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Solving simultaneous homogenous differential equation

  • Thread starter kleyton
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
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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
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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
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3,277
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
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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
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3,277
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
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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
22,097
3,277
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
HallsofIvy
Science Advisor
Homework Helper
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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
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@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.
 

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