Question

Write the given linear system without the use of matrices.

$\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{\left(}\begin{array}{c}\mathbf{x}\\ \mathbf{y}\\ \mathbf{z}\end{array}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{2}\\ \mathbf{3}& \mathbf{-}\mathbf{4}& \mathbf{1}\\ \mathbf{-}\mathbf{2}& \mathbf{5}& \mathbf{6}\end{array}\mathbf{\right)}\mathbf{\left(}\begin{array}{c}\mathbf{x}\\ \mathbf{y}\\ \mathbf{z}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{c}\mathbf{1}\\ \mathbf{2}\\ \mathbf{2}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathbf{-}\mathbf{\left(}\begin{array}{c}\mathbf{3}\\ \mathbf{-}\mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right)}\mathit{t}$

Short answer

Matrix form of the given linear system is$\left\{\begin{array}{l}\frac{dx}{dt}=x-y+2z+e^{-t}-3t\\ \frac{dy}{dt}=3x-4y+z+2e^{-t}+t\\ \frac{dz}{dt}=-2x+5y+6z-2e^{-t}-t\end{array}\right.$

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;

$\frac{\mathrm{d}}{\mathrm{dt}}\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)=\left(\begin{array}{ccc}1& -1& 2\\ 3& -4& 1\\ -2& 5& 6\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)+\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)e^{-\mathrm{t}}-\left(\begin{array}{c}3\\ -1\\ 1\end{array}\right)t$

The given system of differential equation, without the use of matics, can be represented as,

$\{\begin{array}{l}\frac{dx}{dt}=x-y+2z+e^{-t}-3t\\ \frac{dy}{dt}=3x-4y+z+2e^{-t}+t\\ \frac{dz}{dt}=-2x+5y+6z-2e^{-t}-t\end{array}$

Hence, the final answer is$\{\begin{array}{l}\frac{dx}{dt}=x-y+2z+e^{-t}-3t\\ \frac{dy}{dt}=3x-4y+z+2e^{-t}+t\\ \frac{dz}{dt}=-2x+5y+6z-2e^{-t}-t\end{array}$