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Problems with integrating and Deferential equation

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data
    give the general solution of the following equation

    x' = tx + 6te-t2



    2. Relevant equations

    for x'+p(t)x=q(t)

    xeI=[tex]\int[/tex]q(t)eIdt where I=[tex]\int[/tex]p(t)dt

    integration by parts
    [tex]\int[/tex]f'g = [fg] - [tex]\int[/tex]fg'

    3. The attempt at a solution

    x'-tx=6te-2t

    I=[tex]\int[/tex]-t dt = -t2[tex]/[/tex]2

    xe-t2[tex]/[/tex]2dt = [tex]\int[/tex]6te-2te-t2[tex]/[/tex]2dt

    using integration by parts i get

    [tex]\int[/tex]f'g = [fg] - [tex]\int[/tex]fg'
    f'=e-t2[tex]/[/tex]2
    g=6t

    [tex]\int[/tex]6te-2te-t2[tex]/[/tex]2dt = [[tex]\frac{6t}{-2-t}[/tex]e-2t-t2[tex]/[/tex]2] - [tex]\int[/tex][tex]\frac{6}{-2-t}[/tex]e-2t-t2[tex]/[/tex]2 dt

    I've tried to integrate the second part of this integral i.e.
    [tex]\int[/tex][tex]\frac{6}{-2-t}[/tex]e-2t-t2[tex]/[/tex]2 dt
    using integration by parts but it seems to be a very difficult integral to solve. I also have my suspicions that this method may go on forever.

    can anyone help? am i missing some kind of identity that i should know? any help would be appreciated.
     
  2. jcsd
  3. Jun 3, 2009 #2
    i just noticed a typo where i stated what value i used for f' it should read

    f' = e-2-t2[tex]/[/tex]2
     
  4. Jun 3, 2009 #3
    You seem to have stated the question twice, but differently both times. first you have

    x' = tx + 6texp(-t^2)
    and then you write
    x'-tx=6texp(-2t)

    Which is the correct form?
     
  5. Jun 3, 2009 #4
    Hi Physics Math

    The second one was just me re-arranging the first one to get it into the form of
    x'+p(t)x=q(t)
    so that i could apply the relevant formula (under relevant equations).
     
  6. Jun 3, 2009 #5

    Cyosis

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    You didn't just rearrange you changed the argument of the exponent as well. It would also be nice if you would put [tex] tags around the entire expression and not just the occasional symbol.
     
  7. Jun 3, 2009 #6
    oh yes, i see what you mean

    exp(-t^2) does not equal exp(-2t).

    I'll give it another go.

    thanks guys.
     
  8. Jun 3, 2009 #7

    Cyosis

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    Using the correct argument will make the integration easier.
     
  9. Jun 3, 2009 #8
    Okay I've had another go and got an answer and want to check whether its correct or not.

    x'-tx=6te-t2

    I = [tex]\int[/tex]-t dt = -t2[tex]/[/tex]2

    xe-t2[tex]/[/tex]2=[tex]\int[/tex]6te-t2e-t2[tex]/[/tex]2 dt

    xe-t2[tex]/[/tex]2 = [tex]\int[/tex]6te-(3/2)t2dt

    now using integration by parts where
    f'=e-(3/2)t2 and
    g=6t
    and recalling that [tex]\int[/tex]e-ax2 dx = [tex]\sqrt{pi/a}[/tex]

    i get

    [[tex]\sqrt{2pi/3}[/tex] 6t] - [tex]\int[/tex][tex]\sqrt{2pi/3}[/tex]6 dt

    = 2[[tex]\sqrt{2pi/3}[/tex] 6t] + C

    so

    x=(2[[tex]\sqrt{2pi/3}[/tex] 6t] + C) / (e-t2[tex]/[/tex]2)

    However, in all the other examples i have done the exp term has disappeared by the time i got to the final answer, so i just wanted to check that i have the correct answer here or not.
     
  10. Jun 3, 2009 #9

    Cyosis

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    This is only true if you're integrating from -infinity to +infinity, which you are not. If you want to use partial integration you will have to resort to the error function. That said you do not want to use partial integration, but instead use a substitution u=-t^2.
     
  11. Jun 3, 2009 #10
    Thank you cyosis for all your help. My integration skills are quite weak so please bare with me here.


    f' = exp(-3t2/2)
    using
    u= t2
    f' = exp((-3/2)u)

    f = [tex]\int[/tex]exp((-3/2)u) (dt/du) du
    f = (-2/3)exp((-3/2)u) (1/2t)
    f = (-1/3t)exp((-3/2)t2)
    g = 6t
    g'=6

    now using integration by parts

    [tex]\int[/tex]6t exp((3/2)tt) dt = [(-6t/3t)exp((-3/2)t2)] + [tex]\int[/tex](6/3t)exp((-3/2)t2) dt

    integrating the second term by parts

    f' = exp(-3t2/2)
    f = (-1/3t)exp((-3/2)t2)
    g = 6/3t
    g' = -6/3t2

    [tex]\int[/tex](6/3t)exp((-3/2)t2) dt = [(6/9t2)exp((-3/2)t2)] - [tex]\int[/tex](-6/9t3)exp((-3/2)t2) dt

    i seem to be going in circles...

    it seems there is always going to be an integral that needs to be solved...

    help
     
  12. Jun 3, 2009 #11

    Cyosis

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    Why are you using partial differentiation again? The original integral is already cast in a very easy form.

    Example integral: Using the substitution u=t^2 du=2tdt.
    [tex]
    \begin{align*}
    \int t e^{t^2}dt & =\int \frac{1}{2}e^u du\\
    & =\frac{1}{2}e^u\\
    & =\frac{1}{2}e^{t^2}
    \end{align*}
    [/tex]
     
  13. Jun 3, 2009 #12
    i have no idea why i didnt see that. its been a really long day

    so now i have

    [tex]\int[/tex] 6t exp((-3/2)t2) dt

    u = t2
    du = 2t dt

    [tex]\int[/tex]3 exp((-3/2)u) du = 3(-2/3) exp((-3/2)u) = -2 exp((-3/2)t2) + C

    x exp(-t2/2) = -2 exp((-3/2)t2) + C

    x = (1/exp(-t2/2))C - 2 exp(-t2/2)

    does this look correct?
     
  14. Jun 3, 2009 #13
    there is a typo in my answer. it should read

    x = (1/exp(-(t^2)/2))C - 2 exp(-t^2)

    Thank you for your help Cyosis
     
  15. Jun 3, 2009 #14

    Cyosis

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    You're welcome, and your answer is correct.
     
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