Solving a System of Differential Equations

In summary: 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.The work shown is correct.
  • #1
KevinD6
10
0

Homework Statement


Y' =
[ -14 0 12
24 -2 24
16 0 14 ] Y
Subject to the condition [itex] y_1 (0) = -6, y_2 (0) = 11, y_3 (0) = 8, find: y_1(t), y_2(t), y_3(t) [/itex]

Homework Equations

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 :
[itex] v_1 = [-1, 0 ,1], v_2 = [0, 1, 0], v_3 = [-3, 6, 4] [/itex]
Therefore I have the general solution: [itex] Y = C_1 v_1 e^{-2t} + C_2 v_2 t e^{-2t} + C_3 v_3 e^{2t} [/itex]

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

However, the question wants me to find [itex] y_1 (t), y_2 (t), and y_3 (t) [/itex]
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.
 
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  • #2
KevinD6 said:

Homework Statement


Y' =
[ -14 0 12
24 -2 24
16 0 14 ] Y
Subject to the condition [itex] y_1 (0) = -6, y_2 (0) = 11, y_3 (0) = 8, find: y_1(t), y_2(t), y_3(t) [/itex]

Homework Equations

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 :
[itex] v_1 = [-1, 0 ,1], v_2 = [0, 1, 0], v_3 = [-3, 6, 4] [/itex]
Therefore I have the general solution: [itex] Y = C_1 v_1 e^{-2t} + C_2 v_2 t e^{-2t} + C_3 v_3 e^{2t} [/itex]

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

However, the question wants me to find [itex] y_1 (t), y_2 (t), and y_3 (t) [/itex]
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.
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:
  • #3
How are y1, y2 and y3 related to Y? (Mark beat me to it!)
 
  • #4
KevinD6 said:

Homework Statement


Y' =
[ -14 0 12
24 -2 24
16 0 14 ] Y
Subject to the condition [itex] y_1 (0) = -6, y_2 (0) = 11, y_3 (0) = 8, find: y_1(t), y_2(t), y_3(t) [/itex]

Homework Equations

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.

Are you sure that those are the eigenvalues?
 
  • #5
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
What is your characteristic equation?
It is also possible that you did not copy the matrix correctly. Check, please.
 
Last edited:
  • #7
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:
  • #8
KevinD6 said:
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.
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
Hmm, I just tried that but it seems that the variable 'y' cannot be used in my answer.
 
  • #10
y is a vector. I'm pretty sure you need to specify y1, y2, and y3 separately.
 
  • #11
Mark44 said:
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.

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
Mark44 said:
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.

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
ehild said:
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?)
Yes. I didn't notice the factor of t in the second term of the solution the OP showed in post 1.
 
  • #14
@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.
 

What is a system of differential equations?

A system of differential equations is a set of equations that involve derivatives and are used to model real-world phenomena. It typically consists of multiple equations that are interrelated, and the goal is to find a set of functions that satisfy all of the equations simultaneously.

What is the process for solving a system of differential equations?

The general process for solving a system of differential equations involves first identifying the type of system (linear, nonlinear, etc.), then using various techniques such as substitution, elimination, or matrix methods to reduce the system to a simpler form. Finally, the solution is found by integrating the resulting equations and applying initial or boundary conditions.

What are some common techniques for solving a system of differential equations?

Some common techniques for solving a system of differential equations include separation of variables, the method of undetermined coefficients, and the method of variation of parameters. Other techniques, such as Laplace transforms and power series, may also be used depending on the specific characteristics of the system.

What are the applications of solving a system of differential equations?

Solving a system of differential equations has a wide range of applications in various fields such as physics, engineering, economics, and biology. It can be used to model and predict the behavior of complex systems, analyze the stability of a system, and optimize processes in industries such as aerospace, chemical, and environmental engineering.

What are some challenges in solving a system of differential equations?

Some challenges in solving a system of differential equations include dealing with complex or nonlinear systems, finding appropriate initial or boundary conditions, and handling singularities or discontinuities in the equations. It can also be time-consuming and computationally intensive, especially for larger and more complex systems.

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