Integrating Factor Method for Solving y' + y = e^{-2t}

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Discussion Overview

The discussion revolves around solving the first-order linear differential equation \(y' + y = e^{-2t}\). Participants explore different methods for finding the solution, including the integrating factor method and separation of variables.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants present the equation in the form \(y' = -y + e^{-2t}\) and inquire about finding the integrating factor.
  • One participant suggests that the equation can also be solved by separating variables, leading to the homogeneous equation \(y' = -y\) and integrating to find \(y = C'e^{-t}\).
  • Another participant proposes a solution of the form \(y = Ae^{-2t}\) based on the method of undetermined coefficients, deriving a particular solution and ultimately the general solution \(y(t) = Ce^{-t} - \frac{1}{2}e^{-2t}\).
  • There is a reiteration of the initial question about the integrating factor, indicating a potential lack of clarity or consensus on the preferred method.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to be used for solving the differential equation, with multiple approaches discussed and no agreement on a singular solution method.

Contextual Notes

Some assumptions about the methods used, such as the applicability of the method of undetermined coefficients and the conditions under which separation of variables is valid, are not fully explored. The discussion also reflects differing perspectives on the best approach to take.

hiyum
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\[ y'=-y+e^{(-2)t} \]
 
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hiyum said:
\[ y'=-y+e^{(-2)t} \]
As in the other thread:
[math]y' + y = e^{-2t}[/math]

How do you find the integrating factor here?

-Dan
 
Since this, while a first order differential equation, is also linear we can also separate it. The "associated homogeneous equation" is y'= -y which we can write as \(\frac{dy}{dx}= -y\) and separate as \(\frac{dy}{y}= -dt\). Integrating, \(ln{y}= -t+ C\). Taking the exponential of both sides, \(y=C'e^{-t}\) where \(C'= e^C\).

Since the "non-homogeneous part", \(e^{-2t}\), is of the type of function we expect as a solution to a "linear differential equation with constant coeffcients" we try a solution of the form \(y= Ae^{-2t}\) (this is the "method of undetrmined coefficients". A is the coefficient to be determined.

If \(y= Ae^{-2t}\) then \(y'= -2Ae^{-2t}\) and putting those into the differential equation, \(-2Ae^{-2t}= -Ae^{-2t}+ e^{-2t}\). \(-Ae^{-2t}= e^{-2t}\). Dividing by \(e^{-2t}\) we have -A= 1 so A= -1/2.

The general solution to this differential equation is \(y(t)= Ce^{-t}- \frac{1}{2}e^{-2t}\).
 
topsquark said:
As in the other thread:
[math]y' + y = e^{-2t}[/math]

How do you find the integrating factor here?

-Dan
Though you could try the integrating factor approach as I showed you in the other Forum.

-Dan
 

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