# Thread: using matrix-differential eqns-help exam quite soon

1. ## using matrix-differential eqns-help exam quite soon

Hello,
find the particular solution to the system of differential eqns.
dx/dt=x+3z
dy/dt=y-5z
dz/dt=2z
given the initial condition x(o)=y(0)=z(0)=1
once I chuck the coefficients of the differential eqns into the matrix im not sure what to do?

2. ## Re: using matrix-differential eqns-help exam quite soon

I presume you have written that as a matrix differential equation as you title this "using matrix differential equations":
$\displaystyle \frac{d\begin{bmatrix}x \\ y \\ z \end{bmatrix}}{dt}=$$\displaystyle \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -5\\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$

Surely you have some instruction in this?

Start by finding the eigenvalues and eigenvectors of the matrix. Then $\displaystyle B^{-1}AB= D$ where A is the matrix, B is a matrix having the eigenvectors of A as its columns, and D is a diagonal matrix having the eigenvalues of A on its diagonal. That is dX/dt= AX, which, after multiplying by $\displaystyle B^{-1}$ gives $\displaystyle d(A^{-1}X)/dt= B^{-1}AX= B^{-1}ABB^{-1}X$. Let $\displaystyle Y= A^{-1}X$ and the equation becomes dY/dt= DY where D is that diagonal matrix. Because D is diagonal, that is easy to solve and, once you have Y, X= AY.

HOWEVER, in this particular problem, there is no point in doing that or even writing this in terms of a matrix. The third equation is dz/dt= 2z which is easily solvable: $\displaystyle z= Ce^t$ and z(0)= C= 1. Then dy/dt= y- 5z= y- 5 which has general solution $\displaystyle y= Ce^t+ 5$ and y(0)= C+ 5= 1 so C= -4. Finally, dx/dt= x+ 3z= x+ 3 has general solution $\displaystyle z= Ce^t- 3$ and z(0)= C- 3= 1 so C= 4.

(Okay, not all matrices are diagonalizable, and this particular one is not.)