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

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SUMMARY

The discussion focuses on solving the initial value problem represented by the second-order linear differential equation: cos(t/3)y'' + (6t^2)y' + ((t-5)^-3)y = 0. The characteristic equation derived is r^2 + [(6t^2)/(cos(t/3))]r + [(t-5)^-3]/(cos(t/3)). The user expresses confusion regarding the quadratic equation's roots, which are dependent on the variable t rather than being constant. Ultimately, the solution requires identifying the interval of validity, determined by the singularity at t=5 and continuity of other terms, resulting in the interval [1,5).

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Tricky557
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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.
 
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Since (t-5)^-3 has a singularity at t=5 & other terms are continuous,the answer is [1,5).
 

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