Nonhomogeneous Differential Equations

Exercise 3.3

1. Solve the following differential equations.

(a) $\ \left\{\begin{array}{rl} x_{1}^{\prime} =& 2x_{1} + x_{2} - e^{t}\\ x_{2}^{\prime} =& 3x_{1} + 4x_{2} - 7e^{t} \end{array} \right .$

(b) $\ {\bf X}^{\prime} = \left(\begin{array}{cc} 0&4\\ -1&0 \end{array}\right){\bf X} + \left(\begin{array}{c} -3\cos{t}\\ 0 \end{array}\right)$
(c) $\ \left\{\begin{array}{rl} x_{1}^{\prime} =& x_{1} + x_{2} + x_{3} + 1\\ x_{2}... ..._{2}\\ x_{3}^{\prime} =& -2x_{1} -x_{2} -2x_{3} + 3e^{-t} \end{array} \right .$
(d) $\ {\bf X}^{\prime} = \left(\begin{array}{rrrr} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ ... ...\right){\bf X} + \left(\begin{array}{c} t^{2}\\ 0\\ 0\\ 0 \end{array}\right)$
2. Rewrite the following differential equation into the system of linear differential equation. Then solve.
$y^{\prime\prime} + 4y^{\prime} + 3y = t$
3. Solve the following differential equation using the elimination method.

(a) $\ \left\{\begin{array}{rl} x_{1}^{\prime} + x_{2}^{\prime} - 2x_{1} - 4x_{2} &= e^{t}\\ x_{1}^{\prime} + x_{2}^{\prime} - x_{2} &= e^{4t} \end{array}\right .$
(b) $\ \left\{\begin{array}{rl} x_{1}^{\prime\prime} + x_{2}^{\prime} - x_{1} + x_{2... ... x_{1}^{\prime} + x_{2}^{\prime\prime} - x_{1} + x_{2} &= 0 \end{array}\right .$
(c) $\ \left\{\begin{array}{rl} 2x_{1}^{\prime} + x_{2}^{\prime} + x_{1} + 5x_{2} &= 4t\\ x_{1}^{\prime} + x_{2}^{\prime} + 2x_{1} + 2x_{2} &= 2 \end{array}\right .$
(d) $\ \left\{\begin{array}{rl} x_{1}^{\prime\prime} - x_{2}^{\prime} &= t+1 \\ x_{1}^{\prime} + x_{2}^{\prime} - 3x_{1} + x_{2} &= 2t - 1 \end{array}\right .$

Answer
1. (a)
${\bf X}^{\prime} = \left(\begin{array}{cc} 2&1\\ 3&4 \end{array}\right){\bf X} - \left(\begin{array}{c} e^{t}\\ 7e^{t} \end{array}\right)$. Then we find the solution $\Phi$ of ${\bf X}^{\prime} = A{\bf X}$.

$$\det(A - \lambda I) = \left\vert\begin{array}{cc} 2-\lambda&1\\ 3&4-\lambda \end{array}\right\vert = \lambda^2 - 6\lambda + 5 = 0$$

Then the eigenvalues are $\lambda = 1,5$.

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

$$A - I = \left(\begin{array}{cc} 1&1\\ 3&3 \end{array}\right) \longrightarrow \left(\begin{array}{cc} 1&1\\ 0&0 \end{array}\right)$$

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

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

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

We find the eigenvector ${\bf C}$ corresponds to $\lambda = 5$.

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

Thus we let $c_{2} = 3\beta$. Then

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...a \end{array}\right) = \beta \left(\begin{array}{c} 1\\ 3 \end{array}\right).$$

Thus, ${\bf X}_{2} = \left(\begin{array}{c} 1\\ 3 \end{array}\right)e^{5t}$. From this, the fundamental matrix $\Phi$ is given by

$$\Phi(t) = \left(\begin{array}{cc} -e^{t}&e^{5t}\\ e^{t}&3e^{5t} \end{array}\right)$$

Next we find the particular solution ${\bf U}$ by solving $\Phi {\bf U}^{\prime} = {\bf F}$.

$$\left(\begin{array}{cc} -e^{t}&e^{5t}\\ e^{t}&3e^{5t} \end{array... ...\end{array}\right) = \left(\begin{array}{c} e^{t}\\ 7e^{t} \end{array}\right)$$

Using Cramer's rule, we have

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} e^{t}&e^{5t}\\... ...^{5t}\\ e^{t}&3e^{5t} \end{array}\right\vert} = \frac{-4e^{6t}}{-4e^{6t}} = 1$$

Thus,

$$u_{1} = \int 1 dt = t \ ($$it is particular solution. So, no constanat$$)$$

Also,

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} -e^{t}&e^{t}\\... ... e^{t}&3e^{5t} \end{array}\right\vert} = \frac{-8e^{2t}}{-4e^{6t}} = 2e^{-4t}$$

よって

$$u_{2} = \int 2e^{-4t} dt = -\frac{1}{2}e^{-4t} \ ($$it is particular solution. So, no constant$$)$$

Therefore, the general solution is given by

$${\bf X} = \Phi{\bf C} + \Phi {\bf U}$$

$$= \left(\begin{array}{cc} -e^{t}&e^{5t}\\ e^{t}&3e^{5t} \end{array... ...array}\right)\left(\begin{array}{c} t\\ -\frac{1}{2}e^{-4t} \end{array}\right)$$

(b)
${\bf X}^{\prime} = \left(\begin{array}{cc} 0&4\\ -1&0 \end{array}\right){\bf X} - \left(\begin{array}{c} -3\cos{t}\\ 0 \end{array}\right)$. Then we find the solution $\Phi$ satisfying ${\bf X}^{\prime} = A{\bf X}$.

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

Then the eigenvalues are $\lambda = \pm 2i$.

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

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

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

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

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

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

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

Thus the fundamental solution $\Phi$ is given by

$$\Phi(t) = \left(\begin{array}{cc} 2\sin{2t}&-2\cos{2t}\\ \cos{2t}&\sin{2t} \end{array}\right)$$

We next find the particular solution ${\bf U}$ satisfying $\Phi {\bf U}^{\prime} = {\bf F}$.

$$\left(\begin{array}{cc} 2\sin{2t}&-2\cos{2t}\\ \cos{2t}&\sin{2t}... ... \end{array}\right) = \left(\begin{array}{c} -3\cos{t}\\ 0 \end{array}\right)$$

Solve this using Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} -3\cos{t}&-2\c... ...\sin{2t}\cos{t}}{2\sin^{2}{2t}+2\cos^{2}{2t}} = \frac{-3}{4}(\sin{3t}+\sin{t})$$

Thus,

$$u_{1} = \frac{-3}{4}\int (\sin{3t}+\sin{t}) dt = \frac{3}{4}(\frac{\cos{3t}}{4} + \cos{t}) \ ($$no constant term$$)$$

Also,

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} 2\sin{2t}&-3\c... ...}\right\vert} = \frac{3\cos{2t}{\sin{t}}}{2} = \frac{3}{4}(\cos{3t} + \cos{t})$$

Thus,

$$u_{2} = \frac{3}{4} \int (\cos{3t} + \cos{t}) dt = \frac{3}{4}(\frac{\sin{3t}}{3} + \sin{t}).$$

Therefore, the general solution is given by

$${\bf X} = \Phi{\bf C} + \Phi {\bf U}$$

$$= \left(\begin{array}{cc} 2\sin{2t}&-2\cos{2t}\\ \cos{2t}&\sin{2t}... ...rac{\cos{3t} + 3\cos{t}}{4}\\ \frac{\sin{3t} + 3\sin{t}}{4} \end{array}\right)$$

(c)
${\bf X}^{\prime} = \left(\begin{array}{ccc} 1&1&1\\ 0&-1&0\\ -2&-1&-2 \end{array}\right){\bf X} + \left(\begin{array}{c} 1\\ 0\\ 3e^{-t} \end{array}\right)$. Then find the solution $\Phi$ of ${\bf X}^{\prime} = A{\bf X}$.

$$\det(A - \lambda I) = \left\vert\begin{array}{ccc} 1-\lambda&1&1\... ...ay}\right\vert = (-1-\lambda)(\lambda^2 + \lambda) = -\lambda(\lambda + 1)^{2}$$

Then the eigenvalues are $\lambda = -1,0$.

We find the eigenvector ${\bf C}$ corresponds to $\lambda = 0$ using Gaussian elimination.

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

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

We let $c_{3} = \alpha$. Then

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

Thus we have a solutoin ${\bf X}_{1} = \left(\begin{array}{c} -1\\ 0\\ 1 \end{array}\right)e^{0t} = \left(\begin{array}{c} -1\\ 0\\ 1 \end{array}\right)$.

Find the eigenvector ${\bf C}$ corresponds to $\lambda = -1$.

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

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

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

Then, we have a solution ${\bf X}_{2} = \left(\begin{array}{c} -\frac{1}{2}\\ 1\\ 0 \end{array}\right)e... ...X}_{3} = \left(\begin{array}{c} -\frac{1}{2}\\ 0\\ 1 \end{array}\right)e^{-t}$. Thus the fundamental matrix $\Phi$ is given by

$$\Phi(t) = \left(\begin{array}{ccc} -1& - \frac{e^{-t}}{2}&- \frac{e^{-t}}{2}\\ 0&e^{-t}&0\\ 1&0&e^{-t} \end{array}\right)$$

We find the particular solution ${\bf U}$ of $\Phi {\bf U}^{\prime} = {\bf F}$.

$$\left(\begin{array}{ccc} -1& - \frac{e^{-t}}{2}&- \frac{e^{-t}}{2... ...nd{array}\right) = \left(\begin{array}{c} 1\\ 0\\ 2e^{-t} \end{array}\right)$$

Solve this using Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{ccc} 1& - \frac{e^... ...t\vert} = \frac{e^{-2t} + e^{-3t}}{-e^{-2t} + \frac{e^{-2t}}{2}} = -2 -2e^{-t}$$

Thus,

$$u_{1} = \int (-2 -2e^{-t}) dt = -2t + 2e^{-t} \ ($$no constant term$$)$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{ccc} -1& 1&- \frac... ...-t}&0\\ 1&0&e^{-t} \end{array}\right\vert} = \frac{0}{-\frac{e^{-2t}}{2}} = 0$$

Thus

$$u_{2} = \int 0 dt = 0 \ ($$no constant term$$)$$

$$u_{3}^{\prime} = \frac{\left\vert\begin{array}{ccc} -1& - \frac{e... ...{array}\right\vert} = \frac{-2e^{-t}-1}{-\frac{e^{-2t}}{2}} = 4e^{t} + 2e^{2t}$$

Thus

$$u_{3} = \int (4e^{t} + 2e^{2t}) dt = 4e^{t} + e^{2t} \ ($$no constant term$$)$$

Therefore, the general solution is given by

$${\bf X} = \Phi{\bf C} + \Phi {\bf U}$$

$$= \left(\begin{array}{ccc} -1& - \frac{e^{-t}}{2}&- \frac{e^{-t}}{2... ...\left(\begin{array}{c} -2t + 2e^{-t}\\ 0\\ 4e^{t} + e^{2t} \end{array}\right)$$

(d)

$$\det(A - \lambda I) = \left\vert\begin{array}{rrrr} -\lambda&1&0&0\\ 0&-\lambda&1&0\\ ... ...&-\lambda&1\\ 1&0&0&-\lambda \end{array}\right\vert = -\lambda(-\lambda^3) - 1$$

$$= \lambda^4 - 1 = (\lambda^2 + 1)(\lambda + 1)(\lambda - 1) = 0$$

Then the eigenvalues are $\lambda = -1,1,\pm i$.

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

$$A + I = \left(\begin{array}{rrrr} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 1&0&0&... ...begin{array}{rrrr} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&-1&0&1 \end{array}\right)$$

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

$$\longrightarrow \left(\begin{array}{rrrr} 1&1&0&0\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0... ...begin{array}{rrrr} 1&0&0&1\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0&0 \end{array}\right)$$

Then no leading one in the 4th row. So, we let $c_{4} = \alpha$. Then

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

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

We find the eigenvector ${\bf C}$ corresponds to $\lambda = 1$.

$$A - I = \left(\begin{array}{rrrr} -1&1&0&0\\ 0&-1&1&0\\ 0&0&-1&1\\ 1&0... ...in{array}{rrrr} 1&-1&0&0\\ 0&-1&1&0\\ 0&0&-1&1\\ 0&1&0&-1 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{rrrr} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0... ...gin{array}{rrrr} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0&0&0 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{rrrr} 1&-1&0&0\\ 0&1&0&-1\\ 0&0&1&-1\\ 0&0... ...gin{array}{rrrr} 1&0&0&-1\\ 0&1&0&-1\\ 0&0&1&-1\\ 0&0&0&0 \end{array}\right)$$

Thus, we let $c_{4} = \beta$. Then

$${\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}... ...ay}\right) = \beta \left(\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right).$$

Then we have ${\bf X}_{2} = \left(\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right)e^{t}$.

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

$$A - i I = \left(\begin{array}{rrrr} -i&1&0&0\\ 0&-i&1&0\\ 0&0&-i&1\\ 1&0... ...egin{array}{rrrr} 1&i&0&0\\ 0&1&i&0\\ 0&0&1&i\\ 0&-i&0&-i \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{rrrr} 1&i&0&0\\ 0&1&i&0\\ 0&0&1&i\\ 0&0&-1... ...\begin{array}{rrrr} 1&i&0&0\\ 0&1&i&0\\ 0&0&1&i\\ 0&0&0&0 \end{array}\right)$$

$$\longrightarrow \left(\begin{array}{rrrr} 1&i&0&0\\ 0&1&0&1\\ 0&0&1&i\\ 0&0&0&... ...\begin{array}{rrrr} 1&0&0&i\\ 0&1&0&1\\ 0&0&1&i\\ 0&0&0&0 \end{array}\right)$$

Thus, we let $c_{4} = \gamma$. Then，

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

Thus,

$${\bf X}_{3} = \Re {\bf C}e^{it} = \left(\begin{array}{c} -\sin{t}... ...\begin{array}{c} \cos{t}\\ -\sin{t}\\ -\cos{t}\\ \sin{t} \end{array}\right)$$

Then the fundamental matrix $\Phi$ is given by

$$\Phi(t) = \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t... ...{-t}&e^{t}&\sin{t}&-\cos{t}\\ e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right)$$

Next we find the particular solution ${\bf U}$ satisfying $\Phi {\bf U}^{\prime} = {\bf F}$.

$$\left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t}\\ e^{-t... ...d{array}\right) = \left(\begin{array}{c} t^2\\ 0\\ 0\\ 0 \end{array}\right)$$

Solve this using Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{rrrr} t^2&e^{t}&-\... ...\\ e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{2t^2 e^t}{-8}$$

Thus,

$$u_{1} = -\frac{1}{4} \int t^2 e^{t} dt = -\frac{1}{4}(t^2 e^{t} - 2t e^{t} + 2e^{t})$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{rrrr} -e^{-t}&t^2&... ...e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{-2t^2 e^{-t}}{-8}$$

Thus,

$$u_{2} = \frac{1}{4} \int t^2 e^{-t} dt = -\frac{1}{4}(t^2 e^{-t} + 2t e^{-t} + 2e^{-t})$$

$$u_{3}^{\prime} = \frac{\left\vert\begin{array}{rrrr} -e^{-t}&e^{t... ...e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{4t^2 \sin{t}}{-8}$$

Thus,

$$u_{3} = -\frac{1}{2} \int t^2 \sin{t} dt = \frac{1}{2}(t^2 \cos{t} - 2t \sin{t} - 2\cos{t})$$

$$u_{4}^{\prime} = \frac{\left\vert\begin{array}{rrrr} -e^{-t}&e^{t... ...^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{-4t^2 \cos{t}}{-8}$$

Thus,

$$u_{4} = \frac{1}{2} \int t^2 \cos{t} dt = \frac{1}{2}(t^2 \sin{t} - 2t \cos{t} + 2\sin{t})$$

Therefore, the general solution is given by

$${\bf X} = \Phi{\bf C} + \Phi {\bf U}$$

$$= \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t}\\ e^{-t... ...ight)\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4} \end{array}\right)$$

$$+ \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t}\\ e^{-t... ...\cos{t})\\ \frac{1}{2}(t^2 \sin{t} - 2t \cos{t} + 2\sin{t}) \end{array}\right)$$

2.
Let $y_{1}^{\prime} = y_{2}$. Then $y_{2}^{\prime} = y^{\prime\prime} = -4y^{\prime} - 3y + t$. Thus,

$$\left\{\begin{array}{l} y_{1}^{\prime} = y_{2}\\ y_{2}^{\prime} = -4y_{2} - 3y_{1} + t \end{array}\right.$$

Now let ${\bf Y} = \left(\begin{array}{c} y_{1}\\ y_{2} \end{array}\right)$. Then we can write

$${\bf Y}^{\prime} = \left(\begin{array}{cc} 0&1\\ -3&-4 \end{array}\right){\bf Y} + \left(\begin{array}{c} 0\\ t \end{array}\right).$$

$$\det(A - \lambda I) = \left\vert\begin{array}{cc} -\lambda & 1\\ -3 & -4 - \lambda \end{array}\right\vert = \lambda^2 + 4\lambda + 3 = 0$$

Then the eigenvalues are $\lambda = -3,-1$.

We find the eigenvector ${\bf C}$ corresponds to $\lambda = -3$ using Gaussian elimination.

$$A + 3I = \left(\begin{array}{cc} 3&1\\ -3&-1 \end{array}\right) ... ...rightarrow \left(\begin{array}{cc} 1 & \frac{1}{3}\\ 0 & 0 \end{array}\right)$$

Then we let $c_{2} = 3\alpha$.

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

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

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

$$A + I = \left(\begin{array}{cc} 1&1\\ -3&-3 \end{array}\right) \longrightarrow \left(\begin{array}{cc} 1 & 1\\ 0 & 0 \end{array}\right)$$

Then let $c_{2} = 3\beta$. Then

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

Thus we have ${\bf Y}_{2} = \left(\begin{array}{c} -1\\ 1 \end{array}\right)e^{-t}$. Thus the fundamental matrix $\Phi$ is given by

$$\Phi(t) = \left(\begin{array}{cc} -e^{-3t} & -e^{-t}\\ 3e^{-3t} & e^{-t} \end{array}\right)$$

Next we find the particular solution ${\bf U}$ satisfying $\Phi {\bf U}^{\prime} = {\bf F}$.

$$\left(\begin{array}{cc} -e^{-3t} & -e^{-t}\\ 3e^{-3t} & e^{-t} \... ...{\prime} \end{array}\right) = \left(\begin{array}{c} 0\\ t \end{array}\right)$$

Using Cramer's rule, we have

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0&-e^{-t}\\ t... ...^{-t} \end{array}\right\vert} = \frac{te^{-t}}{2e^{-4t}} = \frac{1}{2}te^{3t}.$$

Then

$$u_{1} = \frac{1}{2}\int te^{3t} dt = \frac{1}{2}(\frac{te^{3t}}{3} - \frac{e^{3t}}{9}).$$

Also,

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} -e^{-3t} & 0\\... ...-t} \end{array}\right\vert} = \frac{-te^{-3t}}{2e^{-4t}} = -\frac{1}{2}te^{t}.$$

Thus,

$$u_{2} = -\frac{1}{2}\int te^{t} dt = -\frac{1}{2}(te^{t} - e^{t}).$$

Therefore, the general solution is given by

$${\bf Y} = \Phi{\bf C} + \Phi {\bf U}$$

$$= \left(\begin{array}{cc} -e^{-3t} & -e^{-t}\\ 3e^{-3t} & e^{-t} \... ...{3t}}{3} - \frac{e^{3t}}{9})\\ -\frac{1}{2}(te^{t} - e^{t}) \end{array}\right)$$

3.
(a)

$$\left\{\begin{array}{ll} (D-2)x_{1} + (D-4)x_{2} & = e^{t}\\ Dx_{1} + (D-1)x_{2} & = e^{4t} \end{array}\right.$$ (3.3)

Then we have

$$\left\vert\begin{array}{cc} D-2 & D-4\\ D & D-1 \end{array}\righ... ...\left\vert\begin{array}{cc} e^{t} & D-4\\ e^{4t} & D-1 \end{array}\right\vert$$

Now expand this, we have

$$(D^2 - 3D + 2 - D^2 + 4D)x_{1} = (D-1)e^{t} - (D-4)e^{4t}$$

$$= e^{t} - e^{t} - 4e^{4t} + 4e^{4t} = 0$$

or

$$(D+2)x_{1} = 0$$

Thus, $x_{1} = c_{1}e^{-2t}$. Similarly, we can find $x_{2}$. To find $x_{2}$, eliminate $Dx_{2}$ frpm the equation 3.3. Then substitute $x_{1}$. In fact, subtract the 2nd equation from the 1st equation in 3.3. Then $-2x_{1}-3x_{2} = e^{t} - e^{4t}$. Thus,

$$x_{2} = -\frac{1}{3}(2x_{1} + e^{4t} - e^{t}) = -\frac{1}{3}(2c_{1}e^{-2t} + e^{4t} - e^{t}).$$

Therefore, the general solution is given by

$$\left\{\begin{array}{l} x_{1} = c_{1}e^{-2t}\\ x_{2} = -\frac{1}{3}(2c_{1}e^{-2t} + e^{4t} - e^{t}) \end{array}\right.$$

(b)

$$\left\{\begin{array}{ll} (D^2 - 1)x_{1} + (D + 1)x_{2} & = 1\\ (D-1)x_{1} + (D^2 + 1)x_{2} & = 0 \end{array}\right.$$ (3.4)

Then

$$\left\vert\begin{array}{cc} D^2 - 1 & D + 1\\ D-1 & D^2 + 1 \end... ...x_{1} = \left\vert\begin{array}{cc} 1 & D+1\\ 0&D^2 +1 \end{array}\right\vert$$

Expand this, we have

$$(D^4 - 1 - D^2 + 1)x_{1} = (D^2 + 1)1 - (D+1)0 = 1$$

or

$$D^2(D^2 - 1)x_{1} = 1.$$

The caracteristic equation is $m^2(m^2 - 1) = 0$. Then $m = -1,0,0,1$. Thus , the complementary function is

$$x_{1c} = c_{1}e^{-t} + c_{2} + c_{3}t + c_{4}e^{t}.$$

Next find the particular solution $x_{1p}$ using the method of undetermined coefficient. $D1 = 0$. Then

$$D^3(D^2 - 1)x_{1} = D1 = 0.$$

Thus, we can set

$$x_{1p} = At^2 + Bt + C$$

$D^2(D^2-1)x_{1p} = 1$ implies that $-2A = 1$. Thus, $A = -\frac{1}{2}$. Therefore,

$$x_{1p} = -\frac{1}{2}t^2$$

Thus,

$$x_{1} = x_{1c} + x_{1p} = c_{1}e^{-t} + c_{2} + c_{3}t + c_{4}e^{t} - \frac{1}{2}t^2.$$

Next we find $x_{2}$. Eliminate $x_{2}^{\prime\prime}$ from the equation 3.4. Then

$$(D^3 - 2D + 1)x_{1} + Dx_{2} - x_{2} = 0$$

Now subtract the first equation in 3.4

$$(D^3 - D^2 - 2D + 2)x_{1} -2x_{2} = 0$$

Therefore,

$$x_{2} = \frac{1}{2}(D^3 - D^2 - 2D + 2)x_{1}$$

$$= \frac{1}{2}(D^2 - 2)(D - 1)x_{1}$$

$$= \frac{1}{2}(D^2 - 2)(-2c_{1}e^{-t} -c_{2} + c_{3} - c_{3}t - t + \frac{1}{2}t^2)$$

$$= \frac{1}{2}(-2c_{1}e^{-t} + 1 -2(-2c_{1}e^{-t} -c_{2} + c_{3} - c_{3}t - t + \frac{1}{2}t^2))$$

$$= \frac{1}{2}(2c_{1}e^{-t} + 2c_{2} - 2c_{3} + 2c_{3}t + 2t - t^2).$$

Then the general solution is given by

$$\left\{\begin{array}{l} x_{1} = c_{1}e^{-t} + c_{2} + c_{3}t + c_... ...= c_{1}e^{-t} + c_{2} - c_{3} + c_{3}t + t - \frac{1}{2}t^2 \end{array}\right.$$

(c)

$$\left\{\begin{array}{ll} (2D + 1)x_{1} + (D + 5)x_{2} & = 1\\ (D + 2)x_{1} + (D + 2)x_{2} & = 0 \end{array}\right.$$ (3.5)

Then we have

$$\left\vert\begin{array}{cc} 2D + 1 & D + 5\\ D + 2 & D + 2 \end{... ...x_{1} = \left\vert\begin{array}{cc} 4t & D+5\\ 2&D + 2 \end{array}\right\vert$$

Now expand this,

$$(2D^2 + 5D + 2 - D^2 - 7D - 10)x_{1} = (D+2)4t - (D+5)2$$

$$= 4+ 8t -10$$

$$= 8t - 6$$

Then

$$(D^2 - 2D - 8 )x_{1} = 8t - 6.$$

The characteristic equation is $m^2 - 2m - 8 = 0$. Then $m = -2, 4$. Thus the complementary function is

$$x_{1c} = c_{1}e^{-2t} + c_{2}e^{4t}.$$

Next we find the particular solution $x_{1p}$ using the method of undermined coefficient. $D^2(8t-6) = 0$ implies

$$D^2(D + 2)(D - 4)x_{1} = D^2(8t-6) = 0.$$

Thus, we can set

$$x_{1p} = At + B$$

Then $(D^2 -2D - 8)x_{1p} = 8t-6$. Thus $A = -1, B = 1$.

$$x_{1p} = -t + 1$$

Therfore,

$$x_{1} = x_{1c} + x_{1p} = c_{1}e^{-2t} + c_{2}e^{4t} - t + 1.$$

We next find $x_{2}$. eliminate $x_{2}^{\prime}$ from the equation 3.5. Then we have

$$(D - 1)x_{1} + 3x_{2} = 4t - 2$$

Thus,

$$x_{2} = \frac{1}{3}(4t - 2 -(D - 1)x_{1})$$

$$= \frac{1}{3}(4t -2 - Dx_{1} + x_{1})$$

$$= \frac{1}{3}(4t - 2 + 2c_{1}e^{-2t} - 4c_{2}e^{4t} + 1)$$

$$= \frac{1}{3}(3c_{1}e^{-2t} - 3c_{2}e^{4t} + 3t)$$

$$= c_{1}e^{-2t} - c_{2}e^{4t} + t .$$

Therefore, the general solution is given by

$$\left\{\begin{array}{l} x_{1} = c_{1}e^{-2t} + c_{2}e^{4t} - t + 1\\ x_{2} = c_{1}e^{-2t} - c_{2}e^{4t} + t \end{array}\right.$$

(d)

$$\left\{\begin{array}{ll} D^{2}x_{1} - Dx_{2} & = t +1\\ (D - 3)x_{1} + (D + 1)x_{2} & = 2t - 1 \end{array}\right.$$ (3.6)

Then we have

$$\left\vert\begin{array}{cc} D^2 & -D\\ D - 3 & D + 1 \end{array}... ... = \left\vert\begin{array}{cc} t+1 & -D\\ 2t - 1&D + 1 \end{array}\right\vert$$

Expand this,

$$(D^3 + D^2 + D^2 - 3D)x_{1} = (D+1)(t+1) + D(2t-1)$$

$$= 1 + t+1 + 2$$

$$= t + 4$$

Thus,

$$D(D-1)(D+3)x_{1} = t + 4.$$

Then the characteristic equation is $m(m-1)(m+3) = 0$ and $m = 0,-3,1$. Thus, the complementary function is given by

$$x_{1c} = c_{1} + c_{2}e^{-3t} + c_{3}e^{t}.$$

We next find the particular solution $x_{1p}$ using the mathod of undetermined coefficient. $D^2(t + 4) = 0$ implies $D^3(D-1)(D+3)x_{1} = D^2(t + 4) = 0$. Note that $m = -3,0,1$ are already used in the complementary function. So we let

$$x_{1p} = At^2 + Bt$$

Then $D(D-1)(D+3)x_{1p} = t + 4$. Thus we have $-6A = 1, 3B = -\frac{14}{3}$. Then

$$x_{1p} = -\frac{t^2}{6} - \frac{14t}{9}$$

Therefore,

$$x_{1} = x_{1c} + x_{1p} = c_{1} + c_{2}e^{-3t} + c_{3}e^{t} -\frac{t^2}{6} - \frac{14t}{9} .$$

We find $x_{2}$. Eliminate $x_{2}^{\prime}$ from the equation 3.6. Then we have

$$(D^2 + D - 3)x_{1} + x_{2} = 3t$$

Then

$$x_{2} = 3t - (D^2 + D - 3)x_{1}$$

$$= 3t - (9c_{2}e^{-3t} + c_{3}e^{t} - \frac{1}{3}$$

$$+ -3c_{2}e^{-3t} + c_{3}e^{t} - \frac{t}{3} - \frac{14}{3}$$

$$- 3(c_{1} + c_{2}e^{-3t} + c_{3}e^{t} - -\frac{t^2}{6} - \frac{14t}{9}))$$

$$= 3c_{1} - 3c_{2}e^{-3t} + c_{3}e^{t} - \frac{t^2}{2} - \frac{4t}{3} + \frac{17}{9} .$$

Therefore, the general solution is given by

$$\left\{\begin{array}{l} x_{1} = c_{1} + c_{2}e^{-3t} + c_{3}e^{t}... ... + c_{3}e^{t} - \frac{t^2}{2} - \frac{4t}{3} + \frac{17}{9} \end{array}\right.$$