# How to solve: AX+XB=C, X=? I tried many times but fail...

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This problem origins from Mathematics for Engineer and Scientists 2nd ed - Alan Jeffrey P670 Chapter15.6 System of first order equations

$\displaystyle \dot{y}+2\begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix} y = 5\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} e^t \\ e^{-t} \end{pmatrix}$

Try particular solution $\displaystyle y_p=X\begin{pmatrix} e^t \\ e^{-t} \end{pmatrix}$

$\displaystyle \dot{y_p}=X\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix}e^t \\ e^{-t}\end{pmatrix}$

substitute $\displaystyle \dot{y_p}$ and $\displaystyle y_p$ into the original diff eqt...

$\displaystyle X\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}+2\begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix}X= 5\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$

$\displaystyle X=\begin{pmatrix} 3 & 3 \\ -5 & -4 \end{pmatrix}$

Edited by Tapeworm
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• 5 months later...

Consider X=$\begin{pmatrix}a & b\\ c & d\end{pmatrix}$

Then do the multiplications :

$\begin{pmatrix}a & b\\ c & d\end{pmatrix}$$\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$=$\begin{pmatrix}a & -b\\ c & -d\end{pmatrix}$ (1)

$\begin{pmatrix}-6 & -4\\ 10 & 6\end{pmatrix}$$\begin{pmatrix}a & b\\ c & d\end{pmatrix}$=$\begin{pmatrix}-6a-4c & -6b-4d\\ 10a+6c & 10b+6d\end{pmatrix}$ (2)

You add (1) and (2), that equals to $\begin{pmatrix}5 & -5\\ -5 & 10\end{pmatrix}$, then you find the a,b,c,d

Edited by mathmari

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