Intergrating Factor - dy/dt = -2ty + 4e^-t^2

  • #1

Homework Statement


Solve the differential equations [Find the general solution y(t)=KYh(t) + Yp (t)]. Use the method of finding the integrating factor.

dy/dt = -2ty + 4e^-t^2


The Attempt at a Solution



S = intergrat

dy/dt = -2ty + 4e^-t^2
dy/dt + 2ty = 4e^-t^2

P(t) = 2t
b(t) = 4e^-t^2

U(t) = e^(S 2t dt)
U(t) = e ^t^2

Y(t) = 1 / e^t^2 * S e^t^2 * 4e^-t^2

I guess I just don't know how to intergrat e^t^2 * 4e^-t^2
 

Answers and Replies

  • #2
dextercioby
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[tex] e^{t^2} e^{-t^2} = 1 [/tex]

, as one function is the inverse of the other.
 
  • #3
Would the 4 go down to? Like this:

[tex]
e^{t^2} / 4e^{t^2}
[/tex]

Wouldn't that leave you with 1/4?
 
  • #4
dextercioby
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No, with a 4. The 4 is in the numerator, if you write [itex] 4e^{-t^2} [/itex] as a fraction.
 
  • #5
[tex] int_/ 4*e^{t^2} / e^{t^2} dt[/tex]

Then, the [tex] e^{t^2} [/tex] cancel out and you are left with [tex] int_/ 4 dt[/tex] which is just 4t. Is this right?
 
  • #6
dextercioby
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You should be more certain of yourself. You dropped the integration constant.
 
  • #7
oh, yeah... I meant 4t + c......So the whole answer is [tex]y(t) = {4t + C}/e^{t^2}[/tex] (The [tex]e^{t^2}[/tex] is under both).....Yeah, I second guess myself way too much. Most the time, after I ask for help, I realize how easy a problem was and can't believe I ask for help.
 

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