1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Use Laplace to solve this differential equation

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Obtain the solution of the differential equation

    x'' + w2nx = t

    My use of L refers to the Laplace



    2. Relevant equations



    3. The attempt at a solution

    L{x'' + w2nx = t}

    I decided to do the Laplace of each part individually starting with x''

    L{x''} = sL{x'} - x'(0)

    then

    L{x'} = sL{x} - x(0)

    Putting this together gives

    L{x''} = s[sL{x} - x(0)] - x'(0)

    Also, the L{x} = x/s

    L{x''} = sx - sx(0) - x'(0)

    Then the Laplace of w2nx is:

    L{w2nx} = w2nx∫inf0e-st dt

    because I believe that w2nx is a constant, hence I can move it out of the integral

    evaluating the integral then gives me

    L{w2nx} = w2nx/s

    Then for L{t} I can see from the Laplce tables that this is just 1/s2

    Consequently, my Laplace transformation is

    L{x'' + w2nx = t} = sx - sx(0) - x'(0) + wn2x/s + 1/s2

    After this step I'm pretty sure I need to do the inverse Laplace transform. But I wanted to check if what I had at the moment looked right, and if so, what is the best way to do the inverse of this laplace transformation?
     
  2. jcsd
  3. Sep 11, 2013 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You're confusing the function x with the variable t. L[x] = L[x(t)] = X(s) while L[t] = 1/s. X(s), the Laplace transform of x(t), is what you're solving for.
     
  4. Sep 11, 2013 #3
    I was thinking it looked a bit odd.

    So Should it instead be:

    [sL{x''} - x'(0)] + wn2[X(s) - x(0)] = 1/s2

    This simplifies to:

    s2X(s) - sx(0) - x'(0) + wn2X(s) - wn2x(0) = 1/s2
     
  5. Sep 11, 2013 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yup, looks right.
     
  6. Sep 11, 2013 #5
    So, now I would solve for X(s), but what would I do after this point? Would I need partial fractions?
     
  7. Sep 11, 2013 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    After you solve for X(s), you want to invert it to find x(t). Partial fractions will probably be useful to get X(s) into a form where you can invert it using a table.
     
  8. Sep 12, 2013 #7
    Okay, I've solved it for X(s) and used the initial conditions x(0)=0 and x'(0)=0

    This gives me:

    x(s) = 1/(s2(s2+wn2))

    Which I can break down into:

    s/(s2+ wn2) * 1/s3

    From the tables this, I think, gives me:

    cos(wnt) * t2/2

    How does this look?
     
  9. Sep 12, 2013 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The Laplace transform is a linear operator. That means if you have two transform pairs x(t), X(s) and y(t), Y(s), then L[ax(t)+by(t)] = aL[x(t)]+bL[y(t)] = aX(s) + bY(s). What you can't say is that x(t)y(t) corresponds to X(s)Y(s).

    You want to use partial fractions to split your X(s) up into a sum of terms you can invert using the tables.
    $$X(s) = \frac{1}{s^2(s^2+\omega^2)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s^2+\omega^2} + \frac{Ds}{s^2+\omega^2}$$
     
  10. Sep 12, 2013 #9
    So, as I understand it. I have s=0, 0, -w, w.

    s2*1/(s2(s2+w2)) evaluated at s=0

    So, then I'd have A = 1/w2
     
  11. Sep 12, 2013 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    It looks wrong. The L.T. of a convolution is the product of the individual L.Ts, so the inverse of a product would be a convolution, not a product. You could, of course, actually carry out the convolution operation to get the correct answer.

    Just use partial fractions; it is easier, and it avoids convolutions.
     
  12. Sep 12, 2013 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Multiply both sides by ##s^2(s^2+\omega^2)##. You can then set s=0 in the resulting equation. All terms except for the one with A will vanish, so you can solve for A. The rest of the coefficients will take a little more work. Note that with ##s=\omega##, you get ##s^2+\omega^2 = 2\omega^2##, not 0, so setting ##s=\omega## doesn't help you.
     
  13. Sep 12, 2013 #12
    So, I multiplied both sides by s2(s2+w2) which left me with:

    1 = As2+Aw2

    Solving for A I get

    A = 1/(s2+w2)

    Why was s=0?

    Could I do something similar to find B, like use s(s2+w2) on both sides?
     
  14. Sep 12, 2013 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I meant multiply this equation
    $$\frac{1}{s^2(s^2+\omega^2)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s^2+\omega^2} + \frac{Ds}{s^2+\omega^2}$$ by ##s^2(s^2+\omega^2)##, and then set s=0 to solve for A.
     
  15. Sep 12, 2013 #14
    Yes, I did that. You were right. The left side of the equation became 1 whereas everything else but A turned to 0

    1 = A(s2+w2)

    A = 1/w2

    Did we choose w=0 because it as convenient for us, as it left us with only A?

    What would be the best method to solve for B?

    I had considered multiplying both sides by s(s2+w2) but this results in A being divided by 0 if s=0.
     
  16. Sep 12, 2013 #15

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The other terms don't turn into 0. Why do you think they do?
     
  17. Sep 12, 2013 #16

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Why are you making such heavy weather of this question? You just need the partial-fraction expansion of
    [tex] \frac{1}{s^2(s^2+\omega^2)}.[/tex]
    and surely you must have done lots of those in Calculus 101. If not, just Google 'partial fractions' to see lots of on-line sources showing you how to do it. It really is not that hard; it just takes a bit of practice.
     
  18. Sep 12, 2013 #17
    Because after each term is multiplied by s2(s2+w2)

    we're left with

    1 = A(s2+w2) + Bs(s2+w2) + cs2 + Dss2

    Which simplifies to

    1 = As2 + Aw2 + Bs3 + Bsw2 + cs2 + Ds3

    evaluating at s=0

    We have

    1 = Aw2

    A = 1/w2
     
  19. Sep 12, 2013 #18

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Right. This is a lot different than saying that you get ##1=A(s^2+\omega^2)##. I hope you can see that. (If you're going to say you were just setting s=0 in the other terms, then I'd just ask why didn't you set s=0 in the first term then?)

    To get the rest of the constants, you can collect the terms:
    $$ 1 = (B+D)s^3 + (A+C)s^2 + (B\omega^2)s + A\omega^2$$ and then match coefficients on the two sides. You should eventually get B=D=0 and C=-1/ω2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted