Phrasing 3rd order ODE as a first order problem

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Homework Statement



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)

Homework 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

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 that's equivalent to the ODE with the right coefficients
 
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You need three equations for u's. The first one: u1' = u2. Can you see why? Can you find the other two?
 
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?
 
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.
 
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??