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Differential Equations general solution

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    For the system of differential equations

    [itex]\frac{dx}{dt}=(-3x-y)[/itex]
    [itex]\frac{dy}{dt}=(-2x-2y)[/itex]

    (a) Find the general solution.
    (b) Find the solution if x(0)=1 and y(0)=2.

    2. Relevant equations



    3. The attempt at a solution

    I have absolutely no clue how to do this. I have never seen a problem like this. I was wondering if someone could tell me what topic this type of problem would fall under so that way I could look up similar problems and understand the concept of how to solve these type of problems.

    I know how to solve initial value problems but I have never seen a system of equations like this were both x and y are a function of t and I'm given both their derivatives. Thanks for any help.

    Some ideas that pop into my head right away is that it's a system of linear equations and that I can use Gaussian reduction or something but I'm not sure.
     
  2. jcsd
  3. Feb 7, 2012 #2

    rock.freak667

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    Perhaps you can find dy/dx using the chain rule and then apply a substitution such as y=vx.
     
  4. Feb 7, 2012 #3

    epenguin

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    Do you know how to solve (homogeneous) equations in one variable?

    Differentiate your equations and see what you can get.

    (Later on maybe you'll be doing the opposite but don't worry.)
     
  5. Feb 7, 2012 #4
    Ya I do but I don't see how to apply it to this. What exactly do you mean by differentiate?

    I set up the system of the system of linear differential equations and got that
    x = -7/6 dx/dt + 1/4 dy/dt
    y = 1/2 dx/dt -3/4 dy/dt

    it didn't seem to help
     
  6. Feb 8, 2012 #5
    rewrite your DE's in the vector form [itex]\frac{d}{dt}[x,y]^T=A\cdot[x,y]^T[/itex] and try a solution of the form [itex][x,y]^T=[x_0,y_0]^T\exp(\lambda t)[/itex], you'll end up having a simple eigenvalue problem to solve ...
     
  7. Feb 8, 2012 #6

    epenguin

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    I mean if any equation A = B is identically true for all t, then dA/dt = dB/dt is also true for all t.

    So what do you get when you differentiate your

    [itex]\frac{dx}{dt}=(-3x-y)[/itex]
    [itex]\frac{dy}{dt}=(-2x-2y)[/itex]

    by t?

    (I should mention that underneath this is not really different from sunjin09's suggestion; you may or may not be familiar now with the formalism, but solving a 1st order linear d.e. in n variables and solving an nth order in 1 variable are equivalent and can be translated into each other.)
     
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