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ODE: System of Linear Equations usuing Diff. Operator

  1. Nov 22, 2016 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    This is an ordinary differential equation using the differential operator.

    Given the system:
    d^2x/dt - x + d^2y/dt^2 + y = 0
    and
    dx/dt + 2x + dy/dt + 2y = 0

    find x and y equation

    Answer: x = 5ce^(-2t)
    y = -3ce^(-2t)

    2. Relevant equations


    3. The attempt at a solution

    We change it into a differential operator equation...

    [D^2-1]x+[D^2+1]y = 0
    [D+2]x +[D+2]y = 0

    We simply cancel values out until we are left with

    3x+5y = 0
    [D+2]x +[D+2]y = 0

    Here we have x = -5y/3 and y = -3x/5

    We substitute into the second equation and we can easily find that

    x = ce^(-2t)
    y = ce^(-2t)

    My question is: How does the answer book know that there is a 5 for the x equation and a -3 for the y equation. It seems to me that the constant "c" handles this for me.
    Am I doing something wrong to get a less precise answer? Or did I work out this problem correctly.

    Thank you.
     
  2. jcsd
  3. Nov 22, 2016 #2



    What do you mean by "cancel values". This is a bit sloppy notation. Remember that ##D## is an operator.
    Your first equation is telling you that ##3x=-5y##, which is explaining the factors 5 and -3.

    No, this is wrong, since it violates the first equation.

    I also recommend that you use Latex to write your equations. It makes them much easier to read.
     
    Last edited by a moderator: Nov 22, 2016
  4. Nov 22, 2016 #3

    Mark44

    Staff: Mentor

    Yeah, I'm not following what you did, either.How did you go from the 2nd-order DE to 3x + 5y = 0?
    I hope it's obvious that these two equations are equivalent.
     
  5. Nov 22, 2016 #4

    lurflurf

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    Homework Helper

    $$(\mathrm{D}^2-1)x+(\mathrm{D}^2+1)y=0\\
    (\mathrm{D}^2-4)(x+y)+3x+5y=0\\
    (\mathrm{D}-2)(\mathrm{D}+2)(x+y)+3x+5y=0\\
    (\mathrm{D}-2)0+3x+5y=0\\
    3x+5y=0$$
    since the second equation is
    $$(\mathrm{D}+2)(x+y)=0$$

    you should have two constants
    x = c1e^(-2t)
    y = c2e^(-2t)

    related by
    3x+5y=0
     
  6. Nov 22, 2016 #5

    RJLiberator

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    Gold Member

    The key distinction for me in this thread was the relationship of 3x+5y=0. I now see how my answer can be corrected to the final answer as presented in the book.

    Thank you guys for the help.

    I was very close :p.
     
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