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System of Differential Equations

  1. Dec 3, 2007 #1
    System of Differential Equations (Urgent)

    1. The problem statement, all variables and given/known data
    Solve for y(t). You need not find x(t).
    [tex]2x' + y' - y = t[/tex]
    [tex]x' + y' = t^2[/tex]
    x(0) = 1, y(0) = 0

    3. The attempt at a solution
    [tex]2Dx + (D - 1)[y] = t[/tex]
    [tex]Dx + Dy = t^2[/tex]
    [tex]2D^2x + (D^2 - D)[y] = 1[/tex]
    [tex]2D^2x + 2D^2y = 4t[/tex]
    [tex](D^2 + D)[y] = 4t - 1[/tex]
    [tex]y'' + y' = 4t - 1[/tex]
    I solved the above equation using undetermined coefficients. It's a lot of writing, so I'll just put the answer here:
    [tex]y(t) = 2t^2 - 5t + C_{1} + C_{2}E^{-t}[/tex]
    Using the initial value y(0) = 0
    [tex]0 = C_{1} + C_{2}[/tex]
    [tex]C_{1} = -C_{2}[/tex]

    I'm stuck at applying the other initial value. It says you don't need to find x(t), so I was wondering if there is a way to do this without going through and solving for x(t) then solving for the constants and such.
     
    Last edited: Dec 3, 2007
  2. jcsd
  3. Dec 3, 2007 #2

    HallsofIvy

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    Integrating both sides of x'+ y'= t2, you get x(t)+ y(t)= (1/3)t3. In particular, x(0)+ y(0)= 0.
     
  4. Dec 3, 2007 #3
    When doing that integration, isn't a constant introduced into the equation?
    [tex]x + y = \frac{t^3}{3} + C_{3}[/tex]
    [tex]x = \frac{t^3}{3} + C_{3} - y[/tex]
    [tex]x = \frac{t^3}{3} + C_{3} - 2t^2 + 5t - C_{1} - C_{2}E^{-t}[/tex]


    I also tried
    [tex] x + y = \frac{t^3}{3} + C_{3}[/tex]
    let t = 0
    [tex]x(0) + y(0) = C_{3}[/tex]
    [tex]1 + 0 = C_{3}[/tex]
    [tex]x(t) = \frac{t^3}{3} + 1 - 2t^2 + 5t - C_{1} - C_{2}E^{-t}[/tex]
    x(0) = 1
    [tex]1 = 1 - C_{1} - C_{2}[/tex]
    [tex]C_{1} = -C_{2}[/tex]
    Which is the same thing I got before.

    *EDIT* sorry, the initial condition for x was x(0) = 1.
     
  5. Dec 3, 2007 #4

    HallsofIvy

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    Oh, blast! You are right!
     
  6. Dec 3, 2007 #5
    So, is this solvable?
     
  7. Dec 3, 2007 #6

    Dick

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    If x(t) solves your equations, then so does x(t)+C. A value for x(0) doesn't constrain the solution in any way. Are you sure the boundary condition isn't for x'(0)? BTW you could avoid introducing an extra derivative and just written a first order equation for y.
     
  8. Dec 4, 2007 #7

    HallsofIvy

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    Oh, very good, Dick. I'm embarassed I didn't see that. (Don't worry, in few minutes I will have convinced myself that I did!)

    From the second equation, x'= t2- y^2 so you can eliminate x' from the first equation.
     
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