1. Limited time only! Sign up for a free 30min personal 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!

Science and engineering math: non-homogeneous differential equation

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation
    y(iv)(t) - 16y(t) = 30sint
    subject to y(0) = 0, y'(0) = 2, y"(∏) = 0, y'"(∏) = -18

    2. Relevant equations

    There is a Laplace transform table attached if needed :)

    3. The attempt at a solution

    I tried making it homogeneous and then taking the Laplace. But I don't think that was right because wouldn't the Laplace of y(t) just be the definition of a Laplace ∫ y(t)e-stdt? So that's where I got confused
     

    Attached Files:

  2. jcsd
  3. Mar 19, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi chatterbug219! :smile:

    just solve x4 - 16 = 0 :wink:
     
  4. Mar 19, 2012 #3
    (x2-4)(x2+4)=0
    x=2

    But how do I use that to solve the differential?
     
  5. Mar 19, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    there are four roots

    haven't you covered how to solve polynomial differential equations?
     
  6. Mar 19, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you take the Laplace Transform of the DE, you need y''(0) = a and y'''(0) = b, which you are not given. However, you can solve for the transform Y(s) as a function of the unknown parameters a and b. Then you can get the solution y(t) in terms of a and b; then you can impose the boundary conditions at t = π to finally determine a and b.

    RGV
     
  7. Mar 19, 2012 #6
    So I did that and I got
    Y= 30[1/(s2+1)(s4-16)] + (2s2+as+b)/(s4-16)
    Do I do partial fractions for this now to get a and b? Then take the inverse Laplace?
     
  8. Mar 20, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Try it and see!

    RGV
     
  9. Mar 20, 2012 #8
    I've been trying to do the partial fractions for both...so far all I have is
    1/(s2+1)(s4-16)
    where I get
    [As+B/(s2+1)] + [(Cs+D)/(s4-16)]
    which simplifies to
    [As(s4-16)]+[B(s4-16)]+[Cs(s2+1)]+[D(s2+1)]
    and D=1 & C=-2/5...but I can't seem to get A or B to cancel so I can solve for them

    the second one is (2s2+as+b)/(s4-16)
    but I don't know how to do partial fractions for that
     
    Last edited: Mar 20, 2012
  10. Mar 20, 2012 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Write [itex] (s^4 - 16) = (s^2+4)(s^2-4) = (s-2)(s+2)(s^2+4)[/itex] and go on from there. Alternatively, if you want to avoid quadratic denominators altogether, you could take complex roots and write [itex] (s^4 - 16) = (s-2)(s+2)(s-2i)(s+2i), [/itex] but that is not really necessary, since the inverse Laplace of [itex] 1/(s^2 + w^2)[/itex] is tabulated.

    RGV
     
  11. Mar 20, 2012 #10
    So I got the partial fractions for the first one, but not the second one becaue the a and b are throwing me off...I don't know what to do with them in order to get the partial fractions of
    [2s2+as+b]/[(s-2)(s+2)(s2+4)]
     
  12. Mar 20, 2012 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Figure out [tex] \frac{1}{(s-2)(s+2)(s^2+4)} = \frac{A}{s-2} + \frac{B}{s+2} + \frac{Cs + D}{s^2+4},[/tex] then multiply by the numerator.

    RGV
     
  13. Mar 20, 2012 #12
    Okay...
    -1/8(s2+4) - 1/32(s+2) + 1/32(s-2)

    So do I multiply 2s2+as+b by -1/8(s2+4) - 1/32(s+2) + 1/32(s-2)? Is that you were saying?
    But how do I take the inverse Laplace of 2s2+as+b? That's where I'm really confused
     
    Last edited: Mar 21, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Science and engineering math: non-homogeneous differential equation
Loading...