Chem Differential Eq HW Solutions Fall 2011 78

# Chem Differential Eq HW Solutions Fall 2011 78 - 78 Chapter...

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78 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates On the left, we have the desired integral times ( b 2 - a 2 ) and, on the right, we have c J p ( bc ) aJ p ( ac ) - bJ p ( ac ) J p ( bc ) - c aJ p (0) J p (0) - bJ p (0) J p (0) . Since J p (0) = 0 if p > 0 and J 0 ( x ) = - J 1 ( x ), it follows that J p (0) J p (0) - J p (0) J p (0) = 0 for all p> 0. Hence the integral is equal to I = c 0 J p ( ax ) J p ( bx ) xdx = c b 2 - a 2 aJ p ( bc ) J p ( ac ) - bJ p ( ac ) J p ( bc ) . Now using the formula J p ( x ) = 1 2 J p - 1 ( x ) - J p +1 ( x ) , we obtain I = c 2( b 2 - a 2 ) aJ p ( bc ) ( J p - 1 ( ac ) - J p +1 ( ac ) ) - bJ p ( ac ) ( J p - 1 ( bc ) - J p +1 ( bc ) ) . Simplify with the help of the formula J p +1 ( x ) = 2 p x J p ( x ) - J p
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