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 definite integral in a simpler form, just leave it as it is and write the Bessel series expansion as 1 = j =1 2 α 2 1 ,j J 2 ( α 1 ,j ) 2 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 ). For y to be bounded near 0, we must take c 2 = 0. For 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 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 ) satisfies x 2 y 1 + xy 1 + ( a 2 x 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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