Dif.eq.system; complex eigenvalues

[prehisto]

Homework Statement

Given system:
dx/dt=-x-5y
dy/dt=x+y

Homework Equations

The Attempt at a Solution

So I calculated that \lambda_1=-2i and \lambda_2=2i
Generaly \lambda=+-qi

next i know that general solution is in form:
x=C₁cos(qt)+C₂sin(qt)
y=C*₁cos(qt)+C*₂sin(qt)

So to calculate constsnts C*_1 and C*_2 I can choose C_1 and C_2 how i want.
Then from systems second eq.(dy/dt=x+y) i calculate C*_1 and C*_2 .


My question is- which q value I put in solution,is it
x=C₁cos(2t)+C₂sin(-2t)
or
x=C₁cos(2t)+C₂sin(2t)
?

 

[vanhees71]

You have two different complex eigenvalues \lambda_1, \lambda_2 and thus also two linearly independent eigenvectors \vec{e}_1,\vec{e}_2. Then the general solution of the linear ODE system reads
\vec{r}(t)=[\vec{x}(t),\vec{y}(t)]^t=A_1 \vec{e}_1 \exp(\lambda_1 t) + A_2 \vec{e}_2 \exp(\lambda_2 t).
Since your eigenvalues are conjugate complex and the original matrix is real, you have \vec{e}_2=\vec{e}_1^*. In order to get the general real solutions you have to set A_2=A_1^*.

 

[D H]

Homework Statement

Given system:
dx/dt=-x-5y
dy/dt=x+y

Have you learned about the matrix exponential? If you have, this problem is nicely formulated to have a very simple matrix exponential solution.If you haven't, you'll have to use the techniques you have been taught.

My question is- which q value I put in solution,is it
x=C₁cos(2t)+C₂sin(-2t)
or
x=C₁cos(2t)+C₂sin(2t)
?

Does it make any difference? That's a rhetorical question. Since C₂ is an arbitrary constant and since sin(-x) = -sin(x), your two forms are equivalent.

 

[prehisto]

Have you learned about the matrix exponential? If you have, this problem is nicely formulated to have a very simple matrix exponential solution.


If you haven't, you'll have to use the techniques you have been taught.

Does it make any difference? That's a rhetorical question. Since C₂ is an arbitrary constant and since sin(-x) = -sin(x), your two forms are equivalent.

OK,I did not see sin(-x) = -sin(x). So it means that i can put either one of q values and it does not make a difference?

My goal for this system to draw ( i really don't know how its called in english meiby integral lines) integral lines, which should look like ellipse (so the form of elipse is dependent from q)

 

[ehild]

It is easy to solve the system of equations in the OP by transforming them into a second order equation:
d^2y/dt^2=dx/dt+dy/dt--> d^2y/dt^2+4y=0. The solutions are y1=sin(2t ) and y2=cos (2t ), and the corresponding x-s are obtained as x1(t)=y1-dy1/dt, x2(t)=y2-dy2/dt. The general solution is x=c1x1+c2x2, y=c1y1+c2y2, or it can be written as Y=Asin(2t+θ), X=2Acos(2t+ θ)-Asin(2t+θ)
You get the ellipses iin the x,y plane by eliminating t.

ehild