Solving simultaneous homogenous differential equation

In summary, the homework statement is to solve the simultaneous differential equation. The attempt at a solution is to (from eq 2), find dy/dx= y- z and substitute eq 3 in eq 1, yielding dz/dx= z- 4y . But dy/(y- z) = dx and dz= \frac{(z - 4y)}{(y - z)} dy . Finally, finding the eigenvalues and checking whether the matrix is diagonalizable is a good first step.
  • #1
kleyton
7
0

Homework Statement


Solv the simultaneous homogenous differential equation
[tex]\begin{cases} dy/dx + dz/dx + 3y = 0 \\ dy/dx - y + z = 0 \end{cases}[/tex]

Homework Equations





The Attempt at a Solution


(from eq 2),
[tex]dy/dx = y - z [/tex] --- eq (3)
(substituting eq 3 in eq 1),
[tex]\therefore dz/dx = z - 4y [/tex]
[tex]\therefore dz = (z - 4y)dx[/tex]
[tex]but, dy/(y - z) = dx[/tex]
[tex]\therefore dz = \frac{(z - 4y)}{(y - z)} dy[/tex]
[tex]\therefore dz = \frac{(z - 4y)}{(y - z)} dy[/tex]

I am now having trouble separating the variables. Help would be appreciated.
 
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  • #2
Hi Kleyton! :smile:

First, you'll need to put this in matrix form, so you've deduced that you have the system

[tex]\left\{\begin{array}{c}
\frac{dy}{dx}=y-z\\
\frac{dz}{dx}=-4y+z
\end{array}\right.[/tex]

So, putting this in matrix form yields

[tex]\left(\begin{array}{c}\frac{dy}{dx}\\ \frac{dz}{dx}\\ \end{array}\right) = \left(\begin{array}{cc} 1 & -1\\ -4 & 1\\ \end{array}\right)\left(\begin{array}{c} y\\ z\\ \end{array}\right)
[/tex]

Now, what do you know about solving these kinds of systems, that is differential equations of the form [itex]y^\prime=Ay[/itex] (with A a matrix and y is multi-valued function)?
 
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  • #3
Thanks for pointing that out. Never looked at the problem that way.
I think it is a eigenvalue problem.
[tex]\begin{vmatrix}\begin{bmatrix} 1 & -1 \\ -4 & 1 \end{bmatrix}- \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\end{vmatrix} = 0[/tex]
And I solve for lambda and get [tex]\lambda_{1} = 3 [/tex]and[tex]\lambda_{2} = -1[/tex]
How would I proceed then ?
 
  • #4
So, I take it you want to take the matrix exponential of the matrix?? Finding the eigenvalues and checking whether the matrix is diagonalizable is a good first step. so check that first!
 
  • #5
I am not sure what you meant by 'diagonizable'.
But, I tried solving for the eigenvectors as follows (wiki):
[tex]\begin{pmatrix} 1 & -1 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \lambda \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/tex]
This gave me,
[tex]\hat{v_{1}} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}[/tex]
and
[tex]\hat{v_{2}} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}[/tex]
Thus, I get the solution as,
[tex]y = Ae^{3x} + Be^{-x}[/tex]
[tex]z = -2Ae^{3x} + 2Be^{-x}[/tex]
Since there is no initial condition given, how would I solve for A and B ?
edit: Actually I am not sure how far I am expected to solve this. The question just said "Solve the simultaneous differential equation". Also the next question replaces the RHS with sinx and cosx. Would that change the way I solve the problem ?
 
  • #6
Uuh, ok, that is correct. I'm kind of surprised on how you did that without diagonalizing...

But anyway, since there is no initial condition given, this means that there will not be a unique solution. Thus our solution will always contain parameters A and B. Your solution is the best you can do.
 
  • #7
Thanks a ton micromass !
Your pointers were very helpful. Just one last question. If the equations were not homogeneous, which would be the simplest way to solve it, undetermined coefficients or variation of parameters ?

PS: I tried to follow the example on matrix differential equation on http://en.wikipedia.org/wiki/Matrix_differential_equation" [Broken] (thanks to you for pointing out it was matrix diff equation o:))
 
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  • #8
kleyton said:
Thanks a ton micromass !
Your pointers were very helpful. Just one last question. If the equations were not homogeneous, which would be the simplest way to solve it, undetermined coefficients or variation of parameters ?

Both could work out, probably. But I think variation of parameters is easiest, and I see that the wiki page actually gives an example of this.
 
  • #9
If you are not comfortable with matrix methods (although they are a wonderful way of treating problems like this) you can do this problem this less "sophisticated" way:
dy/dx+ dz/dx+ 3y= 0, dy/dx- y+ z= 0.

The second equation is the same as dy/dx= y- z. You can replace dy/dx in the first equation by y- z to get y- z+ dz/dx+ 3y= 0 or dz/dx= z- 2y.

That is, your two equations are dy/dx= y- z and dz/dx= -2y+ z as micromass said. Now, differentiate the second equation again: d^2z/dx^2= -2 dy/dx+ dz/dx. We can substitute dy/dx= y- z into that to get d^2z/dx^2= -2(y- z) +dz/dx= -2y+ 2z+ dz/dx.

From dz/dx= -2y+ z, -2y= dz/dx- z so that becomes d^2z/dx^2= dz/dx- z+ 2z+ dz/dx= 2 dz/dx+ z or d^2z/dx^2- dz/dx- z= 0.

Solve that equation for z(x) and then use -2y= dz/dx- z to find y.

(Since the equation for z is a second order equation you will get two constants. You don't want to solve another differrential equation for y since that would introduce another constant- and we only want two.)
 
  • #10
@HallsofIvy
I cannot believe I did not see that. I learned that in differential equations class and even used it often in adv engg math. Thanks a ton. This just makes solving the problem a whole lot easier (might I say, cleaner too). Kudos to you ! I will post the completed solution once I finish making a latex doc. Thanks for the help micromass and HallsofIvy.
 

What is a simultaneous homogeneous differential equation?

A simultaneous homogeneous differential equation is a set of two or more differential equations that involve the same variables and can be solved together to find a general solution.

How do you solve a simultaneous homogeneous differential equation?

To solve a simultaneous homogeneous differential equation, you can use various methods such as substitution, elimination, or using matrices. The specific method used will depend on the specific equations and variables involved.

What is the importance of solving simultaneous homogeneous differential equations?

Solving simultaneous homogeneous differential equations is important in many areas of science, particularly in physics and engineering. It allows us to model and predict the behavior of systems with multiple variables and helps us understand how different factors interact with each other.

Can simultaneous homogeneous differential equations have multiple solutions?

Yes, simultaneous homogeneous differential equations can have multiple solutions. This is because there are often multiple ways to combine the equations and find a general solution, and these solutions can differ depending on the initial values given.

What are some real-world applications of simultaneous homogeneous differential equations?

Simultaneous homogeneous differential equations have many real-world applications, such as predicting the motion of a pendulum, modeling population growth, and analyzing electrical circuits. They are also used in fields like economics, biology, and chemistry to model complex systems and phenomena.

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