Let
Differentiating twice gives
When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.
Substitute these into the given differential equation:
Then the coefficients in the power series solution are governed by the recurrence relation,
Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.
• If n is even, then n = 2k for some integer k ≥ 0. Then
It should be easy enough to see that
• If n is odd, then n = 2k + 1 for some k ≥ 0. Then
so that
So, the overall series solution is