Guessing a power series for the answer

Mathematics · Differential Equations · ages 20-21

Name ______________________ Date ____________

y' = 3y with y = Σ a_n x^n and a_0 = 2. The recurrence is (n+1) a_{n+1} = 3 a_n. What is a_1?
Answer: ______________
With y as a series and a0 equals 2, the recurrence n plus 1 times a sub n plus 1 equals 3 a sub n holds. Type a1.
Answer: ______________
For y prime equals 3 y in series form, which recurrence is correct?
- n times a sub n equals 3 a sub n plus 1
- a sub n plus 1 equals 3 times n plus 1 times a sub n
- n plus 1 times a sub n plus 1 equals 3 a sub n
Same equation with a1 now known to be 6. Type 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: ______________
Same equation: y' = 3y, a_0 = 2. What is a_2?
Answer: ______________
Why shift indices before matching coefficients?
- So every sum speaks in the same power of x
- To change the radius of convergence
- To remove the initial conditions
A student matches coefficients without shifting indices and gets a wrong recurrence. What broke?
- Nothing, shifting is optional decoration
- Unlike powers were equated, mixing unrelated coefficients
- The initial conditions were applied twice
For y double prime minus x y equals 0 about x equals 0, what do the initial conditions determine?
- The radius of convergence directly
- The starting coefficients that seed the whole recurrence
- Whether the point is ordinary

Answer key

For grown-ups. Fold this page away before handing over the rest.

6 · At n = 0: 1·a_1 = 3·2 = 6.
6 · At n equals 0 the recurrence reads 1 times a1 equals 3 times 2.
n plus 1 times a sub n plus 1 equals 3 a sub n · Differentiating shifts the index, leaving n plus 1 on the new coefficient.
9 · At n equals 1 the recurrence reads 2 times a2 equals 3 times 6.
True · Beyond that distance the sum may diverge and stop representing the solution.
0.5 · Matching x^0: 2·a_2 = a_0 = 1, so a_2 = 0.5.
9 · At n = 1: 2·a_2 = 3·a_1 = 18, so a_2 = 9.
So every sum speaks in the same power of x · Coefficients can only be compared once like powers sit side by side.
Unlike powers were equated, mixing unrelated coefficients · Unshifted sums pair x to the n with x to the n plus 1, scrambling the relation.
The starting coefficients that seed the whole recurrence · Seed values like a0 and a1 generate every later coefficient through the recurrence.