# Using Characteristic equations to solve 2nd order linear DEQ's

1. Sep 20, 2011

### Tricky557

The problem statement:

Find the largest interval in which the solution of the following initial value problem is valid:

cos(t/3)y'' +(6t^2)y' + ((t-5)^-3)y = 0

Initial conditions:

y(1) = 1
y'(1)= 3

I have a few questions concerning this problem.

I've converted it to it's characteristic equation:

r^2 + [(6t^2)/(cos(t/3)]r + [(t-5)^-3]/(cos(t/3))

This isn't something I'm readily able to factor. So I move to the quadratic equation... Which is what I'm confused about.

When I write down the quadratic equation for this problem, I have a feeling that I've gone down the wrong path to the solution because it turns out to be something that I don't really know what to do with. I can get 2 different and unique roots, but they're both going to be in terms of t. Every example I've seen of this being done always ends with the 2 r's as numbers. Never in terms of t.

I'm not sure what to do here.

Do I even need to solve this equation? I've been thinking about it and all it asks is for the interval where the solutions will be valid, and I can do that relatively easily. All I would need to do, if that were the case for this problem, is to solve these few equations:

cos(t/3) = 0

t-5 = 0

Wherever these equations are true are where the interval boundaries are going to be.

A long post.. almost a wall of text, any help would be greatly appreciated.

2. Sep 22, 2011

### Eynstone

Since (t-5)^-3 has a singularity at t=5 & other terms are continuous,the answer is [1,5).