Please help with this question!

Question

Use the method of Power Series to solve the following nonlinear ODE:

Please only use Power series method.

Answer 1

Power series methods means we let y = ∑[0,∞) a_n xⁿ, where the bracket indicates the upper and lower limits of the sum.

Then, our problem is xy'' + 2y = 0.

If y = ∑[0,∞) a_n xⁿ, then y' = ∑[1,∞) na_n x^n-1 , and y'' = ∑[2,∞) n(n-1)a_n x^n-2

Then, as xy'' + 2y = 0, we have

x∑[2,∞) n(n-1)a_n x^n-2 + 2 ∑[0,∞) a_n xⁿ = 0

Bringing x inside, we have

∑[2,∞) n(n-1)a_nx^n-1 + 2 ∑[0,∞) a_n xⁿ = 0

Let k = n-1 in the first expression, and k = n in the second expression. Then,

∑[1,∞) (k+1)(k)a_k+1x^k + 2 ∑[0,∞) a_k x^k = 0

The only x₀ term is from the second expression, so 2a₀ = 0, or a₀ = 0

Thus, ∑[1,∞)( (k+1)(k)a_k+1 + 2a_k)x^k = 0

Then, (k+1)(k)a_k+1 + 2a_k= 0, k>= 1

a_k+1 = -2/(k(k+1)) a_k, k >= 1

As we want a solution for a_n, though, it might make more sense to write

a_n = -2/((n-1)(n)) a_n-1, n >= 1

Then, let a₁ = c

a₂ = -2/((2-1)(2) c

a₃ = -2/((3-1)(3) a2 = -2/((3-1)3) (-2/((2-1)(2) c) = (-2)^2/((3-1)(2-1)(3)(2)) c = (-2)^^n-1 /(n-1!n!) c n = 3

We thus have the solution

y = ∑[1,∞) c(-2)^n-1/(n!(n-1)!) xⁿ

Answer 2

I recorded the video, but I forgot to post it. Maybe this can serve as a complementary answer to the one above.