Results 1 to 12 of 12
Like Tree5Thanks
  • 1 Post By SlipEternal
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy

Thread: System of differential equations

  1. #1
    Newbie
    Joined
    Apr 2017
    From
    Sweden
    Posts
    10

    Question System of differential equations

    x'(t)=-ax(t)
    y'(t)=ax(t)-by(t)

    The question is to find x(t) and y(t).
    I managed to find x(t):

    x(t)=exp(-at)

    but I have trouble finding y(t).

    I know the solution is y(t)=(a/(b-a)) (exp(-at)-exp(-bt))

    I would appreciate very much your help. Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,216
    Thanks
    860

    Re: System of differential equations

    First find the homogeneous solution, then find the specific solution.
    Thanks from mathfn
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,948
    Thanks
    2737

    Re: System of differential equations

    First, your solution for the first equation is wrong. It should be x(t)= C exp(-at) where C can be any constant. With that x(t), the second equation becomes y'(t)= aCexp(-at)- by. That is the same y'- by= aC exp(-at). That is a linear first order differential equation. There is simple formula for an "integrating factor". Do your textbook have that formula? (Look up "integrating factor" in the index.)

    Or you can use the method that SlipEternal refers to. The "associated homogeneous differential equation" is y'- by= 0. That is the same as dy/y= bdx. Integrating both sides, ln(y)= bx+ C. Taking the exponential of both sides gives y= e^(bx+ D)= D'e^(bx) where D'= e^D.

    Then, to find the general solution to the entire equation we only have to add any specific equation to the differential equation. Since the "non-homogeneous" part is aC e^(-at), and I know that the derivative of an exponential is just that exponential again, I would try a function of the form y= He^{-at} where H is a constant to be determined. y'= -aHe^{-at}. Put those into the equation and you will have a simple equation to solve for H.
    Thanks from mathfn
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2017
    From
    Sweden
    Posts
    10

    Re: System of differential equations

    I used the integrating factor model and obtained this, I am not sure whether I'm correct or not.

    y'(t)=aCexp(-at)-by(t)
    y'(t)+by(t)=aCexp(-at) (1)
    I=exp(integral(b)dt
    I=exp(bt) (2)
    Multiply I on both sides of (1)

    Iy'(t)+Iy(t)= IaCexp(-at)

    exp(bt) y'(t)+exp(bt) by(t)=exp(bt) aCexp(-at)

    exp(bt) y(t)= integral (exp(bt) aCexp(-at)dt)

    exp(bt) y(t)=aC/(b-a) exp(bt-at)
    y(t)=aC/(b-a) exp(-at)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,948
    Thanks
    2737

    Re: System of differential equations

    If you put that x(t) and y(t) into the original equations, are they satisfied?
    Thanks from mathfn
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2017
    From
    Sweden
    Posts
    10

    Re: System of differential equations

    Unfortunately they are not satisfied. I can't really see the mistake I made.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,948
    Thanks
    2737

    Re: System of differential equations

    Alright. You are attempting to solve the differential equation y'- by= aCe^{-at}. An integrating factor is [tex]\phi(t)= e^{-bt}[tex]. e^{-bt}\left(y'+ by)= (e^{-bt}y)'= aCe^{-(a+b)t}. Integrating both sides, e^{-bt}y= -\frac{aC}{a+ b}e^{-(a+b)t}+ D so that y(t)= -\frac{aC}{a+b}e^{-at}+ De^{bt}.

    You keep forgetting the "constant of integration"!
    Thanks from mathfn
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2017
    From
    Sweden
    Posts
    10

    Re: System of differential equations

    I understand but I'm having trouble understanding how y(t)= -aC/(b-a) exp(-at) + Dexp(-bt) is equal to

    y(t)=(a/(b-a)) (exp(-at)-exp(-bt)). (This is the solution obtained from the book).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,948
    Thanks
    2737

    Re: System of differential equations

    Quote Originally Posted by mathfn View Post
    I understand but I'm having trouble understanding how y(t)= -aC/(b-a) exp(-at) + Dexp(-bt) is equal to

    y(t)=(a/(b-a)) (exp(-at)-exp(-bt)). (This is the solution obtained from the book).
    It isn't. However, it is if you take C= 1 and D= -1. Those "undetermined constants", C and D, have to be determined by some additional conditions. What were those additional conditions.
    Thanks from mathfn
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2017
    From
    Sweden
    Posts
    10

    Re: System of differential equations

    I think there are boundary conditions at t=0. One should therefore integrate from 0 to t.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,948
    Thanks
    2737

    Re: System of differential equations

    We can't help with a problem when you refuse to tell us what the problem is!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Apr 2017
    From
    Sweden
    Posts
    10

    Re: System of differential equations

    This is the original problem. It's a Markov process problem and I asked about part (a). I replaced α and β by a and b respectively, p_0(t) and p_1(t) by x(t) and y(t) respectively for simplicity.

    System of differential equations-pbm.jpg
    Last edited by mathfn; May 17th 2017 at 08:47 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. {x'=x(1-x-y), y'=y(1.5-x-y)} System of Differential equations
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Dec 16th 2014, 12:02 PM
  2. System of differential equations
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Aug 11th 2012, 12:36 AM
  3. Replies: 0
    Last Post: Mar 21st 2012, 01:38 AM
  4. system of differential equations
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Apr 28th 2010, 01:37 AM
  5. System of Differential Equations
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 21st 2007, 05:36 PM

/mathhelpforum @mathhelpforum