Multiple Roots and Complex Roots

Exercise 3.2

1. Solve the following differential equations.

(a) $\ \left\{\begin{array}{rc} x_{1}^{\prime} =& 6x_{1} + 8x_{2}\\ x_{2}^{\prime} =& -x_{1} + 2x_{2} \end{array} \right .$
(b) $\ \left\{\begin{array}{rc} x_{1}^{\prime} =& 2x_{1} - x_{2}\\ x_{2}^{\prime} =& 4x_{1} + 6x_{2} \end{array} \right .$
(c) $\ \left\{\begin{array}{rc} x_{1}^{\prime} =& 5x_{1} + 2x_{2} + 2x_{3}\\ x_{2}^... ...{2} - 4x_{3}\\ x_{3}^{\prime} =& 2x_{1} - 4x_{2} + 2x_{3} \end{array} \right .$
(d) $\ {\bf X}^{\prime} = \left(\begin{array}{ccc} 1&0&1\\ 1&1&0\\ -2&0&-1 \end{array}\right){\bf X}$
(e) $\ \left\{\begin{array}{rc} 4x_{1}^{\prime} + x_{1} + 2x_{2}^{\prime} + 7x_{2} =& 0\\ x_{1}^{\prime} - x_{1} + x_{2}^{\prime} + x_{2} =& 0 \end{array} \right .$
(f) $\ \left\{\begin{array}{rc} x_{1}^{\prime} + x_{1} + 2x_{2}^{\prime} + 3x_{2} =& 0\\ x_{1}^{\prime} - 2x_{1} + 5x_{2}^{\prime} =& 0 \end{array} \right .$

Answer
1. (a) $\det(A - \lambda I) = \left(\begin{array}{cc} 6-\lambda&8\\ -1&2-\lambda \end{array}\right) = \lambda^2 - 8\lambda + 20 = 0$. Then the eigenvalues are $\lambda = 4 \pm 2i$. Now we find the eigenvector ${\bf C}$ corresponds to $\lambda = 4 + 2i$.

$$(A - (4 + 2i)I){\bf C} = \left(\begin{array}{cc} 2 - 2i & 8\\ -1... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Note that the matrix $\left(\begin{array}{cc} 2 - 2i & 8\\ -1 & -2 - 2i \end{array}\right) \Longrightarrow \left(\begin{array}{cc} 1&2 + 2i\\ 0&0 \end{array}\right)$ is row reduced echelon form and no leading one in the 2nd row. So, we let $c_{2} = \alpha$. Then $c_{1} = -2(1 + i)\alpha$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} -2(1 + i) \\ 1 \end{array}\right)$. Now we find the real part and the imaginary part of ${\bf C}e^{\lambda t}$.

$${\bf C}e^{\lambda t} = \left(\begin{array}{c} -2(1 + i) \\ 1 \end{array}\right) e^{(4 +... ...\begin{array}{c} -2(1 + i) \\ 1 \end{array}\right)e^{4t}(\cos{2t} + i\sin{2t})$$

$$= \left(\begin{array}{c} -2(1+i)e^{4t}(\cos{2t} + i\sin{2t}) \\ e^{4t}(\cos{2t} + i\sin{2t}) \end{array}\right)$$

$$= \left(\begin{array}{c} -2e^{4t}\cos{2t} + 2e^{4t}\sin{2t}) \\ e^... ...y}{c} -2e^{4t}\cos{2t} - 2e^{4t}\sin{2t}) \\ e^{4t}\sin{2t} \end{array}\right)$$

Therefore, we can express the general solution as follows:

$${\bf X} = c_{1}\left(\begin{array}{c} -2\cos{2t} + 2\sin{2t}\\ \... ...egin{array}{c} -2\cos{2t} - 2\sin{2t}\\ \sin{2t} \end{array}\right)e^{4t} \ \$$

(b) $\det(A - \lambda I) = \left(\begin{array}{cc} 2-\lambda&-1\\ 4&6-\lambda \end{array}\right) = \lambda^2 - 8\lambda + 16 = (\lambda - 4)^{2}$ Then the eigenvalues are $\lambda = 4$. Now we find the eigenvector ${\bf C}$ corrsponds to $\lambda = 4$.

$$(A - 4I){\bf C} = \left(\begin{array}{cc} -2 & -1\\ 4 & 2 \end{a... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Then

$$\left(\begin{array}{cc} -2&-1\\ 4&2 \end{array}\right) \Longrightarrow \left(\begin{array}{cc} 1&\frac{1}{2}\\ 0&0 \end{array}\right)$$

Now there is no leading one in the 2nd row. So, we let $c_{2} = -2\alpha$とおくと， $c_{1} = \alpha$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} 1 \\ -2 \end{array}\right)$. We next find the eigenvector corresponds to $\lambda = 4$ and linearly independent from $C$. Note that if we can find ${\bf C}$ which satisfies

$$(A - 4I)^2 {\bf C} = {\bf0}, \ (A - 4I){\bf C} \neq {\bf0}$$

Then

$$(A - 4I)^2 = \left(\begin{array}{cc} -2&-1\\ 4&2 \end{array}\right)^2 = \left(\begin{array}{cc} 0&0\\ 0&0 \end{array}\right)$$

Here no leading one in either the 2st row or the 2nd row. So, we let ${\bf C} = \left(\begin{array}{c} \alpha\\ \beta \end{array}\right)$. Then this matrix satifies $(A - 4I)^2 = {\bf0}$. Furthermore, we choose $\alpha,\beta$ so that $(A - 4I){\bf C} \neq {\bf0}$. Then ${\bf C} = \left(\begin{array}{c} 1\\ 1 \end{array}\right)$. and the 2nd solution is

$$e^{At}{\bf C} = e^{4t}e^{(A-4I)t}{\bf C} = e^{4t}[{\bf C} + t(A - 4I){\bf C} + \frac{t^2}{2!}(A - 4I)^2 {\bf C}]$$

$$= e^{4t}\{\left(\begin{array}{c} 1\\ 1 \end{array}\right) + \left(... ...array}\right)t\} = e^{4t}\left(\begin{array}{c} 1-3t\\ 1+6t \end{array}\right)$$

Note that $\left(\begin{array}{c} 1\\ -2 \end{array}\right), \left(\begin{array}{c} 1-3t\\ 1+6t \end{array}\right)$ are linearly independent. The general solution can be expressed as follows:

$${\bf X} =e^{4t}[c_{1}\left(\begin{array}{c} 1\\ -2 \end{array}\right) + c_{2}\left(\begin{array}{c} 1-3t\\ 1+6t \end{array}\right)] \$$

(c)
${\bf X} = \left(\begin{array}{ccc} 5&2&2\\ 2&2&-4\\ 2&-4&2 \end{array}\right){\bf X}$. Then

$$\det(A - \lambda I) = \left(\begin{array}{ccc} 5-\lambda&2&2\\ 2&2-\lambda&-4\\ 2&-4&2-\lambda \end{array}\right)$$

$$= (5- \lambda)(\lambda^2 - 4\lambda -12) -2(-2\lambda + 12) + 2(2\lambda -12)$$

$$= (5-\lambda)(\lambda+2)(\lambda-6) + 4\lambda -24 + 4\lambda - 24$$

$$= (5-\lambda)(\lambda+2)(\lambda-6) + 8(\lambda - 6)$$

$$= (\lambda - 6)(-\lambda^2 + 3\lambda + 18)$$

$$= -(\lambda - 6)(\lambda^2 - 3\lambda - 18)$$

$$= - (\lambda - 6)(\lambda + 3)(\lambda - 6)$$

Then the eigenvalues are $\lambda = -3,6$.
The eigenvextor ${\bf C}$ corresponds to $\lambda = -3$ is nonzero solution of the equation $(A + 3I){\bf C} = {\bf0}$. Using Gaussian elimination,

$$A + 3I = \left(\begin{array}{ccc} 8&2&2\\ 2&5&-4\\ 2&-4&5 \end{array}\ri... ...in{array}{ccc} 1&\frac{1}{4}&\frac{1}{4}\\ 2&5&-4\\ 2&-4&5 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&\frac{1}{4}&\frac{1}{4}\\ 0&\frac{9}{... ...gin{array}{ccc} 1&\frac{1}{4}&\frac{1}{4}\\ 0&1&-1\\ 0&0&0 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&0&\frac{1}{2}\\ 0&1&-1\\ 0&0&0 \end{array}\right)$$

Then no leading one in the 3rd row. So, we let $c_{3} = 2\alpha$.

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...d{array}\right) = \alpha\left(\begin{array}{c} -1\\ 2\\ 2 \end{array}\right)$$

Thus, ${\bf X}_{1} = \left(\begin{array}{c} -1\\ 2\\ 2 \end{array}\right)e^{-3t}$.

Next we find the eigenvector corresponds to $\lambda = 6$.

$$A - 6I = \left(\begin{array}{ccc} -1&2&2\\ 2&-4&-4\\ 2&-4&-4 \e... ...ghtarrow \left(\begin{array}{ccc} 1&-2&-2\\ 0&0&0\\ 0&0&0 \end{array}\right)$$

Then the degree of freedom is 2. So, we let $c_{2} = \beta, c_{3} = \gamma$.

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...{array}\right) + \gamma \left(\begin{array}{c} 2\\ 0\\ 1 \end{array}\right).$$

Thus, we have

$${\bf X}_{2} = \left(\begin{array}{c} 2\\ 1\\ 0 \end{array}\righ... ...6t}, {\bf X}_{3} = \left(\begin{array}{c} 2\\ 0\\ 1 \end{array}\right)e^{6t}$$

Note that ${\bf X}_{1},{\bf X}_{2},{\bf X}_{3}$ are linearly independent. Thus , the general solution is given by

$${\bf X} = c_{1}\left(\begin{array}{c} -1\\ 2\\ 2 \end{array}\ri... ...ght)e^{6t} + c_{3}\left(\begin{array}{c} 2\\ 0\\ 1 \end{array}\right)e^{6t}.$$

(d)

$$\det(A - \lambda I) = \left\vert\begin{array}{ccc} 1-\lambda&0&1\\ 1&1-\lambda&0\\ -2&0&-1-\lambda \end{array}\right\vert$$

$$= (1-\lambda)(\lambda^2 - 1) + (-2\lambda + 2)$$

$$= (1-\lambda)(\lambda^2 -1) + 2(1-\lambda)$$

$$= (1-\lambda)(\lambda^2 + 1)$$

Then we have eigenvalues $\lambda = 1, \pm i$.

Now we find the eigenvector ${\bf C}$ corresponds to $\lambda = 1$ using Guassian elimination.

$$A - I = \left(\begin{array}{ccc} 0&0&1\\ 1&0&0\\ -2&0&-2 \end{array}\ri... ...rightarrow \left(\begin{array}{ccc} 1&0&0\\ 0&0&1\\ 0&0&-2 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&0&0\\ 0&0&1\\ 0&0&0 \end{array}\right)$$

Note no leading one in the 2nd row. So, we let $c_{2} = \alpha$. Then

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...nd{array}\right) = \alpha \left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)$$

Thus, $X_{1} = \left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)e^{t}$.

We next find the eigenvector ${\bf C}$ corresponds to $\lambda = i$.

$$A - i I = \left(\begin{array}{ccc} 1-i&0&1\\ 1&1-i&0\\ -2&0&-1-i \end{arr... ...row \left(\begin{array}{ccc} 1&1-i&0\\ 1-i&0&1\\ -2&0&-1-i \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&1-i&0\\ 0&2i&1\\ 0&2-2i&-1-i \end{ar... ...eft(\begin{array}{ccc} 1&1-i&0\\ 0&1&- \frac{i}{2}\\ 0&0&0 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{ccc} 1&0&\frac{1+i}{2}\\ 0&1&- \frac{i}{2}\\ 0&0&0 \end{array}\right)$$

Note that no leadin one in the 3rd row. Then we let $c_{3} = 2\alpha$.，

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...rray}\right) = \alpha \left(\begin{array}{c} -1-i\\ i\\ 2 \end{array}\right)$$

Note that the solutions are given by ${\bf X}_{2} = \Re {\bf C}e^{it}, {\bf X}_{3} = \Im {\bf C}e^{it}$. Thus, we need to find the real part and the imaginary part of ${\bf C}e^{it}$.

$${\bf C}e^{it} = \left(\begin{array}{c} -1-i\\ i\\ 2 \end{array}\right)e^{it} = \left(\begin{array}{c} -1-i\\ i\\ 2 \end{array}\right)(\cos{t} + i \sin{t})$$

$$= \underbrace{\left(\begin{array}{c} -\cos{t} + \sin{t} \\ -\sin{t... ...+ \sin{t} \\ -\cos{t}\\ -2\sin{t} \end{array}\right)}_{\mbox{imaginary part}}$$

Thus, the general solution is given by

$${\bf X} = c_{1}\left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\rig... ...begin{array}{c} \cos{t} + \sin{t} \\ -\cos{t}\\ -2\sin{t} \end{array}\right)$$

(e)

$$\left\{\begin{array}{rl} 4x_{1}^{\prime} + x_{1} + 2x_{2}^{\prime... ... -2x_{1}^{\prime} + 2x_{1} - 2x_{2}^{\prime} - 2x_{2}& = 0 \end{array}\right.$$

Solve for $x_{1}^{\prime}$. Then $2x_{1}^{\prime} + 3x_{1} + 5x_{2} = 0$ and

$$x_{1}^{\prime} = - \frac{3}{2}x_{1} - \frac{5}{2}x_{2}$$ (3.1)

Solve for $x_{2}^{\prime}$. Then $-2x_{2}^{\prime} + 5x_{1} + 3x_{2} = 0$ and

$$x_{2}^{\prime} = \frac{5}{2}x_{1} + \frac{3}{2}x_{2}$$ (3.2)

Thus,

$${\bf X}^{\prime} = \left(\begin{array}{cc} -\frac{3}{2}&-\frac{5}{2}\\ \frac{5}{2}&\frac{3}{2} \end{array}\right){\bf X}$$

$$\det(A - \lambda I) = \left\vert\begin{array}{cc} -\frac{3}{2} - \lambda& - \frac{5}{2}... ...3}{2} - \lambda \end{array}\right\vert = \lambda^2 - \frac{9}{4} + \frac{25}{4}$$

$$= \lambda^2 + 4 = (\lambda + 2i)(\lambda - 2i)$$

The eigenvalues are $\lambda = \pm 2i$.

We find the eigenvector corresponds to $\lambda = 2i$ using Gaussian elimination.

$$A - 2i I = \left(\begin{array}{cc} -\frac{3}{2} - 2i & - \frac{5}{2}\\ \fra... ...}{2}& \frac{3}{2} - 2i \\ -\frac{3}{2} - 2i & - \frac{5}{2} \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{cc} 1&\frac{1}{5}(3-4i)\\ 0&0 \end{array}\right)$$

Then let $c_{2} = \alpha$.

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...right) = -\frac{\alpha}{5}\left(\begin{array}{c} 3-4i\\ -5 \end{array}\right)$$

Thus, we find ${\bf X}_{1} = \Re {\bf C}e^{2it}, {\bf X}_{2} = \Im {\bf C}e^{2it}$.

$${\bf C}e^{2it} = \left(\begin{array}{c} 3-4i\\ -5 \end{array}\right)e^{2it} = \left(\begin{array}{c} 3-4i\\ -5 \end{array}\right)(\cos{2t} + i\sin{2t})$$

$$= \underbrace{\left(\begin{array}{c} 3\cos{2t} + 4\sin{2t}\\ -5\co... ...3\sin{2t} - 4\cos{2t}\\ -5\sin{2t} \end{array}\right)}_{\mbox{imaginary part}}$$

Therefore, the general solution is given by

$${\bf X} = c_{1}\left(\begin{array}{c} 3\cos{2t} + 4\sin{2t}\\ -5... ... \left(\begin{array}{c} 3\sin{2t} - 4\cos{2t}\\ -5\sin{2t} \end{array}\right)$$

(f)
Solve for $x_{2}^{\prime}$. Then $3x_{2}^{\prime} - 3x_{1} - 3x_{2} = 0$ and

$$x_{2}^{\prime} = x_{1} + x_{2}$$

Also $x_{1}^{\prime} - 2x_{1} + 5x_{2}^{\prime} = 0$. Thus,

$$x_{1}^{\prime} = 2x_{1} - 5x_{2}^{\prime} = - 3x_{1} - 5x_{2}$$

From this, we have ${\bf X}^{\prime} = \left(\begin{array}{cc} -3&-5\\ 1&1 \end{array}\right){\bf X}.$. Now we find the eigenvalues and eigenvectors.

$$\det(A - \lambda I) = \left\vert\begin{array}{cc} -3-\lambda&-5\\ 1&1-\lambda \end{array}\right\vert = \lambda^2 + 2\lambda + 2 = 0$$

Then the eigenvalues are $\lambda = -1 \pm \sqrt{1-2} = -1 \pm i$.

We find the eigenvector ${\bf C}$ corresponds to $\lambda = -1+i$ using Gaussian elimination.

$$A + (1-i)I = \left(\begin{array}{cc} -2-i&-5\\ 1&2-i \end{array}... ...right) \longrightarrow \left(\begin{array}{cc} 1&2-i\\ 0&0 \end{array}\right)$$

Then we let $c_{2} = \alpha$.

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...end{array}\right) = \alpha\left(\begin{array}{c} -2+i\\ 1 \end{array}\right).$$

Thus we find ${\bf X}_{1} = \Re {\bf C}e^{(-1+i)t}, {\bf X}_{2} + \Im {\bf C}e^{(-1+i)t}$.

$${\bf C}e^{(-1+i)t} = \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)e^{-t}e^{it} = \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)e^{-t}(\cos{t} + i\sin{t})$$

$$= e^{-t}\underbrace{\left(\begin{array}{c} -2\cos{t}-\sin{t}\\ \co... ...ray}{c} \cos{t}-2\sin{t}\\ \sin{t} \end{array}\right)}_{\mbox{imaginary part}}$$

Therefore, the general solution is given by

$${\bf X} = e^{-t}[c_{1}\left(\begin{array}{c} -2\cos{t}-\sin{t}\\ ... ...+ c_{2} \left(\begin{array}{c} \cos{t}-2\sin{t}\\ \sin{t} \end{array}\right)]$$