Troubleshooting Reduction of Order Equations: Where Did I Go Wrong?

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Homework Statement


t^2*y^'' -t(t+2)y^' + (t+2)y=0 , t>0; ysub1(t) = t

Homework Equations


Reduction of order:

A second solution is assumed to be of the form:

ysub2(t) = v(t)*ysub1(t)

The Attempt at a Solution



work.jpg


So, the algebra in the first part of the process seemed to be correct, as it canceled out all the "v" terms and left only derivatives of v. The problem seems to happen somewhere around when I turn it into a first order linear equation and attempt to multiply through by an integrating factor. As far as I, and a CAS are concerned, that integral at the end is not possible to take. This leaves the assumption that I completely goofed up my integrating factor. Where could I have gone wrong?
 
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RyanTAsher said:

Homework Statement


t^2*y^'' -t(t+2)y^' + (t+2)y=0 , t>0; ysub1(t) = t

Homework Equations


Reduction of order:

A second solution is assumed to be of the form:

ysub2(t) = v(t)*ysub1(t)

The Attempt at a Solution



work.jpg


So, the algebra in the first part of the process seemed to be correct, as it canceled out all the "v" terms and left only derivatives of v.
I don't think so. After substituting y2(t) = tv into the differential equation, and simplifying, I ended up with this equation:
##t^3v'' - t^3v' = 0##, which is pretty easy to solve.

I'm pretty confident that this is correct, as I went ahead to find ##y_2## and found that it satisfied the diff. eqn.

RyanTAsher said:
The problem seems to happen somewhere around when I turn it into a first order linear equation and attempt to multiply through by an integrating factor. As far as I, and a CAS are concerned, that integral at the end is not possible to take. This leaves the assumption that I completely goofed up my integrating factor. Where could I have gone wrong?
 
Okay, I re-did all my algebra and I made the mistake somewhere distributing in the middle term. Thank you for your help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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