Results 1 to 3 of 3

Math Help - Initial conditions

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    11

    Initial conditions

    Find the particular solution for the differential equation \dfrac{d^2x}{dt^2}-\dfrac{dx}{dt}-20x = 0 that satisfies the initial conditions x = 0, \dfrac{dx}{dt} = 2 when t = 0.

    Substituting x = Ae^{n_{1}x}+Be^{n_{2}x}, I got n = 5 and n = -4, so the general solution is x = Ae^{5t}+Be^{-4t}.

    What I don't know is how to make use of the initial conditions to find A and B.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member slider142's Avatar
    Joined
    May 2009
    From
    Brooklyn, NY
    Posts
    72
    Find (dx/dt)(t): \frac{dx}{dt}(t) = 5Ae^{5t} - 4Be^{-4t}
    Now you can get two equations in two unknowns using the two initial conditions: (dx/dt)(0) = 2 and x(0) = 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    11
    Quote Originally Posted by slider142 View Post
    Find (dx/dt)(t): \frac{dx}{dt}(t) = 5Ae^{5t} - 4Be^{-4t}
    Now you can get two equations in two unknowns using the two initial conditions: (dx/dt)(0) = 2 and x(0) = 0.
    5A-4B = 2, 5A = 2+4B, A = \dfrac{2+4B}{5}.

    A = -B

    \dfrac{2+4B}{5} = -B

    2+4B = -5B

    B = -\dfrac{2}{9}

    \therefore \;\;\; A = \dfrac{9}{2}<br /> <br />
and B = -\dfrac{9}{2}.

    Therefore the particular equation is x = -\dfrac{2}{9}e^{5t}+\dfrac{2}{9}e^{-4t}.
    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, 04:47 PM
  2. Does it matter when you use the initial conditions?
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 24th 2010, 01:52 PM
  3. Summation with initial conditions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 23rd 2009, 03:05 PM
  4. Power Law Solution/Initial conditions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 16th 2009, 03:35 PM
  5. PDE initial conditions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 29th 2008, 09:42 PM

Search Tags


/mathhelpforum @mathhelpforum