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!

Differential equation using laplace transforms

  1. Oct 17, 2008 #1
    1. The problem statement, all variables and given/known data

    y'' + y = delta(t-2*pi)

    3. The attempt at a solution

    I have solved this using laplace transforms.

    Y = e^(-2*pi*s)/(s^2 + 1) + 10s/(s^2 + 1)

    so that

    y = sint*u(t - 2*pi) + 10 *cost

    what I don't understand is this function is not definied for t<0. According to my book, the answer is:

    y = 10 cos t if 0<t<2*pi and y = 10cost + sint if t>2*pi

    Why is this?
    Last edited: Oct 17, 2008
  2. jcsd
  3. Oct 17, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    u(t-2*pi) IS defined for t<0. It's equal to zero. In fact, it's equal to 0 for t<2*pi and 1 for t>2*pi.
  4. Oct 17, 2008 #3
    So I'm right to say that the answer here is:

    y = 10*cos(t) if t<2*pi, and
    y = 10*cos(t) + sin(t) if t>2*pi

  5. Oct 17, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    Absolutely right.
  6. Oct 17, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Since I have made it clear many times that I dislike the "Laplace transform" method (in only works for linear equations with constant coefficients that you can do more easily anyway):

    The general solution to the associated homogeneous equation is y"+ y= 0 which has y(t)= C cos(t)+ D sin(t) as general solution. Use "variation of parameters" to find a specific solution to the entire equation:

    We seek a solution of the form y(t)= u(t) cos(t)+ v(t) sin(t). Differentiating, y'= u' cos(t)- u sin(t)+ v' sin(t)+ v cos(t). Since we need only one out of the infinitely many solutions, we can simplify by requiring that u' cos(t)+ v' sin(t)= 0. Then y'= -u sin(t)+ v cos(t) and, differentiating again, y"= -u' sin(t)- u cos(t)+ v' cos(t)- v sin(t). y"+ y = -u' sin(t)+ v'cos(t)= [itex]\delta(t- 2\pi)[/itex].

    That, together with u' cos(t)+ v' sin(t)= 0 gives two equations to solve for u' and v'. Multiplying the first equation by cos(t) and the second by sin(t) and adding eliminates u':
    v'= [itex]cos(t) \delta(t- 2\pi)[/itex] and the integral of that, by definition of the delta function, is v(t)= 0 if x< [itex]2\pi[/itex], cos(2\pi)= 1 if [itex]x\le 2\pi[/itex], in other words, [itex]H(x- 2\pi)[/itex] where H is the Heaviside step function.
    Multiplying the first equation by sin(t) and the second by -cos(t) and adding eliminates v': u'= -sin(t)[/itex]\delta(x-2\pi)[/itex] and integrating that u(t)= 0 is x< 2[itex]\pi[/itex], -sin(2\pi)= 0 if [itex]x\ge 2\pi[/itex]. In other words, v(t)= 0 for all t.

    The specific function, y(t)= u(t) cos(t)+ v(t) sin(t) is [itex]H(x- 2\pi) sin(x)[/itex] so the general solution to the entire equation is [itex]y(t)= C cos(t)+ D sin(t)+ H(x-2\pi)\delta(x-2\pi)[/itex].

    I notice that the initial conditions for the problem were never stated.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?