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#### DrWahoo

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First, some may ask why would do we care that we can convert a 3rd order or higher ODE into a system of equations? Well there are quite a few reasons.

1- Almost all first order systems are easier to solve numerically using computer systems (matlab, maple, etc). Yes, it takes some working out by hand first, but the compiling time is much less. Even wolfram limits computation time, so use these notes to your advantage.

2- Nearly all moving (rates of change) problems in real life don't offer direct/exact solutions, especially when approximating natural systems of movement with respect to time or other variables.

3- Some higher order systems can be solved by using simple linear algebra techniques (once converted to a first order ode) by finding eigenvalues and eigenvectors by using $\bf{x'}$$=$ $A\bf{x}$

Examples;

a) Suppose we have the following 3rd order ODE of the form:

$y''' -y'-2y=-4y''$

This may look challenging, but try making it homogeneous and moving the right hand side to the left hand side using simple algebra techniques.

Thus adding $4y''$ to both both sides of the equations yields;

$y''' +4y''-y'-2y=0$

This looks much neater and its homogeneous!

So now we want to convert it into a First Order ODE by the following method;

Let $x_{1} = y $

Let $x_{2} = y'$

Let $x_{3} = y''$

So now we are ready to start making the higher order ODE in the opening statement to a first order system. It may seem daunting at this point, but it is super simply and really just relies on manipulating and denoting variables in a different manner. I will show this now,

Recall, we defined the following;

$x_{1} = y $

$x_{2} = y'$

$x_{3} = y''$

Hence, we can take the derivative of $x_{1}= y$ which is simply $x'_{1} = y'$. Here, all we did was differentiate both sides of what we defined above.

So now we have $x'_{1} = y'$ or equivalently, $ y'= x'_{1} $. If we do this for the other variables we defined we have,

$ y'= x'_{1} = x_{2}$

$ y''= x'_{2} =x_{3}$

$ y'''= x'_{3} $

So with the notation above, we substitute into our original higher order ODE.

Recall our higher order ode was;

$y''' +4y''-y'-2y=0$

So we need to replace $y''' , y'', y' $ with the definitions we gave them above with the denoted sub-scripted $x's$.

Replacing we have the ,

$x'_{3} + 4 x_{3} -x_{2} -2 x_{1}=0$

Or equivalently,

$x'_{3}= - 4 x_{3} +x_{2} +2 x_{1}$

Congratulations, you have converted the higher order ODE into a linear homogeneous ODE!

Can you solve it now?

See below for hints and solutions.

Because its a homogeneous first order system, YES we can solve the system using $\bf{x'}$$=$ $A\bf{x}$, we can do this because our linear homogeneous equation was converted to a linear homogeneous system of the matrix form above. A is defined below;

Note that boldface letters represent vectors in this case (usually in all cases dealing with DE's).

So in our matrix A, we simply have the coefficients of the corresponding system;$x'_{3}= - 4 x_{3} +x_{2} +2 x_{1}$

$A$= $\begin{pmatrix}

0 & 1 & 0 \\

0 & 0 & 1 \\

2 & 1 & -4

\end{pmatrix} $

Now we use the method for finding eigenvalues of the given 3 x 3 matrix by the following;

det$(A-\lambda I)$= characteristic polynomial, where I is the 3 x 3 identity matrix.

If you set the corresponding characteristic polynomial equal to $0$, you will have at most 3 possible eigenvalues. These are the same as the roots of the characteristic equation for the original 3rd order ODE.

Now try to finish the problem by finding the characteristic equation, the three eigenvalues and vectors, and a solution to the system. If you need help or have a solution, please post it up. Love seeing peoples logic and helping others.

Dr. Wahoo

1- Almost all first order systems are easier to solve numerically using computer systems (matlab, maple, etc). Yes, it takes some working out by hand first, but the compiling time is much less. Even wolfram limits computation time, so use these notes to your advantage.

2- Nearly all moving (rates of change) problems in real life don't offer direct/exact solutions, especially when approximating natural systems of movement with respect to time or other variables.

3- Some higher order systems can be solved by using simple linear algebra techniques (once converted to a first order ode) by finding eigenvalues and eigenvectors by using $\bf{x'}$$=$ $A\bf{x}$

Examples;

a) Suppose we have the following 3rd order ODE of the form:

$y''' -y'-2y=-4y''$

This may look challenging, but try making it homogeneous and moving the right hand side to the left hand side using simple algebra techniques.

Thus adding $4y''$ to both both sides of the equations yields;

$y''' +4y''-y'-2y=0$

This looks much neater and its homogeneous!

So now we want to convert it into a First Order ODE by the following method;

Let $x_{1} = y $

Let $x_{2} = y'$

Let $x_{3} = y''$

So now we are ready to start making the higher order ODE in the opening statement to a first order system. It may seem daunting at this point, but it is super simply and really just relies on manipulating and denoting variables in a different manner. I will show this now,

Recall, we defined the following;

$x_{1} = y $

$x_{2} = y'$

$x_{3} = y''$

Hence, we can take the derivative of $x_{1}= y$ which is simply $x'_{1} = y'$. Here, all we did was differentiate both sides of what we defined above.

So now we have $x'_{1} = y'$ or equivalently, $ y'= x'_{1} $. If we do this for the other variables we defined we have,

$ y'= x'_{1} = x_{2}$

$ y''= x'_{2} =x_{3}$

$ y'''= x'_{3} $

So with the notation above, we substitute into our original higher order ODE.

Recall our higher order ode was;

$y''' +4y''-y'-2y=0$

So we need to replace $y''' , y'', y' $ with the definitions we gave them above with the denoted sub-scripted $x's$.

Replacing we have the ,

$x'_{3} + 4 x_{3} -x_{2} -2 x_{1}=0$

Or equivalently,

$x'_{3}= - 4 x_{3} +x_{2} +2 x_{1}$

Congratulations, you have converted the higher order ODE into a linear homogeneous ODE!

Can you solve it now?

See below for hints and solutions.

Because its a homogeneous first order system, YES we can solve the system using $\bf{x'}$$=$ $A\bf{x}$, we can do this because our linear homogeneous equation was converted to a linear homogeneous system of the matrix form above. A is defined below;

Note that boldface letters represent vectors in this case (usually in all cases dealing with DE's).

So in our matrix A, we simply have the coefficients of the corresponding system;$x'_{3}= - 4 x_{3} +x_{2} +2 x_{1}$

$A$= $\begin{pmatrix}

0 & 1 & 0 \\

0 & 0 & 1 \\

2 & 1 & -4

\end{pmatrix} $

Now we use the method for finding eigenvalues of the given 3 x 3 matrix by the following;

det$(A-\lambda I)$= characteristic polynomial, where I is the 3 x 3 identity matrix.

If you set the corresponding characteristic polynomial equal to $0$, you will have at most 3 possible eigenvalues. These are the same as the roots of the characteristic equation for the original 3rd order ODE.

Now try to finish the problem by finding the characteristic equation, the three eigenvalues and vectors, and a solution to the system. If you need help or have a solution, please post it up. Love seeing peoples logic and helping others.

Dr. Wahoo

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