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Variation of parameters (Kinda having trouble with the integral)

  • Thread starter atarr3
  • Start date
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1. The problem statement, all variables and given/known data

Solve the problem: 4y'' - y = 8e^(.5t)/(2 + e^(.5t))


2. Relevant equations

Particular solution of Y = X*integral(inverse of X multiplied by G)

Finding eigenvalues and eigenvectors


3. The attempt at a solution

This might be a little too messy for anyone to make sense of, but

I found the eigenvalues first. 4x^2 - 1 = 0, so eigenvalues are +/- 1/2.
Solution for the homogenous is therefore C_1e^(.5t) + C_2e^(-.5t)
The matrix corresponding to the equation X' = AX is:
[0 1]
[.25 0]

The eigenvector for value 1/2 is [1 ] and for -1/2 is [ 1 ]
[.5] [-.5]

So my matrix for the theorem Y = X*integral(inverse of X multiplied by G)
is:
[e^(.5t) e^(-.5t)]
[.5e^(.5t) -.5e^(-.5t)]

So far so good, I think.

The inverse of this is [.5e^(-.5t) e^(-.5t)]
[.5e^(.5t) -e^(.5t) ]

G is [8e^(.5t)/(2 + e^(.5t))]
[ 0 ]

So plugging that into my integral equation thing, I get the integral of

[ 4/(2 + e^(.5t)) ]
[4e^t/(2 + e^(.5t))]

This is where I am stuck and don't know how to integrate.

I'd appreciate any help or advice you can guys can give! Also, please let me know if this problem can be solved a way different from the one I tried to use. Thanks a lot!
 

LCKurtz

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I haven't checked over your work, but for an integral of the form:

[tex] \int \frac 1 {a+e^{bt}}\, dt[/tex]

try letting [itex]u = a + e^{bt},\ du =be^{bt}dt,\ e^{bt} = u-a[/itex]

should give you something to work with. Same idea for the other integral.
 
76
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Thank you so much! I got the first one, and I think I can get the second one.
 

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