Question

Consider the$\mathbf{5}\mathbf{}\mathbf{\times }\mathbf{5}$matrix given in Problem 33. Solve the system X'=AXwithout the aid of matrix methods, but write the general solution using matrix notation. Use the general solution as a basis for a discussion of how the system can be solved using the matrix methods of this section. Carry out your ideas.

Short answer

The eigenvector and a corresponding solution vector is$X=c_1\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right)e^{2t}\right]+c_3\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)e^{2t}+c_4\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)e^{2t}+c_5\left[\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right)e^{2t}\right]$

Step-by-step solution

Determine the complex eigen values

The polynomial of degree n of the variable $\lambda$is the characteristic polynomial of the n x n matrix A.

$p\left(\lambda \right)=det(\lambda I-A)$

Its root is the eigenvalue of A.

Determine the system of equations

$x_2=c_2{e}^{2t}$We have

$A=\left(\begin{array}{ccccc}2& 1& 0& 0& 0\\ 0& 2& 0& 0& 0\\ 0& 0& 2& 0& 0\\ 0& 0& 0& 2& 1\\ 0& 0& 0& 0& 2\end{array}\right)$

Rewriting the system as a system of equations,

$\left\{\begin{array}{l}\frac{d{x}_1}{dt}=2x_1+x_2\\ \frac{d{x}_2}{dt}=2x_2\\ \frac{d{x}_3}{dt}=2x_3\\ \frac{d{x}_4}{dt}=2x_4+x_5\\ \frac{d{x}_5}{dt}=2x_5\\ \end{array}\right.$

Notice that the second, third and fifth equation are separable equations, so we will start with these. So,

$\frac{d{x}_2}{dt}=2x_2$

$\frac{1}{x_2}d{x}_2=2dt$

Integrating both sides,

$\int \frac{1}{x_2}d{x}_2=\int 2dt$

$\ln \left|x_2\right|=2t+c_2$

$e^{\ln \left|x_2\right|}=e^{2t+c_2}$

Solving the third equation,

$\frac{d{x}_3}{dt}=2x_3$

$\frac{1}{x_3}d{x}_3=2dt$

Integrating both sides,

$\int \frac{1}{x_3}d{x}_3=\int 2dt$

$\ln \left|x_3\right|=2t+c_3$

$e^{\ln \left|x_3\right|}=e^{2t+c_3}$

$x_3=c_3{e}^{2t}$

Solving the fifth equation,

$\frac{d{x}_5}{dt}=2x_5$

$\frac{1}{x_5}d{x}_5=2dt$

Integrating both sides,

$\int \frac{1}{x_5}d{x}_5=\int 2dt$

$\ln \left|x_5\right|=2t+c_5$

$e^{\ln \left|x_5\right|}=e^{2t+c_5}$

$x_5=c_5{e}^{2t}$

Now solving the first equations,

$\frac{d{x}_1}{dt}=2x_1+x_2$

we can substitute $x_2=c_2{e}^{2t}$into the equation,

$=2x_1+c_2{e}^{2t}$

$\frac{d{x}_1}{dt}-2x_1=c_2{e}^{2t}$..............(1)

Here we got ourselves a first order linear equation, using the integrating factor method, we get

$\mu \left(x\right)=e^{-\int 2dt}$

$=e^{-2t}$

multiplying both sides of (1) by $\mu \left(x\right)$

$\frac{d{x}_1}{dt}{e}^{-2t}-2e^{-2t}{x}_1=c_2$

$\int \frac{d}{dt}\left[e^{-2t}{x}_1\right]dt=\int c_{2}dt$

$e^{-2t}{x}_1=c_{2}t+c_1$

$x_1=c_{2}t{e}^{2t}+c_1{e}^{2t}$

Solving the fourth equation,

$\frac{d{x}_4}{dt}=2x_4+x_5$$\frac{d{x}_4}{dt}=2x_4+x_5$

We can substitute $x_5=c_5{e}^{2t}$into the equation,

$=2x_4+c_5{e}^{2t}$

$\frac{d{x}_4}{dt}-2x_4=c_5{e}^{2t}$

Here we got ourselves a first order linear equation, using the integrating factor method, we get

$\mu \left(x\right)=e^{-\int 2dt}$

=e^-2t

Multiplying both sides of (2) by $\mu \left(x\right)$,

$\frac{d{x}_4}{dt}{e}^{-2t}-2e^{-2t}{x}_4=c_5$

$\int \frac{d}{dt}\left[e^{-2t}{x}_4\right]dt=\int c_{5}dt$

$e^{-2t}{x}_4=c_{5}t+c_4$

$x_4=c_{5}t{e}^{2t}+c_4{e}^{2t}$

Finally, we have

$$\begin{gathered} x_1=c_{2}t{e}^{2t}+c_1{e}^{2t} \\ {x}_2=c_2{e}^{2t} \\ {x}_3=c_3{e}^{2t} \\ {x}_4=c_{5}t{e}^{2t}+c_4{e}^{2t} \end{gathered}$$

which can be written as

$X=c_1\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right)e^{2t}\right]+c_3\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)e^{2t}+c_4\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)e^{2t}+c_5\left[\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right)e^{2t}\right]$

Notice that there are three solution vectors of the form

$X=K{e}^{2t}$

where

$X=c_1{K}_1{e}^{2t}+c_2\left[K_{1}t{e}^{2t}+\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right)e^{2t}\right]+c_3{K}_2{e}^{2t}+c_4{K}_3{e}^{2t}+c_5\left[K_{3}t{e}^{2t}+\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right)e^{2t}\right]$

and two solution vectors of the form

$X=Kt{e}^{2t}+P{e}^{2t}$

Determine the eigenvector and a corresponding solution vector

Recall

$(A-\lambda I)K=0and(A-\lambda I)P=K$

Here $\lambda =2$, which implies,

$(A-2I)P=K_1\to \left(\begin{array}{ccccc}0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\\ p_4\\ p_5\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)$

Here we have p1=1, as for p_2,,p₃,p₄and p₅, we will choose arbitrary values, say $p_1=p_2=p_3=p_4=0$.

This gives an eigenvector and a corresponding solution vector

$P=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right),X=\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right)e^{2t}$

The eigenvector and a corresponding solution vector is

$X=c_1\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right)e^{2t}\right]+c_3\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)e^{2t}+c_4\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)e^{2t}+c_5\left[\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right)e^{2t}\right]$width="600">X=c110000e2t+c210000te2t+01000e2t+c300100e2t+c400010e2t+c500010te2t+00001e2t