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Solving ODEs with Laplace.

  1. Apr 5, 2014 #1
    I have the following 2 problems to solve using Laplace.

    1) x'' + 3x' +2x=e^(-t); with x=dx/dt=0 when t = 0
    2) x'' - 2x' +10x=e^(2t); with x=0 and dx/dt=1 when t=0
    Where x' is dx/dt and x'' is the second derivative against time.

    My attempts:

    1)Using laplace I get

    s2X(s)-x(0)-x'(0)+3sX(s)-x(0)+2X(s)=1/(s+1)

    with x(0)=0, x'(0)=0 then

    X(s)=1 over (s+1)(s2+3s+2) which is 1/[(s+1)(s+1)(s+2)] or 1/[(s+1)2(s+2)]
    Using partial fraction
    1/[(s+1)2(s+2)]=A/(s+1)+B/(s+1)2+C/(s+2).

    I'll avoid doing the calcs; I may have made a mistake here, but I checked multiple times and didnt find and error.

    A=10/12, B=-1/6 and C=-1 so that, after doing inverse laplace

    x(t)=5/6e-t-1/6t*e-t-e-2t
    But here's the issue. With this, x(0)=/=0. It's -1/6. and dx(0)/dt=1. I can't figure out where I went wrong.

    2)The laplace transform is
    s2 X(s)-x(0)-x'(0)-2sX(s)-x(0)+10X(s)=1/(s-2)
    with x'(0)=1 this rearranges to

    X(s)(s2-2s+10)=1/(s-2)+1=(s-1)/(s-2)
    X(s)=(s-1)/[(s-2)(s2-2s+10)]

    This is where I helplessly run around in circles. The quadratic roots are complex, and whilst they discomfort me only slightly, the issue is that I cannot rearrange this into a suitable form where the inverse laplace can be done.

    Laplace+Transform+Pairs.JPG
    Taken from http://www.therationaltheorist.org/2009/11/fourier-analysis-and-odes.html
     
  2. jcsd
  3. Apr 5, 2014 #2

    Mark44

    Staff: Mentor

    There's nothing that jumps out at me on how you set things up, but your answer is incorrect.

    Using a different method, I got x(t) = (1/3)e-2t - (1/3)e-t - (1/3)te-t.

    First place I'd check is the partial fractions decomposition. Can you verify that
    $$ \frac{10/12}{s + 1} + \frac{-1/6}{(s + 1)^2} + \frac{-1}{s + 2} = \frac{1}{(s + 1)^2(s + 2)} ?$$


     
  4. Apr 5, 2014 #3
    I think I've noticed where I've made the mistake: left hand side of the partial fractions.

    s+1=A(s+1)^2(s+2)+B(s+1)(s+2)+C(s+1)^3, not 1=A(s+1)^2(s+2)+B(s+1)(s+2)+C(s+1)^3 for which I calculated.

    EDIT:

    A=-1
    B=1
    C=1
    Is what at I get now.
    x(t)=e^(-2t)+te^(-t)-e^(-t) which is equal to 0 on t=0, and the derivative is 0 as well.
    However this answer is different from the above.
     
    Last edited: Apr 5, 2014
  5. Apr 5, 2014 #4

    Mark44

    Staff: Mentor

    I agree with your new answer. I checked my previous answer, but apparently not well enough, as it was incorrect.

    I'll take a look at your other question in a bit.
     
  6. Apr 5, 2014 #5

    Mark44

    Staff: Mentor

    For the second problem, look at the eatcos(ωt) and eatsin(ωt) entries in your table. The expression s2 - 2s + 10 can be written as s2 - 2s + 1 + 9, and factored as (s - 1)2 + 32

    Your solution will involve some linear combination of three functions: etsin(3t), etcos(3t), and e2t.
     
  7. Apr 5, 2014 #6
    Many thanks.
     
  8. Apr 5, 2014 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Can I do my usual rant about how I dislike the whole "Laplace Transform" method? Of course, it is much easier to do both of these 'directly'. The "characteristic" equation for x''+ 3x'+ 2= 0 is [itex]r^+ 3r+ 2= (r+ 2)(r+ 1)= 0[/itex]. which has solutions -2 and -1 so the general solution is [itex]x= Ce^{-2t}+ De^{-t}[/itex]. Then find a particular solution to the entire equation by "undetermined coefficients"- try [itex]x= Ate^{-t}[/itex].

    The second equation has characteristic equation [itex]r^2+ 2r-10= (r+ 5)(r- 2)= 0[/itex]. And you can try [itex]x= Ate^{2t}[/itex] for a specific solution to the equation.
     
  9. Apr 5, 2014 #8

    Mark44

    Staff: Mentor

    I agree, and that's the way I did them.

    No, those aren't the factors. As it turns out, the roots of the characteristic equation are complex; namely, r = 1 ##\pm 3i##.
    xp = Ae2t will work as a particular solution, as there are no repeated roots of the third-degree characteristic equation (thinking in terms of annihilators).
     
  10. Apr 6, 2014 #9
    It was a past examination task.

    I'm not particularly impressed by the Laplace transforms. I do see the convenience but I find it is mostly overshadowed by the need to be able to articulate numbers and expressions around very well.
     
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