Phrasing 3rd order ODE as a first order problem

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Homework Help Overview

The discussion revolves around transforming a third-order ordinary differential equation (ODE) into a first-order system. The specific ODE presented is x'''(t) + x''(t) + 2x'(t) + 2x(t) = 2t^2 + 4t - 5, with given initial conditions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss defining new variables to represent the derivatives of x, specifically u1 = x, u2 = x', and u3 = x''. There is uncertainty about how to express the third-order ODE in terms of these new variables and how to formulate the system in the required first-order form.

Discussion Status

The conversation has progressed with participants providing insights on the relationships between the new variables. There is a collaborative effort to derive the necessary equations, with some guidance on how to express the original ODE in terms of the new variables. However, there is still exploration regarding the formulation of the right-hand side of the equation.

Contextual Notes

Participants are working under the constraints of transforming a specific third-order ODE while adhering to the requirement of expressing it as a first-order problem. Initial conditions are provided, but their role in the transformation process is not fully explored in the discussion.

Sonique
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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??
 
Exactly!
 

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