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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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