Results 1 to 2 of 2

Math Help - Solving 2 differential equations

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    26

    Solving 2 differential equations

    dx/dt=ay and dy/dt=bx where x and y are function of t [x(t) and y(t)] and a and b are constant.

    1) show what x and y satisfy the equation for a hyperbola: y^2-(b/a)*x^2=(y_0)^2-(b/a)*(x_0)^2

    2) suppose at some time t_s, the point (x(t_s),y(t_s)) lies on the upper branch of hyperbola, show that: y(t_s)>sqrt(b/a)*x(t_s)

    I dun know whether i am doing it right.

    First, in integrate both equations,

    dx/dt=ay >>> x/y+C_1=at+C_2 >>> x/y+C_5=at

    dy/dt=bx >>> y/x+C_3=bt+C_4 >>> y/x+C_6=bt

    then I say t = 0 and so

    x/y+C_5=at >>> C_5=-x_0/y_0

    y/x+C_6=bt >>> C_6=-y_0/x_0

    then i say this happens only when C_5 and C_6 are 0

    then going back to

    x/y+C_5=at >>> x/y=at

    y/x+C_6=bt >>> y/x=bt and isolating t to yield

    y^2-(b/a)*x^2=0

    and when t=0

    y_0^2-(b/a)*x_0^2=0

    so y^2-(b/a)*x^2=y_0^2-(b/a)*x_0^2

    am i right about it?

    and can somebody give me some hints to deal with the second problem? thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,569
    Thanks
    1410

    Re: Solving 2 differential equations

    Quote Originally Posted by oxxiissiixxo View Post
    dx/dt=ay and dy/dt=bx where x and y are function of t [x(t) and y(t)] and a and b are constant.

    1) show what x and y satisfy the equation for a hyperbola: y^2-(b/a)*x^2=(y_0)^2-(b/a)*(x_0)^2

    2) suppose at some time t_s, the point (x(t_s),y(t_s)) lies on the upper branch of hyperbola, show that: y(t_s)>sqrt(b/a)*x(t_s)

    I dun know whether i am doing it right.

    First, in integrate both equations,

    dx/dt=ay >>> x/y+C_1=at+C_2 >>> x/y+C_5=at

    dy/dt=bx >>> y/x+C_3=bt+C_4 >>> y/x+C_6=bt
    No, this is not at all correct. You have integrated the first equation assuming that y is a constant and integrating the second equation assuming that x is constant and those assumptions are not true.

    There are two standard ways to handling pairs of equations.
    Simplest is to [b]differentiate[b] the first equation as second time with respect to t: d^2x/dt^2= a(dy/dt) and, from the second equation that becomes
    d^2x/dt^2= a(bx) so you have the second degree equation d^2x/dt^2- abx= 0. The general solution to that will involve two undetermined constants. Once you have x, solve for y by using y= (dx/dt)/a.

    \then I say t = 0 and so

    x/y+C_5=at >>> C_5=-x_0/y_0

    y/x+C_6=bt >>> C_6=-y_0/x_0

    then i say this happens only when C_5 and C_6 are 0

    then going back to

    x/y+C_5=at >>> x/y=at

    y/x+C_6=bt >>> y/x=bt and isolating t to yield

    y^2-(b/a)*x^2=0

    and when t=0

    y_0^2-(b/a)*x_0^2=0

    so y^2-(b/a)*x^2=y_0^2-(b/a)*x_0^2

    am i right about it?

    and can somebody give me some hints to deal with the second problem? thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving differential equations
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2010, 10:21 PM
  2. Solving Differential Equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 3rd 2010, 09:18 AM
  3. Solving differential equations..help!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 28th 2010, 04:02 PM
  4. Solving differential equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 10th 2008, 10:36 AM
  5. Help on solving differential equations...
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: October 19th 2008, 03:14 PM

Search Tags


/mathhelpforum @mathhelpforum