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, 201213 yr by Tapeworm

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, 201312 yr by mathmari