# System of 2 2nd order DE's

1. Dec 7, 2004

### Spectre5

I have this DE and I do not know how to solve it. This is a problem in my review questions that my professor handed out, but we have never done a problem like this in class, so I do not know where to start.

here is the problem:
x'' = 4x - 5y
y'' = 2x - 3

I need to solve it using eigenvalues and eigenvectors. I know how to do this for a system of first order DE's, that is easy. But I don't know what to do for a system of second order DE's.

The one method I did come up with for solving it was to convert it to a system of 4 first order DE's...but then the characteristic equation is a fourth degree polynomial. I know there has got to be a better way, but I just dont know it.

2. Dec 7, 2004

### Tide

How about using a basis set that is the product of functions like

$$e^{ik_x x}$$

and

$$e^{i k_y y$$

?

3. Dec 8, 2004

### Spectre5

I need to use eigenvalues/eigenvectors though. To be honest, I am not sure how to appraoch it with your suggestion...remember, I have never looked at equations like this, and my teacher is not very helpful either

4. Dec 8, 2004

### James R

Are you comfortable with solutions of first order systems using eigenvectors?

5. Dec 8, 2004

### HallsofIvy

Staff Emeritus
If you let u= x' and v= y' then you have a system of 4 first order equations:

x'= u
y'= v
u'= 4x- 5y
v'= 2x- 3

Yes, the characteristic equation is 4th degree but it is trivial.
Did you mean y"= 2x- 3y???

6. Dec 8, 2004

### dextercioby

From the loooks of it,Tide's idea contains eigenvalues +eigenvectors.Those 2 functions are two different vectors in the linear space of solutions of a linear differential eq with constant coeficients.
They for a base in that space,so u need to find the coeficients of the expansion of the general solution in terms of eigenvectors and the eigenvalues "kx" and "ky".

Good luck!!

As Hallsofivy pointed out,u may be missing a term.

7. Dec 8, 2004

### Spectre5

yes, there should be a y on that 3
Yes. Very comfortable with them,

I did think of doing the way that HallsofIvy pointed out. That is the way I guess I know best. I just thought there was an easy/better way. This way seems to be the way that tide and dextercioby are pointing out. maybe I will work on figuring out that way a little bit.

only a few hours til the exam now though...and I don't know if that kind of question will even be on it. however, I know I will do fine on everything else becuase the rest of the it is really easy. Thanks for the suggestions.

8. Dec 8, 2004

### Eye_in_the_Sky

a solution

Spectre5, I hope your exam went well!
---------------------
---------------------

Here is one way to solve the problem using eigenvectors and eigenvalues.

Write the functions x(t) and y(t) together as a 1x2 column-vector x(t). You then have a vector-DE of the form

x"(t) = Mx(t) + a ,

where M is a 2x2 matrix.

If, as you point out, "there should be a y on that 3", then the constant vector a vanishes in the above, and the vector-DE is homogeneous.

Now, suppose M has two linearly independent eigenvectors u and v with corresponding eigenvalues α and β. So, write

x(t) = f(t)u + g(t)v ,

and compute the numbers c and d such that

a = cu + dv

(of course, for the homogeneous case, c = d = 0).

Then, the functions f(t) and g(t) satisfy the DE's:

f"(t) = αf(t) + c ,

g"(t) = βg(t) + d .

Solving these solves the problem.

9. Dec 9, 2004

### James R

For the second-order system, you can use a trial solution of the form:

$$\vec{x} = e^{kt}\vec{v}$$.

Differentiating twice gives:

$$\vec{x}'' = k^2 e^{kt}\vec{v}$$

The initial homogeneous d.e. can be written:

$$\vec{x}'' = \vec{A}\vec{x}$$

where $$\vec{A}$$ is a 2x2 coefficient matrix. Substituing the trial solution we get:

$$k^2 \vec{v} = \vec{A}\vec{v}$$

or

$$(\vec{A} - k^2\vec{I})\vec{v} = 0$$

This is an eigenvector equation. To solve it, find the eigenvalues, $$\lambda_i$$ of the matrix A:

$$k = \pm sqrt(\lambda_i)$$.

Use the eigenvalues to find the eigenvectors [tex]\vec{v}[\tex]

For the system of 2 variables, x and y, there will be 4 values of k, which give 4 linearly-independent solutions, two for x and two for y. The positive and negative k values each have the same eigenvector associated with them.

The general solution of the system is a linear combination of the 4 solutions found.

10. Dec 9, 2004

### rayveldkamp

You can also use the Jordan Canonical form approach to solve systems of DE's