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

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
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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