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First order coupled differential equations

  1. Aug 10, 2005 #1
    Hello
    I'm struggling with this concept, cant seem to get my head round it or find any good reference sites or books.

    I have calculated the eigen values and eigen vectors for the following matrix

    5 3
    1 7

    Eigen values 4, 8

    Eigen vectors 4: 3
    -1

    Eigen vectors 8: 1
    1


    Now need to solve these first order coupled differential equations (this is where i just go uhhh?)

    dx/dt = 5x + 3y

    dy/dt = x + 7y

    initial conditions are x(0) = 5 and y(0) = 1


    Any help or pointers would be greatly appreciated, my mind has just gone blank.

    Thanks Rich (Electronic Engineer - If it aint got wires i cant do it!!)
     
  2. jcsd
  3. Aug 10, 2005 #2

    saltydog

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    For the system:

    [tex]\frac{dx}{dt}=5x+3y[/tex]

    [tex]\frac{dy}{dt}=x+7y[/tex]

    You determined the eigenvalues and eigenvectors. Thus the solution is:

    [tex]\left(\begin{array}{c} x(t) \\y(t)\end{array}\right)=k_1e^{4t}
    \left(\begin{array}{c} 3 \\ -1 \end{array}\right)+k_2e^{8t}
    \left(\begin{array}{c} 1 \\ 1 \end{array}\right)
    [/tex]

    Right?

    Substituting the initial conditions:

    [tex]\left(\begin{array}{c} 5 \\ 1 \end{array}\right)=
    k_1\left(\begin{array}{c}3 \\ -1 \end{array}\right)+
    k_2\left(\begin{array}{c}1 \\ 1 \end{array}\right)
    [/tex]

    Now just figure out what the constants are. You know how to read this matrix stuf huh? That last one is just:

    [tex] 5=3k_1+k2[/tex]

    [tex] 1=-k_1+k_2[/tex]

    Same dif for reading the other one.

    Edit: Oh yea:

    Welcome to PF and

    "Differential Equations" by Blanchard, Devaney, and Hall is one DE book I use.
     
    Last edited: Aug 10, 2005
  4. Aug 10, 2005 #3

    Hurkyl

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    If you're comfortable with the linear algebra, you could look at it like this:

    You have the (matrix) differential equation x' = Ax.
    You diagonalized A, giving [itex]A = P^{-1} D P[/itex].

    Since A is constant, you can rewrite the equation as:

    [tex]\vec{x}' = A\vec{x} = P^{-1} D P\vec{x}[/tex]
    [tex]P\vec{x}' = DP\vec{x}[/tex]
    [tex](P\vec{x})' = D (P\vec{x})[/tex]

    Which is a pair of uncoupled differential equations in the components of Px, which is easy to solve. Once you know Px, you can compute x.


    Actually, if you're comfortable with more advanced matrix manipulations, you can go from x' = Ax directly to [itex]\vec{x}(t) = e^{At} \vec{x}_0[/itex]. Of course, that requires you to know how to compute a matrix exponential. :biggrin: You have to do the same amount of work either way, though.
     
  5. Aug 11, 2005 #4
    One other releated question on this, isn't there a way to avoid using matrices and instead represent two coupled first order differential equations as one second order differential equation? I need to start working on a couple problems today and I was advised that would be the best way to attack them. Does anybody know the form for that?

    ~Lyuokdea
     
  6. Aug 11, 2005 #5

    HallsofIvy

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    Yes, there is. I wouldn't consider it the "best way" to handle systems of equations but it is simpler and less sophisticated and so perhaps better for a beginner.

    saltydog has already pointed out that the d.e.s associated with the matrix given are:
    [tex]\frac{dx}{dt}=5x+3y[/tex] and
    [tex]\frac{dy}{dt}=x+7y[/tex]

    Differentiate that first equation again with respect to t to get
    [tex]\frac{d^2x}{dt^2}= 5\frac{dx}{dt}+ 3\frac{dy}{dt}[/tex]
    Replace the [tex]\frac{dy}{dt}[/tex] in that with x+ 7y:
    [tex]\frac{d^2x}{dt^2}= 5\frac{dx}{dt}+ 3(x+ 7y)[/tex]
    [tex]\frac{d^2x}{dt^2}= 5\frac{dx}{dt}+ 3x+ 21y[/tex]

    There is still a "y" in that but from the first equation,
    [tex]3y= \frac{dx}{dt}- 5x[/tex] so we have
    [tex]\frac{d^2x}{dt^2}= 5\frac{dx}{dt}+ 2x+7(\frac{dx}{dt}- 5x)[/tex]
    [tex]\frac{d^2x}{dt^2}= 12\frac{2x}{dt}-32x[/tex].

    The characteristic equation for that is [tex]r^2= 12r- 32[/tex] exactly the same as the eivenvalue equation for the original matrix and has, of course, the same roots: 4 and 8.

    The general solution for x is: x(t)= C1e4t+ C2e8.

    To solve for y use [tex]3y= \frac{dx}{dt}- 5x[/tex]. Notice that you don't have to do any integration to solve that for y so you have the same "C1" and "C2" coefficients in both x and y.
     
    Last edited: May 26, 2010
  7. Aug 13, 2005 #6
    Soooo

    [tex] 5=3k_1+k2[/tex]

    [tex] 1=-k_1+k_2[/tex]

    That is the acctual soloution?

    Thanks for your help
     
  8. Aug 13, 2005 #7

    saltydog

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    Knish . . . no dude . . . no, no. I'm disappointed you'd think that. Not in you but me for presenting the solution in a way that would make you think that's the solution. That's just the two equations used to figure out what the constants k1 and k2 are. The actual solution is:

    [tex]\left(\begin{array}{c} x(t) \\y(t)\end{array}\right)=k_1e^{4t} \left(\begin{array}{c} 3 \\ -1 \end{array}\right)+k_2e^{8t} \left(\begin{array}{c} 1 \\ 1 \end{array}\right)[/tex]

    That reads:

    [tex]x(t)=3k_1e^{4t}+k_2e^{8t}[/tex]

    [tex]y(t)=-k_1e^{4t}+k_2e^{8t}[/tex]

    Now, just figure out what k1 and k2 are from the expression:

    [tex]5=3k_1+k_2[/tex]

    [tex]1=-k_1+k_2[/tex]
     
    Last edited: Aug 13, 2005
  9. Sep 22, 2009 #8
    close to saltydog' solution:

    x = y' - 7y
    x' = y'' - 7y'

    if we put these in x' = 5x + 3y:

    y'' - 12y' +32y = 0

    now we have second order diff eq. a heuristic solution for y could be

    y = exp(r*t)
    y' = r * exp(r*t)
    y'' = r^2 exp(r*t)

    where r is a constant

    then,

    exp(r*t) {r^2 - 12r + 32} = 0

    exp(r*t) cannot be zero unless t or r is minus infinity. therefore r is either 8 or 4. then you can solve the rest I believe.

    one last thing: engineering is not about wires, it's technician's work. an engineer without math skills is like a butcher without a knife.
     
  10. Sep 23, 2009 #9

    LCKurtz

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    The system given was:

    dx/dt = 5x + 3y

    dy/dt = x + 7y

    initial conditions are x(0) = 5 and y(0) = 1

    Another method of solution which hasn't been mentioned is LaPlace transforms. This method has the advantage of "building in" the initial conditions to the solution. Using L for the LaPlace operator, the equations become:

    sL(x) - 5 = 5L(x) + 3L(y)
    sL(y) -1 = L(x) + 7L(y)

    (5 - s)L(x) + 3L(y) = -5
    L(x) + (7-s)L(y) = -1

    Easy to solve by determinants for L(x) and L(y). For example, for L(x) you get:

    L(x) = (-32 + 5s) / (32 - 12s + s2) = 3/(s-4) + 2 /(s-8)

    using partial fractions. So you get

    x = 3e4t + 2e8t

    and similarly easy for y.
     
  11. May 26, 2010 #10
    dx/dt = 5x + 3y ---- (1)
    dy/dt = x + 7y -----(2)

    For simple DEs as above, you can make equation (1) subject of y and substitute into equation (2) to obtain a 2nd order ODE. From there you can solve for x and then y.
     
  12. May 26, 2010 #11

    HallsofIvy

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    That's what I showed in the 5th post.
     
  13. Jun 14, 2010 #12
    Some references about ordinary differential equations

    If you are interested in some references about solving system of linear differential equations, please refer to the post I have written here:

    Solving system of first order linear differential equations with matrix exponential method

    If you would like to learn more about differential equations, you may refer to my article, which teaches how to solve 1st order, and higher order differential equations:

    Introduction to differential equation, and solving linear differential equations using operator method
     
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