MATH 263 Quiz: 2016-05-19

• School: McGill University
• Department: Mathematics and Statistics
• Course: MATH 263
• Professor: Xin Yang Lu

Document Summary

May 19, 2016: assuming y =p k=0 akxk, solve y 2xy + y = 0, with y(0) = 0 and y (0) = 1. Plugging these into the ode we have y = k(k 1)akxk 2. Xk=2 (k + 2)(k + 1)ak+2xk (k + 2)(k + 1)ak+2xk y 2xy + y = 0. Xk=1 akxk = 0 akxk = 0 akxk = 0. 2a2 + a0 + ((k + 2)(k + 1)ak+2 (2k 1)ak) xk = 0. From this we conclude two things. (i) 2a2 + a0 = 0 or a2 = a0/2 and (ii) (k + 2)(k + 1)ak+2 (2k 1)ak = 0 or ak+2 = 2k 1 (k + 2)(k + 1) for n 2. The later is the recursive relation for the coe cients. You can show that from this recursive relation, we get a2n = . 3 7 11 (4n 5) (2n)! a0 and a2n+1 = 1 5 9 (4n 3) (2n + 1)! a1.