# Solving a System of Differential Equations

1. Nov 8, 2014

### KevinD6

1. The problem statement, all variables and given/known data
Y' =
[ -14 0 12
24 -2 24
16 0 14 ] Y
Subject to the condition $y_1 (0) = -6, y_2 (0) = 11, y_3 (0) = 8, find: y_1(t), y_2(t), y_3(t)$
2. Relevant equations

3. The attempt at a solution
This is more just to see if my answers are correct than anything, so I begin by finding the eigenvalues of the matrix (checked with an internet calculator so I know this is correct).
Eigenvalues end up being: -2, -2, and 2.
I solve for the eigenvectors, which end up being :
$v_1 = [-1, 0 ,1], v_2 = [0, 1, 0], v_3 = [-3, 6, 4]$
Therefore I have the general solution: $Y = C_1 v_1 e^{-2t} + C_2 v_2 t e^{-2t} + C_3 v_3 e^{2t}$

Using the initial values given, putting it into the matrix and solving it, I end up with:
$C_1 = -3, C_2 = 13/2, C_3 = 4$

However, the question wants me to find $y_1 (t), y_2 (t), and y_3 (t)$
It also does not allow me to enter a vector as an answer, so does anyone know how my answer is supposed to be formatted? Or am I going about it wrong? Thank you for any assistance you may provide.

2. Nov 8, 2014

### Staff: Mentor

You have this:
$Y = C_1 v_1 e^{-2t} + C_2 v_2 t e^{-2t} + C_3 v_3 e^{2t}$
With the three vectors and the constants now included, it looks like this:
$$Y = -3 \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} e^{-2t} + 13/2 \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} e^{-2t} + 4 \begin{bmatrix} -3 \\ 6 \\ 4\end{bmatrix} e^{2t}$$

You can get your three solutions out of the above.

Edit: I assumed without checking that the work shown was correct.

Last edited: Nov 9, 2014
3. Nov 8, 2014

### Simon Bridge

How are y1, y2 and y3 related to Y? (Mark beat me to it!)

4. Nov 9, 2014

### ehild

Are you sure that those are the eigenvalues?

5. Nov 9, 2014

### KevinD6

Hmm, I don't know, I've been reading the internet and the textbook but I can't find any clues on what it means.

Also - I'm sure those are the eigenvalues, pretty sure at least, I did them by hand and then checked with wolfram.

6. Nov 9, 2014

### ehild

It is also possible that you did not copy the matrix correctly. Check, please.

Last edited: Nov 9, 2014
7. Nov 9, 2014

### KevinD6

Oh yes, I'm sorry, for some reason I missed a negative sign on the 12 in the first row.

I kind of get the relationship too, Y = [y1, y2, y3], where y2 = y1' and y3 = y2' if I'm not mistaken, I just don't see how this could be used to get a solution.

Last edited: Nov 9, 2014
8. Nov 9, 2014

### Staff: Mentor

There's a technique that can be used to convert a second- or higher-order differential equation into a system of linear differential equations, and possibly that's what happened here. If y1 = y, then from the above, y2 = y' and y3 = y'' and y3' = y'''. So an equation in y, y', y'', and y''' can be rewritten as a system of linear equations in y1, y2, and y3.

9. Nov 9, 2014

### KevinD6

Hmm, I just tried that but it seems that the variable 'y' cannot be used in my answer.

10. Nov 9, 2014

### Staff: Mentor

y is a vector. I'm pretty sure you need to specify y1, y2, and y3 separately.

11. Nov 9, 2014

### ehild

I tried to convert the system of equations into a third-order differential equation, but to my surprise, it fell apart to three independent second-order equations: y1"=4y1, y2"=4y2, y3"=4y3. :)

12. Nov 9, 2014

### ehild

I got something different. The general solution does not involve a term with t e-2t. (Mark, you omitted the factor t in the second term. Was it a typo?)

13. Nov 10, 2014

### Staff: Mentor

Yes. I didn't notice the factor of t in the second term of the solution the OP showed in post 1.

14. Nov 10, 2014

### ehild

@Kevin: y1, y2, y3 are the components of the vector Y, and the matrix equation can be expanded into a system of linear differential equation for y1, y2, y3:
y1'=14y1 - 12y3
y2'=24y1 -2y2 +24y3
y3'=16y1+ 14y3.

If you have repeated roots λ of the characteristic equation, take one solution as v1eλt, and the other solution in the form (v2+tv3)eλt.