Power series methods means we let y = ∑[0,∞) a_{n} x^{n}, where the bracket indicates the upper and lower limits of the sum.

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

If y = ∑[0,∞) a_{n} x^{n}, 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} x^{n }= 0

Bringing x inside, we have

∑[2,∞) n(n-1)a_{n}x^{n-1} + 2 ∑[0,∞) a_{n} x^{n } = 0

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

∑[1,∞) (k+1)(k)a_{k+1}x^{k} + 2 ∑[0,∞) a_{k} x^{k} = 0

The only x_{0} term is from the second expression, so 2a_{0} = 0, or a_{0} = 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 a_{1} = c

a_{2} = -2/((2-1)(2) c

a_{3} = -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)!) x^{n}

Joy J.

08/09/20