Chem Differential Eq HW Solutions Fall 2011 94

Chem Differential Eq HW Solutions Fall 2011 94 - second...

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94 Chapter 6 Sturm-Liouville Theory with Engineering Applications Solutions to Exercises 6.1 1. Let f j ( x ) = cos( jπx ), j =0 , 1 , 2 , 3, and g j ( x ) = sin( jπx ), j =1 , 2 , 3. We have to show that ± 2 0 f j ( x ) g k ( x ) dx = 0 for all possible choices of j and k .I f j =0 , then ² 2 0 f j ( x ) g k ( x ) dx = ² 2 0 sin kπxdx = - 1 cos( kπx ) ³ ³ ³ 2 0 =0 . If j ± = 0, and j = k , then using the identity sin α cos α = 1 2 sin 2 α , ² 2 0 f j ( x ) g j ( x ) dx = ² 2 0 cos( jπx ) sin( jπx ) dx = 1 2 ² 2 0 sin(2 jπx ) dx = - 1 4 cos(2 jπx ) ³ ³ ³ 2 0 =0 . If j ± = 0, and j ± = k , then using the identity sin α cos β = 1 2 ( sin( α + β ) + sin( α - β ) ) , we obtain ² 2 0 f j ( x ) g k ( x ) dx = ² 2 0 sin( kπx ) cos( jπx ) dx = 1 2 ² 2 0 ( sin( k + j ) πx + sin( k - j ) πx ) dx = - 1 2 π ´ 1 k + j cos( k + j ) πx + 1 k - j cos( k - j ) πx µ³ ³ ³ 2 0 =0 . 5. Let f ( x )=1 , g ( x )=2 x , and h ( x )= - 1+4 x . We have to show that ² 1 - 1 f ( x ) g ( x ) w ( x ) dx =0 , ² 1 - 1 f ( x ) h ( x ) w ( x ) dx =0 , ² 1 - 1 g ( x ) h ( x ) w ( x ) dx =0 . Let’s compute: ² 1 - 1 f ( x ) g ( x ) w ( x ) dx = ² 1 - 1 2 x 1 - x 2 dx
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Unformatted text preview: second integral, we have ² 1-1 f ( x ) h ( x ) w ( x ) dx = ² 1-1 (-1 + 4 x 2 ) ¶ 1-x 2 dx = ² π (-1 + 4 cos 2 θ ) sin 2 θ dθ ( x = cos θ, dx =-sin θ dθ, sin θ ≥ 0 for 0 ≤ θ ≤ π. ) =-² π sin 2 θ dθ + 4 ² π (cos θ sin θ ) 2 dθ =-= π 2 · ¸¹ º ² π 1-cos 2 θ 2 dθ +4 ² π ( 1 2 sin(2 θ ) ) 2 dθ =-π 2 + = π 2 · ¸¹ º ² π 1-cos(4 θ ) 2 dθ = 0...
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