Results 1 to 5 of 5

Thread: differencial System

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    36

    differencial System

    Suppose you are given the following system of linear dif eq's:

    $\displaystyle \frac{dx}{dt} = -\lambda_1 x$
    $\displaystyle \frac{dy}{dt} = \lambda_1 x - \lambda_2 y$
    $\displaystyle \frac{dz}{dt} = \lambda_2 y$

    Find a solution for this system WITHOUT USING matrices or the method using differential operators for elimination.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by alikation0 View Post
    Suppose you are given the following system of linear dif eq's:

    $\displaystyle \frac{dx}{dt} = -\lambda_1 x$
    $\displaystyle \frac{dy}{dt} = \lambda_1 x - \lambda_2 y$
    $\displaystyle \frac{dz}{dt} = \lambda_2 y$

    Find a solution for this system WITHOUT USING matrices or the method using differential operators for elimination.
    Solve the first, for $\displaystyle x(t)$, then use that solution to solve the second for $\displaystyle y(t)$, then use both solutions to solve the third for $\displaystyle z(t)$. You will have to handle the cases $\displaystyle \lambda_1=\lambda_2$ and $\displaystyle \lambda_1 \ne \lambda_2$ separately.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    36
    Quote Originally Posted by CaptainBlack View Post
    Solve the first, for $\displaystyle x(t)$, then use that solution to solve the second for $\displaystyle y(t)$, then use both solutions to solve the third for $\displaystyle z(t)$. You will have to handle the cases $\displaystyle \lambda_1=\lambda_2$ and $\displaystyle \lambda_1 \ne \lambda_2$ separately.

    RonL
    Ok so I'm assuming it's ok the represent the first as:

    $\displaystyle \frac{dx}{dt} = -\lambda_1 x(t)$

    How would I solve this for x(t):

    $\displaystyle dx = -\lambda_1 x(t)dt$
    $\displaystyle \frac{1}{x(t)}dx = -\lambda_1 dt $

    $\displaystyle ln(x(t)) = -\lambda dt$

    This looks like an extremely hard problem having to deal with if $\displaystyle \lambda_1 = \lambda_2$ etc...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2007
    Posts
    36

    Question

    Okay:

    Apparently your way is correct Captain! I was told that I would want to treat the lambda's separately. However, the TA said it was fine if I only do case where I assumed the lambda's were NOT equal and proceeded there (so I don't have to do the case where they).

    Could you help me? I got to:

    $\displaystyle ln(x(t)) = \lambda_1 dt$
    $\displaystyle ln(x(t)) = \lambda_1 t$

    $\displaystyle ln(x(t))/t = \lambda_1$

    So then according to what you said:

    $\displaystyle \frac{dy}{dt} = \frac{ln(x(t))}{t} x(t) - \lambda_2 y(t)$

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,131
    Thanks
    723
    Awards
    1
    Quote Originally Posted by alikation0 View Post
    Okay:

    Apparently your way is correct Captain! I was told that I would want to treat the lambda's separately. However, the TA said it was fine if I only do case where I assumed the lambda's were NOT equal and proceeded there (so I don't have to do the case where they).

    Could you help me? I got to:

    $\displaystyle ln(x(t)) = \lambda_1 dt$
    $\displaystyle ln(x(t)) = \lambda_1 t$

    $\displaystyle ln(x(t))/t = \lambda_1$

    So then according to what you said:

    $\displaystyle \frac{dy}{dt} = \frac{ln(x(t))}{t} x(t) - \lambda_2 y(t)$

    You misunderstood what CaptainBlack was saying. You are looking for x(t) in the first step, not $\displaystyle \lambda _1$.

    $\displaystyle \frac{dx}{dt} = \lambda _1 x$

    $\displaystyle \frac{dx}{x} = \lambda _1 dt$

    $\displaystyle ln(x) = \lambda _1 t + C$

    $\displaystyle x(t) = Ae^{\lambda _1 t}$
    where $\displaystyle A = e^C$.

    Now plug this into the second equation:
    $\displaystyle \frac{dy}{dt} = \lambda _1 x - \lambda _2 y$

    $\displaystyle \frac{dy}{dt} = \lambda _1 Ae^{\lambda _1 t} - \lambda _2 y$

    Now solve for y(t).

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. differencial equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 28th 2008, 04:52 AM
  2. differencial equation
    Posted in the Calculus Forum
    Replies: 15
    Last Post: Jul 20th 2007, 04:16 PM
  3. differencial equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 26th 2007, 11:25 PM
  4. differencial eq
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 7th 2007, 08:33 PM
  5. differencial equation system
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 22nd 2006, 11:40 AM

Search Tags


/mathhelpforum @mathhelpforum