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Integrating factors

  • Thread starter cue928
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  • #1
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I am working on the following problem but I am apparently not applying the integrating factor correctly:
The DE is:
dy/dt = e^(-t/20) - (1/40)y
I moved the last term to the left, giving dy/dt + (1/40)y = e^(-t/20). I had e^(t/40) as my integrating factor. Going in my (wrong) direction, I was going to
end up with y = 1+ce^(-t/40). The book gets Y(t) = -40e^(-t/20) + Ce^(-t/40).

Where I am confused is the -40e^(/t-20). I thought when multiplying thru by the IF that I would get a "1" on that side.

What am I missing?

Thank you!
 

Answers and Replies

  • #2
dextercioby
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e^{-t/20} * e^{t/40} =/= 1.
 
  • #3
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Doesn't that multiplication yield e^-40?
 
  • #4
dextercioby
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yes, but you actually have

[tex] e^{t/40} y = C - 40 e^{-t/40} [/tex]

so, the answer given in the the book to your problem is correct
 
  • #5
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Not trying to be dense but I don't see it at all.. Here's what I have:
dy/dt = e^(-t/20) - (1/40)y
dy/dt + (1/40)y = e^(-t/20)
Calculated integrating factor to be e^(t/40)
Multiplying that thru the left hand side, I recognize that as the chain rule.
I think the problem for me is the right side.
I show (integral) y*e^(t/40) = (integral) e^(-t/40)
Post integration, I got: y*e^(t/40) = -40*e^(-t/40) + C
The problem is when I clear the left side, I don't get the -40 * e^-t/20. Everything else is fine, but why is it -t/20?
 
  • #6
dextercioby
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Well, e^(-t/40)*e^(-t/40) = e^(-t/20).
 
  • #7
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I think I am going to be sick. I can't believe I missed that. Thanks for your patience.
 

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