*another and hopefully last* linear algebra problem

1. Nov 25, 2008

bluewhistled

1. The problem statement, all variables and given/known data
Okay, I guess I just don't understand what this question is actually asking for. If you could possible explain better what they are looking for, what it means, and possible what direction to go in I'd really appreciate it.

a) Find the general solution to the system:
x1'=x1+2x2
x2'=2x1+x2
b) Find the specific solution that satisfies x1(0)=1; x2(0)=1

3. The attempt at a solution
A) Do I simply put it reduced row echelon form and then show that it comes out as an identity matrix?
B) I'm just confused on what they are asking, do I plug 0 in for x1 and 0 for x2 and then just say x1' and x2' are zero? That seems unlikely to be correct.

Any input would be very helpful, I know you guys do this in your spare time and it's very much appreciated.

2. Nov 25, 2008

Staff: Mentor

No, this is very much more complicated. You are supposed to find functions x_1(t) and x_2(t) that satisfy the differential equations represented by the matrix equation. The solution of a system like yours involves eigenvalues and eigenvectors, which are used to find a diagonal matrix that's similar to the original matrix, and which you can use to find the solution.

Hopefully, some of those terms ring a bell with you. If they don't, you've got a bit of work ahead of you.

3. Nov 25, 2008

bluewhistled

Nope I know how to get a diagonal matrix, eigenvalues/vectors. Problem is I don't know what to do with them :)

4. Nov 25, 2008

bluewhistled

Okay this is what I have so far:

A:

[1 2
2 1]

LI-A:

[L-1 2
2 L-1]

(L-1)^2-4=(L-3)(L+1)

L_1=3; L_2=-1

plug em back into the matrix:

[2 2 [-2 2
2 2] 2 -2]

find the vectors:

x_1=-t x_1=t
x_2=t x_2=t

<-1,1> <1,1>

Put them vectors together and they become the P?? matrix..

[-1 1
1 1]

D=P-1AP

Which I plug into my nifty calculator because I don't want to do it by hand and...

[-1 0
0 3]

Now I have absolutely no idea where to go from here. Anyone got any ideas?

5. Nov 25, 2008

VeeEight

So you made your coefficient matrix and found your diagonal matrix, D. You also have found P^-1 and P which you used to diagonalize the coefficient matrix.

Define a new function y(t) = P^-1x(t). This function is differentiable since x is differentiable, so we can rearrange it to give y'=P^-1 x' (since P^-1 is constant we can leave it as is and just differentiate x). Now this system will be hard to solve so we can rewrite it as y'=Dy(t) where y(t) is just column vector (ie y(t) = <y1, y2>). So multiply D by y to get two equations. Since you are multiplying by a diagonal matrix the differential equations you obtain should be easy to solve - you want to solve for the functions y1 and y2.

Once you have them you need to plug them back into the system to get x. Since y=P^-1 x', we have that x=Py. So multiply y by P to get the result.

6. Nov 25, 2008

bluewhistled

That went over my head. It's been two years since I've taken calculus. And this is the only problem that uses it. I've been searching for an example problem for like 2 hours now and I can't find squat.

7. Nov 25, 2008

Johnson04

You may use the eigen vectors you have obtained to construct a coordinate transformation so that you can simplify your original differential equations to the form:

y1' = a*y1
y2' = b*y2

this would be easy to solve. After you get y1(t) and y2(t), then use the coordinate transformation, you could get the solution in x variables.

8. Nov 25, 2008

Dick

If you have a diagonal matrix A=[[a1,0],[0,a2]] and you want to solve the equation [x1'(t),x2'(t)]=A*[x1(t),x2(t)], it's pretty easy, because if you equate components of the vectors you x1'(t)=a1*x1(t) and x2'(t)=a2*x2(t). You now have two independent ODE's for the two functions. If you know how to diagonalize the matrix A, then you can change the matrix ODE into this easy form. And then change back to solve the original problem.

9. Nov 25, 2008

VeeEight

The method that I posted only requires the use of calculus in one step, which isn't that hard. I posted all the steps so that you can see how everything works out. It may look a little tough but you already did the hard part by figuring out P such that P^-1AP = D, for D diagonal.

To simplify, let y be a column vector, y = <y1 y2> (but written out in column form). Write out y' = Dy (so you are multiplying your column vector <y1 y2> by the diagonal matrix D). You get two equations that look like y' = something and you are trying to solve for y1 and y2. These equations require some calculus to solve, but since D is diagonal, the equations you get are very simple. Once you solve for y1 and y2, solve for x by the equation x=Py.

Hope that helps!

10. Nov 25, 2008

bluewhistled

y' = [[-1,0],[0,3]] [y_1,y_2] = -y_1+3y_2

I really don't know what I'm doing. I apologize for asking this but could someone (if they have time) complete the problem? I honestly learn by example and this isn't HW, I'm studying example problems for a test I have tomorrow. I can't find examples anywhere on the internet. Oh and to Dick: this isn't a diagonal matrix. It's symmetrical, but not diagonal.

I'm kicking myself for missing the one day of notes that went over complicated material.

11. Nov 25, 2008

Dick

I meant to say, "if it were diagonal", it would be easy. Change to a basis where it's diagonal and then change back to the original basis. Test tomorrow is sort of short notice, but I think you can do this.

12. Nov 25, 2008

bluewhistled

Ah I figured it out. Thanks so much for all of your help. But how do I find the specific solution that satisfies X(0)=1?

13. Nov 25, 2008

Johnson04

Once you find the general solution, which is in the form: x(t) = constant * something or something + constant, by using x(0) = 1, you can determine the specific values of the constants. In other words, you can use the initial condition to obtain the specific solutions.

14. Nov 26, 2008

bluewhistled

Ah I see, Thanks a lot!

15. Nov 26, 2008

bluewhistled

I just want to say thanks to all of you for helping me with this, I really appreciate it. I'll try and come on this forum occasionally and answer some lower difficulty questions if I can. My problem is my retention when it comes to math, I aced calc 2 a couple of years ago but I barely even remember calc 1. Too many late nights I suppose.

Take care.