Find the power series solution of the differential equation

Find the power series solution of the differential equation: y^2 - xy=0, about x_0 = 0, and express it in terms of two independent parameters.

Solution

Assume that y = (n = 0 to ) a(n) x^n.

Substituting this into the DE yields
(n = 2 to ) n(n-1)a(n) x^(n-2) - x * (n = 0 to ) a(n) x^n = 0
==> (n = 2 to ) n(n-1)a(n) x^(n-2) - (n = 0 to ) a(n) x^(n+1) = 0

Re-index the sums:
(n = 0 to ) (n+2)(n+1)a(n+2) x^n - (n = 1 to ) a(n-1) x^n = 0
==> [2 a(2) + (n = 1 to ) (n+2)(n+1)a(n+2) x^n] - (n = 1 to ) a(n-1) x^n = 0
==> 2 a(2) + (n = 1 to ) [(n+2)(n+1)a(n+2) - a(n-1)] x^n = 0

Equating like entries (with a(0), a(1) arbitrary):
a(2) = 0

For n = 1, 2, ...
(n+2)(n+1)a(n+2) - a(n-1) = 0 ==> a(n+2) = a(n-1)/[(n+2)(n+1)].

n = 1 ==> a(3) = a(0)/6 = 0

Hence, y = a(0) + a(1)x + a(2) x^2 + a(3) x^3 + ...
..............= a(0) + a(1)x + 0x^2 + (1/6) a(0) x^3 + ...

We can deduce a(0) from using x(0) = -1:
-1 = a(0) + 0
==> a(0) = -1

Relabeling a(1) as C, we have y = -1 + Cx - (1/6) x^3 + ...