What is the Integrating Factor for Solving this Differential Equation?

iamtrojan3
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Homework Statement


Solving this differential equation
ty' + 2y = t^2 - t + 1

Homework Equations


Its linear so i set it up in linear form
y' + y(2/t) = t - 1 +1/t


The Attempt at a Solution



the integrating factor (u) = e ^ integral (ydt) = e^ integral (2dt/t)...
Here's the question, my book says u = t^2, and i just can't figure out how e ^ integral (2dt/t) is t^2

Its probably some stupid thing that I'm getting stuck on, thanks.
I know how to do this afterwards, i just can't for the life of god figure out why u = t^2
 
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u = e^{\int\frac{2dt}{t}} = e^{2\ln t} = (e^{\ln t})^2
 
oh wow i thought u could bring the 2 out front... thanks for clearing that up
 
Thanks Bohrok, I thought that was were is was coming from also but was having trouble typing it out.

Matt
 
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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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