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Phrasing 3rd order ODE as a first order problem

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi,

    Wondering if anyone can give me some help with reducing this 3rd order ODE to a first order problem, so it can be written in the form u' = f(u, t)

    2. Relevant equations

    The 3rd order ODE is: x'''(t) + x''(t) + 2x'(t) + 2x(t) = 2t^2 + 4t - 5;
    The initial values x(0) = -3; x'(0) = 2; x''(0) = 2 are given

    3. The attempt at a solution

    I tried defining new variables u1 = x, u2 = x', u3 = x'' then u = (u1 u2 u3) and u' = (x' x'' x''') which gives the right order of x, but I'm unsure of how to phrase this in a way thats equivalent to the ODE with the right coefficients
     
  2. jcsd
  3. Jan 22, 2013 #2
    You need three equations for u's. The first one: u1' = u2. Can you see why? Can you find the other two?
     
  4. Jan 22, 2013 #3
    I can see that you can u1, u2 and u3 such that u1' = u2, u2' = u3, u1'' = u3 and u3' will give x''', but I'm not sure how you get it into the form in the question where there is a single u. I was thinking you needed to put the individual u's as a vector like u = (u1 u2 u3) and u' = (u1' u2' u3') = (u2 u3 u3'). Is this right?
     
  5. Jan 22, 2013 #4
    That's precisely correct. You have the vector on the left hand side, u', and you have the vector on the right side. The composition of the left-hand side vector is simply (u1', u2', u3'), and you already know two thirds of the right hand side vector: (u2, u3, ?).

    What is '?' Remember, it must be a function of u and t, but not of u'. This is where you should use the original equation.
     
  6. Jan 23, 2013 #5
    So do you substitute in the u's into the original question to find ?, so that x''' = 2t^2 + 4t -5 - u3 - u2 - u1 which gives a function of u and t??
     
  7. Jan 23, 2013 #6
    Exactly!
     
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