y' = -y with y = Σ a_n x^n and a_0 = 4. The recurrence is (n+1) a_{n+1} = -a_n. What is a_2?
Answer: ______________
For y prime equals 2 y with a0 equals 5, the series recurrence is n plus 1 times a sub n plus 1 equals 2 a sub n. Type a1.
Answer: ______________
For y prime equals 5 y in series form, which recurrence is correct?
For y prime equals 2 y the series recurrence is n plus 1 times a sub n plus 1 equals 2 a sub n. If a1 is 10, what is a2?
Answer: ______________
The radius of convergence tells you how far from the centre the series solution stays valid.
Circle one: True False
y'' = y with y(0) = 1, y'(0) = 0 gives an even series with a_0 = 1. What is a_2?
Answer: ______________
y'' = 4y with y(0) = 3 and y'(0) = 0. The recurrence is (n+2)(n+1) a_{n+2} = 4 a_n. What is a_2?
Answer: ______________
Why shift indices before matching coefficients?
A student matches coefficients without shifting indices and gets a wrong recurrence. What broke?
For y double prime minus x y equals 0 about x equals 0, what do the initial conditions determine?