Results 1 to 6 of 6

Math Help - Initial value problems

  1. #1
    Newbie
    Joined
    Sep 2006
    Posts
    23

    Initial value problems

    I have been working on this hw set all weekend and it just isn't clicking!! Any help would be greatly appreciated.

    Solve the given initial value problems by Laplace transforms. Show the details of your work.

    1.
    y" + 9y = {
    6 sin t if 0 <= t <= pi
    0 if t > pi, y(0) = 0, y'(0) = 0
    }

    2.
    y" - 2y' + 2y = 8e^(-t)cos t ; y(0) = 16; y'(0) = -16
    Last edited by CaptainBlack; April 29th 2007 at 01:26 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    [quote=Hollysti;49174]I have been working on this hw set all weekend and it just isn't clicking!! Any help would be greatly appreciated.

    Solve the given initial value problems by Laplace transforms. Show the details of your work.

    1.
    y" + 9y = {
    6 sin t if 0 <= t <= pi
    0 if t > pi, y(0) = 0, y'(0) = 0
    }

    Take the LT of the DE:

    (s^2 + 9)Y = 6(e^{-pi s}+1)/(s^2+1)

    (note the implicit assumption that y(0)=0, and y'(0)=0 here since
    Ly'' = s^2 Y - s y(0) - y'(0) )

    so:

    Y = 6(e^{-pi s}+1)/[(s^2+1)(s^2+9)

    .. = (6/8)(e^{-pi s}+1)/(s^2+1) - (6/8)(e^{-pi s}+1)/(s^2+9)

    Both of the terms on the right are LT's of single sin arches, so:

    y_1(t) = (6/8) f(t) - (6/8) g(t)

    where:

    f(t) = sin(t) 0<=t<=pi

    f(t) = 0 otherwise.

    and:

    ERROR, this is not the inverse LT of (e^{-pi s}+1)/(s^2+9)

    g(t) = (1/3) sin(3 t) 0<= t <= pi/3

    g(t) = 0 otherwise

    This y is a particular integral of the DE (which as I assumed that y(0)=0 and y'(0)=0
    when I tool the LT is actualy the answere required the following is not strikly necessary)
    the general solution is this plus a solution of the homogeneous equation:

    y'' + 9y = 0

    which is Asin(3t) + Bcos(3t).

    Hence the general solution is:

    y(t) = y_1(t) + Asin(3t) + Bcos(3t).

    Now y(0)=0, forces B=0, and:

    y'(0) means that:

    cos(0) - (1/3) 3 cos(0) + A 3 cos(0) = 0

    so A = 0.

    RonL

    (check this very carefully there is almost certainly an error in the algebra)
    Last edited by CaptainBlack; April 30th 2007 at 12:27 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2006
    Posts
    23
    I'm sorry, but would you please show me how you got that very first step?? I can follow what you did after that... Thank you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Hollysti View Post
    I'm sorry, but would you please show me how you got that very first step?? I can follow what you did after that... Thank you.
    That would be:

    Take the LT of the DE:

    (s^2 + 9)Y = 6(e^{-pi s}+1)/(s^2+1)
    The left hand side is the LT of y''+9y (with the implicit assumption that
    y(0)=0 and y'(0)=0, which is why this will be a particular integral of the
    DE rather than a general one, the initial conditions could be left in here
    but it will complicate the subsequent algebra, and its the initial conditions
    that we need any way).

    The right hand side is the LT of the right hand side of the DE, you can
    find this by looking up the LT in a table or by doing the required integration
    yourself.

    Note there is a mistake in my algebra later which I will correct by the time
    you read this.

    Looking at the solution again there is something about it I don't like, I
    will be looking at it again later today

    RonL
    Last edited by CaptainBlack; April 29th 2007 at 09:38 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    I left this problem with the partial result that the Laplace Transform Y(s) of
    the required solution y(t) satisfying the required initial conditions was:

    Y(s) = 6(e^{-pi s}+1)/[(s^2+1)(s^2+9)]

    Now if we put:

    F(s) = 6(e^{-pi s}+1)/(s^2+1),

    and:

    G(s) = 1/(s^2+9)

    we gave that F(s) is the Laplace Transform of:

    f(t) = 6.sin(t) 0<=t<=pi
    f(t) = 0 otherwise.

    and G(s) is the Laplace Transform of:

    g(t) = (1/3) sin(3t).

    Now we may invoke the convolution theorem to conclude that:

    y(t) = integral_{tau=0 to t} f(tau) g(t - tau) dtau.

    ..... = integral_{tau=0 to min(t, pi)} 2 sin(tau)sin(3(t-tau)) dtau

    so:

    y(t) = integral_{tau=0 to t} 2 sin(tau)sin(3(t-tau)) dtau for 0<t<pi

    ..... = integral_{tau=0 to pi} 2 sin(tau)sin(3(t-tau)) dtau for t>=pi

    and I will leave these integrals as an exercise for the reader (at least untill I have time
    to do them myself)

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    I left us with the solution in the form:

    y(t) = integral_{tau=0 to t} 2 sin(tau)sin(3(t-tau)) dtau for 0<t<pi
    ..... = integral_{tau=0 to pi} 2 sin(tau)sin(3(t-tau)) dtau for t>=pi

    Now using QuickMath to do these integrals (QM does not like multi-character
    variable names so we have to use something like u in place of tau there) we get:

    integral_{tau=0 to t} 2 sin(tau)sin(3(t-tau)) dtau = sin^3(t)

    and:

    integral_{tau=0 to pi} 2 sin(tau)sin(3(t-tau)) dtau = 0

    so:

    y(t) = sin^3(t) for 0<t<pi
    ..... = 0 for t>=pi,

    Which is very satisfying as that agrees to a high level of precission with
    the result of my numerical integration of the DE.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integrals and initial-value problems?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 14th 2010, 10:32 PM
  2. Initial Value Problems
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: September 15th 2009, 05:59 PM
  3. initial value problems
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 31st 2009, 11:39 PM
  4. Initial Value Problems
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 2nd 2008, 04:34 PM
  5. initial value problems
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 25th 2007, 07:44 PM

Search Tags


/mathhelpforum @mathhelpforum