Question

Write the given linear system without the use of matrices.

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{7}& \mathbf{5}& \mathbf{-}\mathbf{9}\\ \mathbf{4}& \mathbf{1}& \mathbf{1}\\ \mathbf{0}& \mathbf{-}\mathbf{2}& \mathbf{3}\end{array}\mathbf{\right)}\mathit{X}\mathbf{+}\mathbf{\left(}\begin{array}{c}\mathbf{0}\\ \mathbf{2}\\ \mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{\left(}\begin{array}{c}\mathbf{8}\\ \mathbf{0}\\ \mathbf{3}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}$

Short answer

Matrix form of the given linear system is$\left\{\begin{array}{l}\frac{dx}{dt}=7x+5y-9z-8e^{-2t}\\ \frac{dy}{dt}=4x+y+z+2e^{5t}\\ \frac{dz}{dt}=-2y+3z+e^{5t}-3e^{-2t}\end{array}\right.$

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,

$$\begin{gathered} \frac{d}{dt}\left(x_1\left(t\right) \\ {x}_2\left(t\right) \\ . \\ . \\ . \\ {x}_n\left(t\right)\right)=\left(\begin{array}{ccc}{a}_{11}\left(t\right)& ...& a_{1n}\left(t\right)\\ ⋮& ⋮& ⋮\\ a_{m1}\left(t\right)& ...& a_{mn}\left(t\right)\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ .\\ .\\ .\\ x_n\end{array}\right)+\left(f_1\left(t\right) \\ {f}_2\left(t\right) \\ . \\ . \\ . \\ {f}_n\left(t\right)\right) \end{gathered}$$

Or simply X’=AX+F

If the system is homogenous, its matrix form is then

X’=AX

Changing the linear system:

In the following question we will change the linear;

$X\text{'}=\left(\begin{array}{ccc}7& 5& -9\\ 4& 1& 1\\ 0& -2& 3\end{array}\right)X+\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^{5\mathrm{t}}-\left(\begin{array}{c}8\\ 0\\ 3\end{array}\right)e^{-2\mathrm{t}}$

This can be written as a system of two differential equations,

$\{\begin{array}{l}\frac{dx}{dt}=7x+5y-9z-8e^{-2t}\\ \frac{dy}{dt}=4x+y+z+2e^{5t}\\ \frac{dz}{dt}=-2y+3z+e^{5t}-3e^{-2t}\end{array}$

Hence, the final answer is$\{\begin{array}{l}\frac{dx}{dt}=7x+5y-9z-8e^{-2t}\\ \frac{dy}{dt}=4x+y+z+2e^{5t}\\ \frac{dz}{dt}=-2y+3z+e^{5t}-3e^{-2t}\end{array}$