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.