DE system

**sinichko** · Jul 5th 2010, 11:08 PM

Please help me with this one:

$y'=\begin{pmatrix}<br /> 2 & 2&0 \\ <br /> 2& 2 &0 \\ <br /> 0& 0 & 4<br /> \end{pmatrix} y$

Thanks.

**HallsofIvy** · Jul 6th 2010, 02:14 AM

Find the eigenvalue of the matrix. (Hint: one of the eigenvalues is 4. What are the eigenvalues of $\begin{pmatrix}2 & 2 \\ 2 & 2 \end{pmatrix}$ ?)

Find the corresponding eigenvectors. Fortunately, for this problem there are three independent eigenvectors so the matrix is "diagonalizable" (every symmetric matrix is diagonalizable). That is, if P is the matrix having the eigenvectors as columns, $P^{-1}AP= D$ where A is the given matrix and P is the diagonal matrix having the eigenvalues on the main diagonal.

Multiply the entire equation by P to get Py'= (Py)'= PAy. Let $X=Py$ so that $y= P^{-1}X$ . Then the equation becomes $X'= PAP^{-1}Py= DX$ . X'= DX is easy to solve and then y= P^{-1}X.

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