Why Use an Integrating Factor in Differential Equations?

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The discussion focuses on the use of an integrating factor in solving the first-order linear differential equation \(\frac{dp}{dt}+2tp=p+4t-2\). The integrating factor is derived as \(e^{\int (2t-1)dt}=e^{t^2 -t}\), which simplifies the equation. Participants clarify that multiplying the differential equation by the integrating factor allows for the left side to be expressed as the derivative of the product of the integrating factor and the function \(p\). This method is essential for obtaining the solution to the differential equation.

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

[tex] \frac{d}{dt}[e^{t^2 -t}]=...[/tex]
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.
 

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