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Chem Differential Eq HW Solutions Fall 2011 89

# Chem Differential Eq HW Solutions Fall 2011 89 - Section...

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Section 5.5 Legendre’s Differential Equation 89 Another faster way to see the answer is to simply note that P 3 is an odd function, so its integral over any symmetric interval is 0. There is yet another more important reason for this integral to equal 0. In fact, 1 - 1 P n ( x ) dx = 0 for all n = 0 . This is a consequence of orthogonality that you will study in Section 5.6. 9. This is Legendre’s equation with n ( n + 1) = 30 so n = 5. Its general solution is of the form y = c 1 P 5 ( x ) + c 2 Q 5 ( x ) = c 1 1 8 (63 x 5 - 70 x 3 + 15 x ) + c 2 ( 1 - 15 x 2 + 30 x 4 + · · · ) = c 1 (63 x 5 - 70 x 3 + 15 x ) + c 2 ( 1 - 15 x 2 + 30 x 4 + · · · ) In finding P 5 ( x ), we used the given formulas in the text. In finding the first few terms of Q 5 ( x ), we used (3) with n = 5. (If you are comparing with the answers in your textbook, just remember that c 1 and c 2 are arbitrary constants.) 13. This is Legendre’s equation with n ( n + 1) = 6 or n = 2. Its general solution is y = c 1 P 2 ( x ) + c 2 Q 2 ( x ). The solution will be bounded on [ - 1 , 1] if and only if c 2 = 0; that’s because P 2 is bounded in [ - 1 , 1] but Q 2 is not.
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