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Substitution/Elimination to find General Solution | System of ODE's

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Use substitution/elimination to find the general solution.

    dx/dt = 3x-y+2t^2

    dy/dt = 4x-2y-8t^2

    2. Relevant equations

    I'm practically clueless on how to solve this problem using substitution/elimination, I'm pretty sure the way I'm doing it is completely wrong. If anyone can show me the correct approach or can show an example it would be a tremendous help.

    3. The attempt at a solution

    dx/dt = 3x - y +2t^2

    dx = (3x -y +2t^2)dt

    x = 3xt - yt + 2/3t^3

    x(1-3t) = - yt + 2/3t^3

    x = (-yt + 2/3t^3)/(1-3t)

    I know this is not the correct way but I can't find anything online, so if someone does know what to do please reply.

    Thanks
     
    Last edited: Dec 9, 2012
  2. jcsd
  3. Dec 9, 2012 #2

    pasmith

    User Avatar
    Homework Helper

    x is a function of t. If you integrate it, you get [itex]\int x(t)\,\mathrm{d}t[/itex], not xt. Similarly for y.

    We want to eliminate either x or y and its derivatives to get a second order ODE for whichever variable we're left with.

    Here it seems easiest to eliminate y, since the first equation tells us that
    [tex]y = 3x - \frac{\mathrm{d}x}{\mathrm{d}t} + 2t^2[/tex]
    Differentiating this gives
    [tex]\frac{\mathrm{d}y}{\mathrm{d}t} = 3\frac{\mathrm{d}x}{\mathrm{d}t}
    - \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4t[/tex]
    You can now substitute these expressions for y and dy/dt into the second equation.
     
  4. Dec 9, 2012 #3
    Thanks for the response and help. Alright so here is what I did:

    y = 3x- x' + 2t^2;

    (3x- x' + 2t^2)' = 4x -2(3x- x' + 2t^2) + 2t^2;

    3x' - x'' + 2t^2' = 4x -6x + 2x' -4t^2 + 2t^2

    x'' - x' -2x = 2t^2' - 2t^2

    r^2 - r - 2 = 0

    r = 2, r = -1

    I don't know how to proceed from this point on.
     
  5. Dec 10, 2012 #4

    pasmith

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    Homework Helper

    The second equation is [itex]y' = 4x - 2y - 8t^2[/itex], so you should have
    [tex]
    (3x- x' + 2t^2)' = 4x -2(3x- x' + 2t^2) - 8t^2
    [/tex]
    which yields
    [tex]
    3x' - x'' + 4t = 4x - 6x + 2x' - 4t^2 - 8t^2
    [/tex]
    (because [itex](2t^2)' = 2(2t) = 4t[/itex]) so that
    [tex]
    x'' - x' - 2x = 4t + 12t^2
    [/tex]

    You need to find a solution of
    [tex]
    x'' - x' - 2x = 4t + 12t^2
    [/tex]
    and add to it the general solution of
    [tex]
    x'' - x' - 2x = 0
    [/tex]
    The second part you should know. For the first part, the form of the right hand side suggests taking [itex]f(t) = at^2 + bt + c[/itex] with constants a, b and c, which you should choose so that [itex]f'' - f' - 2f = 4t + 12t^2[/itex].

    That gives you [itex]x(t)[/itex], and you can then find [itex]y(t)[/itex] from [itex]y = 3x- x' + 2t^2[/itex].
     
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