System of Linear Differential equations

[Xyius]

(I do not know how to make matricies with LateX, so where the semi-colon is denotes a new row!)
$$t \vec{x'}=[-4,2;2,-1]\vec{x}$$

The part that throws me off with this problem is the x' vector being multiplied by t! I know if the t wasn't there, all I would need to solve is det(A-(lamda)I)=0

 

[lanedance]

consider what you would try in the single variable case and how you could generalise it to the system

 

[lanedance]

also here's how to do a matrix
$$t \vec{x}'=\begin{pmatrix} -4 & 2 \\ 2& -1 \end{pmatrix}\vec{x}$$

 

[Xyius]

Thanks for showing me how to do a matrix :)
I still do not understand! I tried to plug in the assumption that the solution is..
$$\vec(x)=\vec(u)e^{\lambda t}$$
And I get..
$$(A-\lambda t I)=\vec(0)$$

Don't know where to go. :\

 

[HallsofIvy]

Thats' because the solution isn't an exponential. That's only for "constant coefficient" d.e. What you have here is a Cauchy-Euler equation but you can still reduce it just as you have constant coefficient equations.

This matrix can be diagonalized. (Every symmetric matrix can.) First find the eigenvalues $\lambda_1$ and $\lambda_2$. Find corresponding eigenvectors and use them as columns to form the (invertible) matrix P. Then, using A as this matrix, $P^{-1}AP= D$ where
$$D= \begin{pmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{pmatrix}$$

If you multiply both sides of the differential equation, tx'= Ax, by the matrix $P^{-1}$, you get [itex]P^{-1}tx'= tP^{-1}x'= P^{-1}Ax= P^{-1}A(PP^{-1})x[/tex]$= (P^{-1}AP)P^{-1}x= DP^{-1}x$<br /> <br /> Now Let $y= P^{-1}x$ and the differential equation becomes <br /> $$ty'= Dy$$[/itex][tex]which can be written as the two uncoupled equations <br /> $$ty_1'= \lambda_1y_1$$<br /> and<br /> $$ty_2'= \lambda_2y_2$$<br /> <br /> Those will be easy to solve and then $x= Py$.[/tex]

 

[Xyius]

Thank you very much. I will need to crack open the old linear Algebra book as a refresher on diagonalization. :)

I got this as a solution..
$$x_1=c_1-2c_2t^{-5}$$
$$x_2=2c_1+c_2t^{-5}$$

I plugged them back into the original equation and it is indeed correct. Thanks again :D