Why Use an Integrating Factor in Differential Equations?

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<br /> \frac{dp}{dt}+2tp=p+4t-2<br />
<br /> e^{\int (2t-1)dt}=e^{t^2 -t}<br />
then we do

<br /> \frac{d}{dt}[e^{t^2 -t}]=...<br />
why??
i was told that i will get the original left side of the equation times p.
i didnt get the same resultwhat is the role of integration factor
??
 
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An integrating factor is a function with which you multiply the differential equation to help facilitate the solving of the differential equation.

There was a similar question here a few days ago, where I did part of the calculation.

https://www.physicsforums.com/showthread.php?t=316907

Check post 4.
 
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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