There actually isn't a whole lot to do throughout this case.  We'll find two solutions which will form a basic set of solutions and therefore our general solution will be as,

Example: Solve the following initial value problem

2 x² y′′ + 3xy′ -15 y = 0,

 y (1) = 0

 y′ (1) = 1

 Solution

We first require finding the roots to (3).

2r ( r -1) + 3r -15 = 0

2r² + r -15 = (2r - 5)(r + 3) = 0

⇒         r₁ = 5/2 and                          r₂ = -3

Then the general solution is,

y(x) = c₁x^5/2 + c₂ x^-3

To get the constants we differentiate and plug into the initial conditions where we did back into the second order differential equations section.

y'(x) = (5/2) c₁x^3/2 - 3c₂ x^-4

0 = y(1) = c₁ + c₂

1 = y'(1) = (5/2)c₁ + (-3) c₂

By solving these equations we get:

c₁ = 2/11,

c₂ = -(2/11)

the actual solution is,

y(x) = (2/11) x^5/2  -(2/11) c₂ x^-3