Tapeworm Posted November 8, 2012 Share Posted November 8, 2012 (edited) This problem origins from Mathematics for Engineer and Scientists 2nd ed - Alan Jeffrey P670 Chapter15.6 System of first order equations [latex]\displaystyle \dot{y}+2\begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix} y = 5\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} e^t \\ e^{-t} \end{pmatrix} [/latex] Try particular solution [latex]\displaystyle y_p=X\begin{pmatrix} e^t \\ e^{-t} \end{pmatrix}[/latex] [latex]\displaystyle \dot{y_p}=X\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix}e^t \\ e^{-t}\end{pmatrix}[/latex] substitute [latex]\displaystyle \dot{y_p}[/latex] and [latex]\displaystyle y_p[/latex] into the original diff eqt... [latex]\displaystyle X\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}+2\begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix}X= 5\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}[/latex] [latex]\displaystyle X=\begin{pmatrix} 3 & 3 \\ -5 & -4 \end{pmatrix}[/latex] Edited November 8, 2012 by Tapeworm Link to comment Share on other sites More sharing options...
mathmari Posted May 5, 2013 Share Posted May 5, 2013 (edited) Consider X=[latex]\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/latex] Then do the multiplications : [latex]\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/latex][latex]\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}[/latex]=[latex]\begin{pmatrix}a & -b\\ c & -d\end{pmatrix}[/latex] (1) [latex]\begin{pmatrix}-6 & -4\\ 10 & 6\end{pmatrix}[/latex][latex]\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/latex]=[latex]\begin{pmatrix}-6a-4c & -6b-4d\\ 10a+6c & 10b+6d\end{pmatrix}[/latex] (2) You add (1) and (2), that equals to [latex]\begin{pmatrix}5 & -5\\ -5 & 10\end{pmatrix}[/latex], then you find the a,b,c,d Edited May 5, 2013 by mathmari Link to comment Share on other sites More sharing options...
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