Section 5.5Legendre’s Differential Equation89Another faster way to see the answer is to simply note thatP3is an odd function, soits integral over any symmetric interval is 0. There is yet another more importantreason for this integral to equal 0. In fact,1-1Pn(x)dx= 0for alln= 0.This is a consequence of orthogonality that you will study in Section 5.6.9.This is Legendre’s equation withn(n+ 1) = 30 son= 5. Its general solutionis of the formy=c1P5(x) +c2Q5(x)=c118(63x5-70x3+ 15x) +c2(1-15x2+ 30x4+· · ·)=c1(63x5-70x3+ 15x) +c2(1-15x2+ 30x4+· · ·)In findingP5(x), we used the given formulas in the text. In finding the first fewterms ofQ5(x), we used (3) withn= 5. (If you are comparing with the answers inyour textbook, just remember thatc1andc2are arbitrary constants.)13.This is Legendre’s equation withn(n+ 1) = 6 orn= 2. Its general solutionisy=c1P2(x) +c2Q2(x).The solution will be bounded on [-1,1] if and onlyifc2= 0; that’s becauseP2is bounded in [-1,1] butQ2is not.
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