Phrasing 3rd order ODE as a first order problem

In summary, the conversation discusses reducing a 3rd order ODE to a first order problem and finding the equivalent form of the ODE with the right coefficients. The solution involves defining new variables and using them in a vector form to compose the left and right hand sides of the equation. The missing function in the vector is found by substituting the u's into the original equation.
  • #1
Sonique
3
0

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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  • #2
You need three equations for u's. The first one: u1' = u2. Can you see why? Can you find the other two?
 
  • #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?
 
  • #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.
 
  • #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??
 
  • #6
Exactly!
 

FAQ: Phrasing 3rd order ODE as a first order problem

What is a third order ODE?

A third order ODE (ordinary differential equation) is a type of mathematical equation that involves a function and its derivatives up to the third order. This means that the equation contains the function, its first, second, and third derivatives.

Why do we need to phrase a third order ODE as a first order problem?

Phrasing a third order ODE as a first order problem allows us to represent the equation in a system of first order differential equations. This makes it easier to solve and analyze the equation using numerical methods or other mathematical techniques.

How do we convert a third order ODE to a first order problem?

To convert a third order ODE to a first order problem, we need to introduce two new variables and rewrite the equation as a system of three first order differential equations. This involves substituting the derivatives of the original function with the new variables and adding additional equations to represent the derivatives of those variables.

Are there any advantages to using a first order problem over a third order ODE?

Yes, there are several advantages to using a first order problem over a third order ODE. First, it allows us to use numerical methods to solve the equation, which can be more accurate and efficient. Additionally, it allows us to easily analyze the stability and behavior of the solution using techniques such as phase portraits.

Can any third order ODE be converted to a first order problem?

In general, yes, any third order ODE can be converted to a first order problem. However, the process may become more complicated for certain types of equations, such as those with nonlinear terms. In these cases, additional techniques may be needed to properly phrase the equation as a first order problem.

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