Solving simultaneous homogenous differential equation

[kleyton]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

I am now having trouble separating the variables. Help would be appreciated.

 

[micromass]

Hi Kleyton!

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

$$\left\{\begin{array}{c}<br /> \frac{dy}{dx}=y-z\\<br /> \frac{dz}{dx}=-4y+z<br /> \end{array}\right.$$

So, putting this in matrix form yields

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

Now, what do you know about solving these kinds of systems, that is differential equations of the form $y^\prime=Ay$ (with A a matrix and y is multi-valued function)?

 

[kleyton]

Thanks for pointing that out. Never looked at the problem that way.
I think it is a eigenvalue problem.
$$\begin{vmatrix}\begin{bmatrix} 1 & -1 \\ -4 & 1 \end{bmatrix}- \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\end{vmatrix} = 0$$
And I solve for lambda and get $$\lambda_{1} = 3$$and$$\lambda_{2} = -1$$
How would I proceed then ?

 

[micromass]

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!

 

[kleyton]

I am not sure what you meant by 'diagonizable'.
But, I tried solving for the eigenvectors as follows (wiki):
$$\begin{pmatrix} 1 & -1 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \lambda \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$$
This gave me,
$$\hat{v_{1}} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$$
and
$$\hat{v_{2}} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
Thus, I get the solution as,
$$y = Ae^{3x} + Be^{-x}$$
$$z = -2Ae^{3x} + 2Be^{-x}$$
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 ?

 

[micromass]

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.

 

[kleyton]

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" (thanks to you for pointing out it was matrix diff equation )

 

[micromass]

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.

 

[HallsofIvy]

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.)

 

[kleyton]

@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.