Find the series solutions of the differential equation

Assume x>0
Find the series solutions of 2xy''+y'+xy=0 around the point 0.

Since x = 0 is a regular singular point, assume that y = Σ(n = 0 to ∞) a(n) x^(n+r).

Substituting this into 2xy'' + y' + xy = 0 yields
2x * Σ(n = 0 to ∞) (n+r)(n+r-1) a(n) x^(n+r-2) + Σ(n = 0 to ∞) (n+r) a(n) x^(n+r-1) + x * Σ(n = 0 to ∞) a(n) x^(n+r) = 0

==> Σ(n = 0 to ∞) [2(n+r)(n+r-1) + (n+r)] a(n) x^(n+r-1) + Σ(n = 0 to ∞) a(n) x^(n+r+1) = 0
==> Σ(n = 0 to ∞) (n+r)[2(n+r-1) + 1] a(n) x^(n+r-1) + Σ(n = 2 to ∞) a(n-2) x^(n+r-1) = 0
==> r (2(r-1) + 1) a(0) x^(r-1) + (1+r)(2r+1) a(1) x^r +
Σ(n = 2 to ∞) [(n+r) (2(n+r-1) + 1) a(n) + a(n-2)] x^(n+r-1) = 0

The indicial equation (from the first term) is
r(2r - 1) = 0 ==> r = 0 or r = 1/2.

(i) If r = 0, then we obtain a(1) + Σ(n = 2 to ∞) [n(2n-1) a(n) + a(n-2)] x^(n-1) = 0

Assuming that a(0) is arbitrary, equating like coefficients yields a(1) = 0, and
n(2n-1) a(n) + a(n-2) = 0 for n ≥ 2
==> a(n) = -a(n-2) / [n(2n-1)] for n ≥ 2.

n = 2 ==> a(2) = -a(0) / (2 * 3)
n = 3 ==> a(3) = -a(1) / (3 * 5) = 0 [likewise, a(5) = a(7) = ... = 0]
n = 4 ==> a(4) = -a(2) / (4 * 7) = a(0) / (2 * 3 * 4 * 7).
...

So, one solution (setting a(0) = A) is
y = A(1 - x^2/(2 * 3) + x^4/(2 * 3 * 4 * 7) - ...).
-------------------
(ii) Setting r = 1/2, we obtain
3 a(1) x^(1/2) + Σ(n = 2 to ∞) [(n + 1/2) (2(n - 1/2) + 1) a(n) + a(n-2)] x^(n - 1/2) = 0
==> 3 a(1) x^(1/2) + Σ(n = 2 to ∞) [(2n + 1) n a(n) + a(n-2)] x^(n - 1/2) = 0

Assuming that a(0) is arbitrary, equating like coefficients yields a(1) = 0, and
n(2n+1) a(n) + a(n-2) = 0 for n ≥ 2.
==> a(n) = -a(n-2) / [n(2n+1)] for n ≥ 2.

As before a(1) = 0 ==> a(3) = a(5) = ... = 0.

a(0) = 0
==> a(2) = -a(0) / (2 * 5)
==> a(4) = -a(2) / (4 * 9) = a(0) / (2 * 4 * 5 * 9)
...

Hence, another solution (setting a(0) = B) is
y = B(1 - x^2/(2 * 5) + x^4/(2 * 4 * 5 * 9) - ...).

I hope this helps!