Solve this DE (by series)

It´s simple:

y' - (1/t)y=t^2

The problem is, the teacher asked us to solve this USING POWER SERIES METHOD.

It´s non-homegenous, so i don´t know what to do.

Someone help us?

I'll first clear the denominators.
t y' - y = t^3.

Since t = 0 is a regular singular point, let y = Σ(n = 0 to ∞) a(n) t^(n+r)

Plugging this into the DE yields
t * Σ(n = 0 to ∞) a(n) (n+r) t^(n+r-1) - Σ(n = 0 to ∞) a(n) t^(n+r) = t^3
==> Σ(n = 0 to ∞) a(n) (n+r) t^(n+r) - Σ(n = 0 to ∞) a(n) t^(n+r) = t^3
==> Σ(n = 0 to ∞) (n + r - 1) a(n) t^(n+r) = t^3.

The first term yields the indicial equation (to find r):
n = 0 ==> (r - 1) a(0) t^r = 0 (momentarily ignoring the right side)
........==> r = 1.

Since r = 1, we obtain
Σ(n = 0 to ∞) n a(n) t^(n+1) = t^3

Now, we equate like coefficients (with a(0) arbitrary).
n = 0 ==> 0 = 0.
n = 1 ==> 1 a(1) t^2 = 0 ==> a(1) = 0
n = 2 ==> 2 a(2) t^3 = t^3 (this is where the non-homog stuff comes in) ==> a(2) = 1/2.
For n > 2, n a(n) t^(n+1) = 0 ==> a(n) = 0.

So, the general solution is y = Σ(n = 0 to ∞) a(n) t^(n+1)
==> y = a(0) t + 0t^2 + (1/2)t^3 + 0
==> y = Ct + (1/2) t^3 for some constant C (renaming a(0) = C).

I hope this helps!