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Differential Equations System Solutions

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the initial value problem for the system of first-order differential equations

    y_1' = -2y_2+1, y_1(0)=2

    y_2' = -8y_1+2, y_2(0)=-1

    If the matrix

    [ 0 -2
    -8 0 ]

    has eigenvalues and eigenvectors L_1= -4 V_1= [ 1
    2 ]

    L_2=4 V_2= [ 1

    then its solution will be:

    2. Relevant equations

    3. The attempt at a solution

    e^(-4t) +e^ (4t) from eigenvalues

    multiply by respective eigenvectors and set to initial conditions gives 2 sets of equations and two unknown coefficients




    I am very confident with these values being right for the coefficients, I know need to know how to use these to form a general solution.

    I plug these values back in to get


    Then solution given is


    any help is appreciated! thanks!
  2. jcsd
  3. Dec 7, 2014 #2


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    This is the solution of the homogeneous part only. The system is inhomogeneous, you have to add the particular solution of the equations to the general solution of the homogeneous part.
  4. Dec 7, 2014 #3
    I have not encountered this type of problem before. How do you go about finding the particular solution?
  5. Dec 7, 2014 #4


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    It is simple in this case. Set both y1'=0 and y2' =0. What are y1 and y2 then?
  6. Dec 7, 2014 #5
    Okay, I work this out to y_1=1/4 and y_2=1/2...

    which seems to fit in the equation... but im still not really sure why the coefficients for the e^(-4t) dont seem to match the solution
  7. Dec 7, 2014 #6


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    The general solution is

    Fit these equations to the initial conditions y1(0)=2, y2(0)=-1.
  8. Dec 7, 2014 #7
    Thank you so much! That makes total sense!
  9. Dec 7, 2014 #8


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    You are welcome. :)
  10. Dec 8, 2014 #9


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    iHere is how I would have done this, without using "matrices": differentiate the first equation again to get [itex]y_1''= -2y_2'[/itex]. But from the second equation, [itex]y_2'= -8y_1+ 2[/itex] so that [itex]y_1''= -2(-8y_1+ 2)= 16y_1- 4[/itex] or [itex]y_1''- 16y_1= -4[/itex]. The associated homogeneous equation, [itex]y_1''- 16y= 0[/itex] has characteristic equation [itex]r^2- 16= 0[/itex] which has roots [itex]r= 4[/itex] and [itex]r= -4[/itex] so the general solution to the associated homogeneous equation is [itex]y(t)= C_1e^{4t}+ C_2e^{-4t}[/itex].

    Now new look for a particular solution to the entire equation. Since the right hand, "non-homogeneous", part is the constant -4, we try a solution of the form y= A for A some constant. Then y'= y''= 0 so the equation becomes [itex]0- 16A= -4[/itex] and [itex]A= 1/4[/itex].

    That is, [itex]y_1(t)= C_1e^{4t}+ C_2e^{-4t}+ \frac{1}{4}[/itex]. Now, we can write the first equation, [itex]y_1'= -2y_2+ 1[/itex] is the same as [itex]2y_2= -y_1'+ 1[/itex] or [itex]y_1= -(1/2)y_2'+ 1/2[/itex]. Since [itex]y_1(t)= C_1e^4t+ C_2e^{-4t}+ \frac{1}{4}[/itex], [itex]y_1'= 4C_1e^{4t}- 4C_2e^{-4t}[/itex], [itex]-(1/2)y_2'= -2C_1e^{4t}+ 2e^{-4t}[/itex] and [itex]y_1(t)= -2C_1e^{4t}+ 2C_2e^{-4t}+ 1/2[/itex].
    Last edited by a moderator: Dec 11, 2014
  11. Dec 10, 2014 #10

    Ray Vickson

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    Re-write your system as
    [tex] y_1' = -2y_2 + 1 = -2\left(y_2 - \frac{1}{2} \right) \\
    y_2' = -8y_1+2 = -8\left(y_1 - \frac{1}{4} \right) [/tex]
    Since ##d(y_1 - 1/4)/dt = dy_1/dt## and ##d(y_2 - 1/2)/dt = dy_2/dt##, the variables ##Y_1 = y_1 - 1/4## and ##Y_2 = y_2 - 1/2)## satisfy the homogeneous system
    [tex] Y_1' = -2Y_2\\
    Y_2' = -8 Y_1
    You can easily figure out the boundary values ##Y_1(0), \, Y_2(0)##.

    This type of trick always works whenever the non-homogeneous terms on the right are constants.
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