Question

Write the given linear system in matrix form;

$\left\{\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=-3x+4y+e^{-t}\sin 2t\\ \frac{\mathrm{dy}}{\mathrm{dt}}=5x+9z+4e^{-t}\cos 2t\\ \frac{\mathrm{dz}}{\mathrm{dt}}=y+6z-e^{-t}d\end{array}\right.$

Short answer

Matrix form of the given linear system is$X\text{'}=\left(\begin{array}{ccc}-3& 4& 0\\ 5& 0& 9\\ 0& 1& 6\end{array}\right)X+\left(\begin{array}{c}\sin 2t\\ 4\cos t\\ -1\end{array}\right)e^{-t}$

Step-by-step solution

Matrix form of a Linear System

If X, A(t), and F(t) denote the respective matrices

$$\begin{gathered} X=\left(x_1\left(t\right) \\ {x}_2\left(t\right) \\ . \\ . \\ . \\ {x}_n\left(t\right)\right),A\left(t\right)=\left(\begin{array}{ccc}{a}_{11}\left(t\right)& ...& a_{1n}\left(t\right)\\ ⋮& ⋮& ⋮\\ a_{m1}\left(t\right)& ...& a_{mn}\left(t\right)\end{array}\right),F\left(t\right)=\left(f_1\left(t\right) \\ {f}_2\left(t\right) \\ . \\ . \\ . \\ {f}_n\left(t\right)\right) \end{gathered}$$

Then the system of linear first order differential equations (3) can be written as,

$$\begin{gathered} \frac{d}{dt}\left(x_1\left(t\right) \\ {x}_2\left(t\right) \\ . \\ . \\ . \\ {x}_n\left(t\right)\right)=\left(\begin{array}{ccc}{a}_{11}\left(t\right)& ...& a_{1n}\left(t\right)\\ ⋮& ⋮& ⋮\\ a_{m1}\left(t\right)& ...& a_{mn}\left(t\right)\end{array}\right)\left(x_1 \\ {x}_2 \\ . \\ . \\ . \\ {x}_n\right)+\left(f_1\left(t\right) \\ {f}_2\left(t\right) \\ . \\ . \\ . \\ {f}_n\left(t\right)\right) \end{gathered}$$

Or simply X’=AX+F

If the system is homogenous, its matrix form is then

X’=AX

Changing the linear system in matrix form:

In the following question we will change the linear form in its matrix form;

$$\begin{gathered} \left\{\frac{dx}{dt}=-3x+4y+e^{-t}\sin 2t \\ \frac{dy}{dt}=5x+9z+4e^{-t}\cos 2t \\ \frac{dz}{dt}=y+6z-e^{-t}\right. \\ X\text{'}=\left(\begin{array}{ccc}-3& 4& 0\\ 5& 0& 9\\ 0& 1& 6\end{array}\right)X+\left(\begin{array}{c}\sin 2t\\ 4\cos t\\ -1\end{array}\right)e^{-t} \\ A=\left(\begin{array}{ccc}-3& 4& 0\\ 5& 0& 9\\ 0& 1& 6\end{array}\right) \\ X\text{'}=\left(\begin{array}{ccc}-3& 4& 0\\ 5& 0& 9\\ 0& 1& 6\end{array}\right)X+\left(\begin{array}{c}\sin 2t\\ 4\cos t\\ -1\end{array}\right)e^{-t} \end{gathered}$$

Hence, the final answer is$X\text{'}=\left(\begin{array}{ccc}-3& 4& 0\\ 5& 0& 9\\ 0& 1& 6\end{array}\right)X+\left(\begin{array}{c}\sin 2t\\ 4\cos t\\ -1\end{array}\right)e^{-t}$