MAT244--2019F > Term Test 2

Problem 3 (noon)

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Victor Ivrii:
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} 1 &2\\ 1 &0\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

(b) Find the general solution
$$\mathbf{x}'=\begin{pmatrix} 1 &2\\ 1 &0\end{pmatrix}\mathbf{x}+ \begin{pmatrix} 0 \\[1pt] \dfrac{6 e^{3t }}{e^{2t}+1}\end{pmatrix}.$$

Changhao Jiang:
(a)
To find eigenvalues, $(1-\lambda x)(-\lambda)-2=0$,we can get $\lambda = 2$ or $\lambda = -1$
To find eigenvectors, when $\lambda=2$,
$\begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} ~ \begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix}$
let $x_2=t, x_1=2t$, so the eigenvector is \begin{bmatrix}1 \\ 2\end{bmatrix}
when $\lambda = -1$
$\begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}$
~
$\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$
let $x_2=-t, x_1=t$, so the eigenvector is \begin{bmatrix}1 \\ -1\end{bmatrix}
Therefore, the general solution is y=$c_1 \begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t}$ + $c_2 \begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$

(b)
from (a), we can know $\phi(t)= \begin{pmatrix} e^{2t} & e^{-t} \\ 2e^{2t} & -e^{-t} \end{pmatrix}$
then
$\begin{pmatrix} e^{2t} & e^{-t} \\ 2e^{2t} & -e^{-t} \end{pmatrix} \begin{bmatrix}u_1'\\ u_2'\end{bmatrix} = \begin{bmatrix} 0 \\ \frac{6e^{3t}}{e^{2t}+1}\end{bmatrix}$
By ref form, we can know $u_1'= \frac{6e^t}{e^2t+1}, u_2' = 0$, then by integrating, we can get $u_1=6arctan(e^t)+c_1, u_2=c_2$
then the  general solution is $x(t)=(6arctan(e^t)+c_1)\begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t} + c_2\begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$

xuanzhong:
Here's the solution for sketching.

Jingjing Cui:
$$a)det(A-\lambda I)=0\\ det\begin{vmatrix} 1-\lambda&2\\ 1&-\lambda\\ \end{vmatrix}=0\\ (1-\lambda)(-\lambda)-2=0\\ \lambda_1=-1 \;\; \lambda_2=2\\ (A-\lambda I)x=0\\ \\ when\; \lambda=-1\\ \begin{pmatrix} 2&2\\ 1&1\\ \end{pmatrix}-> \begin{pmatrix} 2&2\\ 0&0\\ \end{pmatrix}\\ so\; 2x_1+2x_2=0\\ let\; x_2=t\;\;\;then\;x_1=-t\\ so\;the\;corresponding\;eigenvector\;is\; (\begin{array}{cc} -1 \\ 1 \end{array})\\ when\; \lambda=2\\ \begin{pmatrix} -1&2\\ 1&-2\\ \end{pmatrix}-> \begin{pmatrix} -1&2\\ 0&0\\ \end{pmatrix}\\ so\; -x_1+2x_2=0\\ let\; x_2=t\;\;\;then\;x_1=2t\\ so\;the\;corresponding\;eigenvector\;is\; (\begin{array}{cc} 2 \\ 1 \end{array})\\ so\;the\;general\;solution\;is\;x=c_1e^{-t}(\begin{array}{cc} -1 \\ 1 \end{array})+c_2e^{2t}(\begin{array}{cc} 2 \\ 1 \end{array})\\$$

Jingjing Cui:
b)
$$\phi(t) U'(t)=g(t)\\ \begin{pmatrix} -e^{-t}&2e^{2t}\\ e^{-t}&e^{2t}\\ \end{pmatrix}(\begin{array}{cc} U_1' \\ U_2' \end{array})=(\begin{array}{cc} 0 \\\frac{6e^{3t}}{e^{2t}+1} \end{array})\\ -U_1'e^{-t}+2U_2'e^{2t}=0\\ U_1'e^{-t}+U_2'e^{2t}=\frac{6e^{3t}}{e^{2t}+1}\\ U_1'=\frac{4e^{4t}}{e^{2t}+1}\\ so\;U_1=2e^{2t}-2ln|e^{2t}+1|+C_1\\ U_2'=\frac{2e^{t}}{e^{2t}+1}\\ so\;U_2=2arctan(e^t)+C_2\\ x=\phi(t)U(t)\\ so\;the\;solution\;is\;:\\ x=(2e^{2t}-2ln|e^{2t}+1|+C_1)(\begin{array}{cc} -e^{-t} \\ e^{-t} \end{array})+(2arctan(e^t)+C_2)(\begin{array}{cc} 2e^{2t} \\ e^{2t} \end{array})\\$$

OK, except LaTeX sucks:

1) text should not be a part of math formulae or included like \text{blah blah}
2)  "operators" should be escaped: \cos, \sin, \tan, \ln