- #1

skrat

- 748

- 8

## Homework Statement

Solve the system

##\dot{x}=-2x+y+5sint## and ##\dot{y}=x-2y+3## for ##x(0)=1## and ##y(0)=1##

## Homework Equations

## The Attempt at a Solution

Well, I tried expressing x from the second equation, derived it and inserted it in the first equation but this process is long. Just to be clear, I will write a few steps..:

from second equation ##x=\dot{y}+2y-3## therefore ##\dot{x}=\ddot{y}+2\dot{y}##

Now inserting these two in the first expression gives me ##\ddot{y}+4\dot{y}+3y=6+5sint##

Homogeneous part is: ##y_h=Ae^{-t}+Be^{-3t}##

particular 1 (for the constant polynom 6): Constant C=2

particular 2 (for sin function): ##y_p=\frac{1}{2}sint-cost##

All together, the solution is than: ##y=Ae^{-t}+Be^{-3t}+\frac{1}{2}sint-cost+2##

and x from ##x=\dot{y}+2y-3= Ae^{-t}-Be^{-3t}+\frac{1}{2}cost+2sint-3##

because ##x(0)=1## and ##y(0)=1##, than ##A=-B=\frac{7}{4}##

HOWEVER, I do know that this can be done using matrices... It goes somehow like this...:

##\begin{bmatrix}

\dot{x}\\

\dot{y}

\end{bmatrix}=\begin{bmatrix}

-2 & 1\\

1& -2

\end{bmatrix}\begin{bmatrix}

x\\

y

\end{bmatrix}+\begin{bmatrix}

5sint\\

3

\end{bmatrix}##

Eigenvalues ##\lambda _1=-1## and ##\lambda _2=-3## give eigenvectors for matrix ##A=\begin{bmatrix}

-2 & 1\\

1& -2

\end{bmatrix}##:

##v_1=(-1,1)## and ##v_2=(1,1)##

SO, now my question:

What will the solution look like? If the system would me homogeneous it would be something like ##\vec{r}=Pe^{Dt}\vec{c}## where P is matrix from eigenvectors and D matrix from eigenvalues. C is vector of constants.

But my system also has 5sint and +3 ... What do I do now?

Thanks for all the help!