System of Homogeneous Linear Differential Equation

Exercise 3.1

1. Solve the following differential equations.

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

Answer
1. (a) $\det(A - \lambda I) = \left(\begin{array}{cc} 1-\lambda&4\\ 1&1-\lambda \end{array}\right) = \lambda^2 - 2\lambda - 3 = (\lambda + 1)(\lambda - 3)$ implies that the eigenvalues are $\lambda = -1, 3$. Then we find the eigenvector ${\bf C}$ corresponding to $\lambda = -1$. Note that $C$ satisfies the follwing equation.

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

Now the matrix $\left(\begin{array}{cc} 2&4\\ 1&2 \end{array}\right)$ is row reduced to echlon form $\left(\begin{array}{ccc} 1&2\\ 0&0 \end{array}\right)$. Then let $c_{2} = \alpha$. Then $c_{1} = - 2\alpha$. Therefore, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} -2 \\ 1 \end{array}\right)$となり， ${\bf X}_{1} = \left(\begin{array}{c} -2\\ 1 \end{array}\right)e^{-t}$.

We next find the eigenvector corresponds to $\lambda = 3$.

$$(A - 3I){\bf C} = \left(\begin{array}{cc} -2&4\\ 1&-2 \end{array... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Note that the matrix $\left(\begin{array}{cc} -2&4\\ 1&-2 \end{array}\right)$ is row reduced echelon form $\left(\begin{array}{ccc} 1&-2\\ 0&0 \end{array}\right)$. Now look at the echelon form. no leading one in 2nd row. So, we let $c_{2} = \beta$. Then $c_{1} = 2\beta$. Therefore, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \beta \left(\begin{array}{c} 2 \\ 1 \end{array}\right)$. Thus, ${\bf X}_{2} = \left(\begin{array}{c} 2\\ 1 \end{array}\right)e^{3t}$. Then the general solution is given by

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

(b) $\det(A - \lambda I) = \left(\begin{array}{cc} 6-\lambda&-3\\ 2&1-\lambda \end{array}\right) = \lambda^2 - 7\lambda + 12 = (\lambda - 3)(\lambda - 4)$ implies that the eigenvalues are $\lambda = 3,4$. Then we find the eigenvector C corresponds to $\lambda = 3$. $C$ satisfies the following equation.

$$(A - 3I){\bf C} = \left(\begin{array}{cc} 3&-3\\ 2&-2 \end{array... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Then the matrix $\left(\begin{array}{cc} 3&-3\\ 2&-2 \end{array}\right)$ is row reduced to echelon form $\left(\begin{array}{ccc} 1&-1\\ 0&0 \end{array}\right)$. Now there is no leading one in 2nd row. So we let $c_{2} = \alpha$. Then $c_{1} = \alpha$. Therefore, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} 1 \\ 1 \end{array}\right)$ and ${\bf X}_{1} = \left(\begin{array}{c} 1\\ 1 \end{array}\right)e^{3t}$.

We next find the eigenvector corresponds to $\lambda = 4$.

$$(A - 3I){\bf C} = \left(\begin{array}{cc} 2&-3\\ 2&-3 \end{array... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Now the matrix $\left(\begin{array}{cc} 2&-3\\ 2&-3 \end{array}\right)$ is row reduced to echelon form $\left(\begin{array}{ccc} 1&-\frac{3}{2}\\ 0&0 \end{array}\right)$. Then there is no leading one in 2nd row. So, we let $c_{2} = 2\beta$. Then $c_{1} = 3\beta$. Thus, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \beta \left(\begin{array}{c} 3 \\ 2 \end{array}\right)$ and ${\bf X}_{2} = \left(\begin{array}{c} 3\\ 2 \end{array}\right)e^{4t}$. Therefore, the general solution is give by

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

(c) $\det(A - \lambda I) = \left(\begin{array}{ccc} 1-\lambda&1&-1\\ 0&2-\lambda&0\\ 0&1&-1 - \lambda \end{array}\right) = -(1 - \lambda)(2 - \lambda)(1+\lambda)$ implies that the eigenvalues are $\lambda = -1, 1, 2$. Then we find the eigenvector ${\bf C}$ correponds to $\lambda = -1$. $C$ satisfies the following equation.

$$(A + I){\bf C} = \left(\begin{array}{ccc} 2&1&-1\\ 0&3&0\\ 0&1&... ...{3} \end{array}\right) = \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)$$

Then the matrix

$$\left(\begin{array}{ccc} 2&1&-1\\ 0&3&0\\ 0&1&0 \end{array}\rig... ... \left(\begin{array}{ccc} 1&0&-\frac{1}{2}\\ 0&1&0\\ 0&0&0 \end{array}\right)$$

is row reduced to echelon form. Note that there is no leading one in the 3rd row. So, we let $c_{3} = 2\alpha$. Then $c_{1} = \alpha, c_{2} = 0$. Thus, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = \alpha \left(\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right)$ and ${\bf X}_{1} = \left(\begin{array}{c} 1\\ 0\\ 2 \end{array}\right)e^{-t}$.

We next find the eigenvector corresponding to $\lambda = 1$.

$$(A - I){\bf C} = \left(\begin{array}{ccc} 0&1&-1\\ 0&1&0\\ 0&1&... ...{3} \end{array}\right) = \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)$$

Then the matrix

$$\left(\begin{array}{ccc} 0&1&-1\\ 0&1&0\\ 0&1&-2 \end{array}\ri... ...grightarrow \left(\begin{array}{ccc} 0&1&0\\ 0&0&1\\ 0&0&0 \end{array}\right)$$

is row reduced echelon form. Now no leading on in the 1st row. So, let $c_{1} = \beta$. Then $c_{2} = 0, c_{3} = 0$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = \beta \left(\begin{array}{c} 1 \\ 0\\ 0 \end{array}\right)$ and ${\bf X}_{2} = \left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)e^{t}$.

We find the eigenvector corresponds to $\lambda = 2$.

$$(A - 2I){\bf C} = \left(\begin{array}{ccc} -1&1&-1\\ 0&0&0\\ 0&... ...{3} \end{array}\right) = \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)$$

Now the matrix

$$\left(\begin{array}{ccc} -1&1&-1\\ 0&0&0\\ 0&1&-3 \end{array}\r... ...ghtarrow \left(\begin{array}{ccc} 1&0&-2\\ 0&1&-3\\ 0&0&0 \end{array}\right)$$

is row reduced echelon form，Now there is no leading one in the 3rd row. Let $c_{3} = \gamma$. Then $c_{2} = 3 \gamma, c_{1} = 2 \gamma$. Thus, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = \gamma \left(\begin{array}{c} 2 \\ 3\\ 1 \end{array}\right)$ and ${\bf X}_{3} = \left(\begin{array}{c} 2\\ 3\\ 1 \end{array}\right)e^{2t}$.

Therefore, the general solution is given by

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

(d) $\det(A - \lambda I) = \left(\begin{array}{ccc} 1-\lambda&-3&2\\ 0&-1-\lambda&0\\ 0&-1&-2 - \lambda \end{array}\right) = (1 + \lambda)(1 - \lambda)(2+\lambda)$ implies that the eigenvalues are $\lambda = -2,-1,1$. Then we find the eigenvector ${\bf C}$ corresponds to $\lambda = -2$. Note that the eigenvector $C$ satisfies the following equation.

$$(A + 2I){\bf C} = \left(\begin{array}{ccc} 3&-3&2\\ 0&1&0\\ 0&-... ...{3} \end{array}\right) = \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)$$

Now the matrix

$$\left(\begin{array}{ccc} 3&-3&2\\ 0&1&0\\ 0&-1&0 \end{array}\ri... ...w \left(\begin{array}{ccc} 1&0&\frac{2}{3}\\ 0&1&0\\ 0&0&0 \end{array}\right)$$

is row reduced echelon form and the leading one exits in 1st row and 2nd row. Thus we let $c_{3} = 3\alpha$. Then $c_{1} = -2\alpha, c_{2} = 0$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = \alpha \left(\begin{array}{c} -2 \\ 0 \\ 3 \end{array}\right)$ and ${\bf X}_{1} = \left(\begin{array}{c} -2\\ 0\\ 3 \end{array}\right)e^{-2t}$.

Next we find the eigenvector corresponding to $\lambda = -1$

$$(A - I){\bf C} = \left(\begin{array}{ccc} 2&-3&2\\ 0&0&0\\ 0&-1... ...{3} \end{array}\right) = \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)$$

Now the matrix

$$\left(\begin{array}{ccc} 2&-3&2\\ 0&0&0\\ 0&-1&-1 \end{array}\r... ...w \left(\begin{array}{ccc} 1&0&\frac{5}{2}\\ 0&1&1\\ 0&0&0 \end{array}\right)$$

is row reduced echelon form. Now there is no leading one in the 3rd row. Let $c_{3} = 2\beta$. Then $c_{2} = -2\beta, c_{3} = -5\beta$. Thus, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = \beta \left(\begin{array}{c} -5 \\ -2\\ 2 \end{array}\right)$ and ${\bf X}_{2} = \left(\begin{array}{c} -5 \\ -2\\ 2 \end{array}\right)e^{-t}$.

We find the eigenvector corresponds to $\lambda = 1$.

$$(A - 2I){\bf C} = \left(\begin{array}{ccc} 0&-3&2\\ 0&-2&0\\ 0&... ...{3} \end{array}\right) = \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right)$$

Then the matrix

$$\left(\begin{array}{ccc} 0&-3&2\\ 0&-2&0\\ 0&-1&-3 \end{array}\... ...grightarrow \left(\begin{array}{ccc} 0&1&0\\ 0&0&1\\ 0&0&0 \end{array}\right)$$

is row reduced echelon form and no leading one in the 1st row. So, we let $c_{1} = \gamma$. Then $c_{2} = 0, c_{3} = 0$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = \gamma \left(\begin{array}{c} 1 \\ 0\\ 0 \end{array}\right)$ and ${\bf X}_{3} = \left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)e^{2t}$.

Therefore the general solution is

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

(e) Solve this for $x_{1}^{\prime},x_{2}^{\prime}$. Then

$$x_{1}^{\prime} = 3x_{1} + 2x_{2}$$

$$x_{2}^{\prime} = -3x_{1} - 4x_{2}$$

Thus, $\det(A - \lambda I) = \left(\begin{array}{cc} 3-\lambda&2\\ -3&-4-\lambda \end{array}\right) = \lambda^2 + \lambda -6 = (\lambda - 2)(\lambda + 3)$. This implies that eigenvalues are $\lambda = -3,2$. Then we find the eigenvector ${\bf C}$ corresponds to $\lambda = -3$

$$(A + 3I){\bf C} = \left(\begin{array}{cc} 6&2\\ -3&-1 \end{array... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Then the matrix

$$\left(\begin{array}{ccc} 6&2\\ -3&-1 \end{array}\right) \Longrightarrow \left(\begin{array}{ccc} 1&\frac{1}{3}\\ 0&0 \end{array}\right)$$

is row reduced echelon form，Now there is no leading one in the 2nd row. So, we let $c_{2} = 3\alpha$. Then $c_{1} = -\alpha$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} -1 \\ 3 \end{array}\right)$ and ${\bf X}_{1} = \left(\begin{array}{c} -1 \\ 3 \end{array}\right)e^{-3t}$.

We next find the eigenvector corresponding to $\lambda = 2$.

$$(A - 2I){\bf C} = \left(\begin{array}{cc} 1&2\\ -3&-6 \end{array... ...\ c_{2} \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Here the matrix

$$\left(\begin{array}{cc} 1&2\\ -3&-6 \end{array}\right) \Longrightarrow \left(\begin{array}{ccc} 1&2\\ 0&0 \end{array}\right)$$

is row reduced echelon form and no leading one in the 2nd row. So, we let $c_{2} = \beta$. Then $c_{1} = -2\beta$. Thus, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \beta \left(\begin{array}{c} -2 \\ 1 \end{array}\right)$ and ${\bf X}_{2} = \left(\begin{array}{c} -2 \\ 1 \end{array}\right)e^{2t}$. THerefore, the general solution is given by

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

(f) Solve the equation for $x_{1}^{\prime},x_{2}^{\prime}$.，

$$2x_{1}^{\prime} + 6x_{1} + 4x_{2} = 0, \ 2x_{2}^{\prime} + 6x_{1} + 2x_{2} = 0$$

Then

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

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

Thus $\det(A - \lambda I) = \left(\begin{array}{cc} -3-\lambda&-2\\ -3&-1-\lambda \end{array}\right) = \lambda^2 + 4\lambda - 3 = 0$. Then the eigenvalues are $\lambda = -2 \pm \sqrt{7}$. Now we find the eigenvector ${\bf C}$ corresponds to $\lambda = -2 + \sqrt{7}$. Now the matrix $\left(\begin{array}{ccc} -1 - \sqrt{7}&-2\\ -3&1 - \sqrt{7} \end{array}\right)... ...ow \left(\begin{array}{ccc} 1&-\frac{1 - \sqrt{7}}{3}\\ 0&0 \end{array}\right)$ is row reduced echelon form and no leading one in the 2nd row. So, we let $c_{2} = 3\alpha$. Then $c_{1} = (1 - \sqrt{7})\alpha$. Thus, the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} 1 - \sqrt{7} \\ 3 \end{array}\right)$ and ${\bf X}_{1} = \left(\begin{array}{c} 1 - \sqrt{7} \\ 3 \end{array}\right)e^{(-2 + \sqrt{7})t}$.

We next find the eigenvector corresponds to $\lambda = -2 - \sqrt{7}$. The matrix $\left(\begin{array}{ccc} -1 + \sqrt{7}&-2\\ -3&1 + \sqrt{7} \end{array}\right)... ...ow \left(\begin{array}{ccc} 1&-\frac{1 + \sqrt{7}}{3}\\ 0&0 \end{array}\right)$ is row reduced echelon form and no leading one in the 2nd row. So, we let $c_{2} = 3\alpha$. Thne $c_{1} = (1 + \sqrt{7})\alpha$. Thus the eigenvector is $\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) = \alpha \left(\begin{array}{c} 1 + \sqrt{7} \\ 3 \end{array}\right)$ and ${\bf X}_{2} = \left(\begin{array}{c} 1 + \sqrt{7} \\ 3 \end{array}\right)e^{(-2 - \sqrt{7})t}$.

Therefore, the general solution is given by

$${\bf X} = c_{1}\left(\begin{array}{c} 1 - \sqrt{7} \\ 3 \end{arr... ...eft(\begin{array}{c} 1 + \sqrt{7} \\ 3 \end{array}\right)e^{(-2 - \sqrt{7})t}$$