Results 1 to 3 of 3

Math Help - Power Law Solution/Initial conditions

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    12

    Power Law Solution/Initial conditions

    Hi,

    I have the following equation as a general solution:

    L = k1t^1/3 + k2t^2/3

    I need to write down the initial values for L and dl/dt at time t=to.

    First question is am i right in thinking that these are: t=0, t=1?

    and then will it be obvious how this leads onto

    L=L0{2(t/t0)^1/3 - (t/t0)^2/3}

    Cheers
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    the initial conditions are at the same time --you want L(t0) and dL/dt(t0)

    If you think about it suppose you were doing projectile motion then

    you would need the position s(t) and velocity v(t) both at t = 0


    For L = k1t^1/3 + k2t^2/3

    L(t0) = L0 = k1(t0)^1/3 + k2(t0)^2/3

    dL/dt(t0) = 1/3k1(t0)^(-2/3) +2/3k2(t0)^-1/3

    do you have any other info on dL/dt(t0) ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2008
    Posts
    12
    no but what u said makes sense thats what I was heading towards just not confident enough to make final step!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. using initial conditions
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: March 17th 2011, 03:47 PM
  2. Initial conditions
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 9th 2010, 05:11 PM
  3. Summation with initial conditions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 23rd 2009, 02:05 PM
  4. finding solution by initial conditions
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 18th 2009, 10:36 AM
  5. PDE initial conditions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 29th 2008, 08:42 PM

Search Tags


/mathhelpforum @mathhelpforum