Question

Use theorem 9.3.13 to solve the initial value problem

$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left(\begin{array}{ccc}2& 3& \mathbf{1}\\ 0& 1& 2\\ \mathbf{0}& 0& 1\end{array}\right)\overrightarrow{\mathbf{x}}$with $\overrightarrow{\mathbf{x}}\left(0\right)\mathbf{=}\left[\begin{array}{c}2\\ 1\\ -1\end{array}\right]$.

Hint: Find first ${\mathit{x}}_{\mathbf{3}}\left(t\right)$, then ${\mathit{x}}_{\mathbf{2}}\left(t\right)$, and finally ${\mathit{x}}_{\mathbf{1}}\left(t\right)$.

Short answer

The solution is$e^n\left[\begin{array}{c}2-3t+t^2\\ -1+2t\\ t\end{array}\right]$

Step-by-step solution

Definition of first order linear differential equation.

Consider the differential equation $\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{af}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$where$\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$is a smooth function and 'a'is a constant. Then the general solution will be$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{a}\mathbf{t}}\mathbf{\int }{\mathit{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathit{d}\mathit{t}$.

Determination of the solution.

Consider the differential equation as follows.

$$\begin{gathered} \frac{d\overrightarrow{x}}{dt}=\left(\begin{array}{ccc}2& 3& \mathbf{1}\\ 0& 1& 2\\ \mathbf{0}& 0& 1\end{array}\right)\overrightarrow{x} \\ A=\left(\begin{array}{ccc}2& 3& \mathbf{1}\\ 0& 1& 2\\ \mathbf{0}& 0& 1\end{array}\right) \\ A=2 \\ f\text{'}\left(t\right)-2f\left(t\right)=0 \end{gathered}$$

Now, the differential equation is in the form as follows.

f'(t)-af(t)=g(t), where g(t) is a smooth function, then the general solution will be as follows.

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$

Compute the calculation of the solution.

Substitute the value 0 for g(t) and 2 for a in$f\left(t\right)=e^{at}\int e^{-at}g\left(t\right)dt$as follows.

role="math" localid="1660805430798" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int {\mathrm{e}}^{-2\mathrm{t}}\times 0\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int 0\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}·\mathrm{C} \end{gathered}$$

Here C is constant

Also it can be written as follows.

$C\text{'}=\left[\begin{array}{c}2-3t+t^2\\ -1+2t\\ t\end{array}\right]$

The solution is$e^{2t}\left[\begin{array}{c}2-3k+t^2\\ -1+2t\\ t\end{array}\right]$