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

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SUMMARY

The discussion focuses on solving the first-order linear differential equation \(y' + y = e^{-2t}\) using the integrating factor method and the method of undetermined coefficients. The associated homogeneous equation is \(y' = -y\), leading to the solution \(y = Ce^{-t}\). The non-homogeneous part \(e^{-2t}\) suggests a particular solution of the form \(y = Ae^{-2t}\). By substituting and solving, the general solution is determined to be \(y(t) = Ce^{-t} - \frac{1}{2}e^{-2t}\).

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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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