# System of Differential Equations

1. Jun 24, 2008

### aeroguy2008

SOLVE THE SYSTEM OF DIFFERENTIAL EQUATIONS:

X'= -x+y
Y'= -x-y

WHEN x(0)=2 och y(0)=3.

3. The attempt at a solution

HELP...I don't know how to go about this. I have used the matrix
A= [-1 1
-1 -1]

and got a lambda1=-1+i
lambda2=-1-i and eigenvectors: [1
i] for lambda1
and [i
1] for lambda2

I need to get the solution in a general form. Please help!

2. Relevant equations

Last edited: Jun 24, 2008
2. Jun 24, 2008

Hey Aeroguy! Welcome

So if you have the Eigenvectors, what is the solution to the system? You should have two arbitrary constants right?

Now just plug in the given conditions and solve for C1 and C2!

3. Jun 24, 2008

### aeroguy2008

Thanks for your reply and for the welcome. I guess my problem is that I don't really know what the system should look like. I can't figure out which of the following four cases I'm dealing with.

Case 1)

[x [a [c
y] = b]exp*(t)+ d]exp*(-t) ?

Case 2)

[x [a [c
y] = b]cos(t)+ d]sin(t) ?

Case 3)

[x [a [c
y] = b]exp*(t)*cos(t)+ d] exp*(t)sin(t) ?

Case 4)

[x [a [c
y] = b]exp*(-t)*cos(t)+ d] exp*(-t)sin(t) ?

Further I know that I should use 2 and 3 as a Left Hand Side in Two Linear Equations since x(0)=2 och y(0)=3 is given. But, unfortunately I don't quite know where to use them:(

4. Jun 24, 2008

one minute AG, I am just looking in my text. BTW, I have a stupid question: what is OCH?

5. Jun 24, 2008

### aeroguy2008

OH... OCH means AND in Swedish. I mixed up the languages:P sorry about that!

6. Jun 24, 2008

Alrighty-then!

Here we go! I believe that in with the case of complex Eigenvalues of lambda, you will need only to establish one Eigenvector. It is due to the fact that complex Eigenvalues will always show up in conjugate pairs

Now lets take your first one and make some use of it:

$$K_1= \left[\begin{array}{c}1\\i\end{array}\right]$$

This column vector can be split up into its real and imaginary parts, that is:
$$\left[\begin{array}{c}1\\i\end{array}\right]=\left[\begin{array}{c}1\\0\end{array}\right]+\left[\begin{array}{c}0\\1\end{array}\right]i$$

Do you follow so far?

7. Jun 24, 2008

### aeroguy2008

Yes I do please go on...

8. Jun 24, 2008

Well, I don't know what kind of approaches your teacher has you take, but there is a pretty straight forward formula to these kinds of problems:

$$X=C_1[B_1\cos\beta t-B_2\sin\beta t]e^{\alpha t}$$

$$Y=C_2[B_2\cos\beta t+B_1\sin\beta t]e^{\alpha t]$$

Where B1 is the column vector that is the REAL part of your Eigenvector i.e $$\left[\begin{array}{c}1\\0\end{array}\right]$$

B2 is the column vector that is the IMAGINARY part i.e. $$\left[\begin{array}{c}0\\1\end{array}\right]$$

$\alpha$ is the REAL term in $\lambda=-1+i$ which is of course -1

and $$\beta$$ is the COEFFICIENT of the imaginary term in lambda which is +1

Now plug it all in and apply those initial conditions and solve for C1 and C2!

9. Jun 26, 2008

### aeroguy2008

Thanks for trying to help me. To be honest I dont not know how to get C1 and C2? Do I put t=0? If you can, could you please tell me what values C1 and C2 will get? Thanks...

10. Jun 27, 2008

From the initial conditions you know that when t=0--->x=2 AND when t=0---->y=3

Plug those values in. Two equations and two unknowns.

11. Jun 27, 2008

### aeroguy2008

Hello. I have tried your method with regards to my earlier attempts. But the problem I am facing is that I can't get C1 and C2 without setting c1=2 and c2 =3 according to the equations. Yet, I don't think my C1 and C2 are correct.

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12. Jun 27, 2008

What makes you think that these are incorrect? Why don't you take the first derivative of the solution and plug it back into the original system and see what happens

13. Jun 28, 2008

### aeroguy2008

I did that. Thanks. So now I have to chose between these options (found as an attachment):
I would say the correct answer to this problem is D with

a=2, b=0, c=3, d=0. But I know that is not correct. Yet, I can't figure out why. Can you pls tell me what I am doing wrong?

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14. Jul 3, 2008

### aeroguy2008

Ok I got it now:

a=2
b=0
c=3
d=-2

What has to be taken into account is to sum of X and Y to a general solution. This problem is thus solved.