MAT244--2019F > Term Test 2

Problem 3 (morning)

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

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

Yiheng Bian:
(a):
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} -2-\lambda & 1 \\ -1 & -\lambda \end{vmatrix}=2\lambda+{\lambda}^2+1=0$$
So
$$(\lambda+1)^2=0\\ \lambda_1=\lambda_2=-1$$
Then:
$$(A-{\lambda}I)x=0$$
$$\begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
RREF
$$\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
$$\text{if let }x_2=t\\ x_1=x_2=t$$
So
$$t*{\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad}$$
Since we just have only one eigenvector:
$$\begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix} \quad= \begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad$$
$$\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix} \quad= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad$$
$$\text{So the other eigenvector is:}= \begin{pmatrix} 0 \\ 1 \end{pmatrix} \quad$$
Therefore:
$$y=c_1e^{-t}*{\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad}+c_2e^{-t}[t*{\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad} +{\begin{pmatrix} 0 \\ 1 \end{pmatrix} \quad}]$$

(b):
$$\phi = \begin{pmatrix} e^{-t} & te^{-t} \\ e^{-t} & e^{-t}(t+1) \end{pmatrix} \quad$$
$$\phi * u' = g(t)$$
$$\begin{pmatrix} e^{-t} & te^{-t} \\ e^{-t} & e^{-t}(t+1) \end{pmatrix} \quad *{\begin{pmatrix} u_1' \\ u_2' \end{pmatrix} \quad}=\begin{pmatrix} 0 \\ \frac{e^{-t}}{t^2 + 1 } \end{pmatrix} \quad$$
Simplify we can get:
$$u_1'=-\frac{t}{t^2 + 1}\\ u_2'=\frac{1}{t^2+1}$$
Therefore:
$$u_1=-0.5ln(t^2+1)+c_1\\ u_2=arctant+c_2$$
Finally:
$$x=\phi * u=(-0.5ln(t^2+1)+c_1)*{\begin{pmatrix} e^{-t} \\ e^{-t} \end{pmatrix} \quad} +(arctant+c_2)*\begin{pmatrix} te^{-t} \\ e^{-t}(t+1) \end{pmatrix} \quad$$

xuanzhong:
HERE'S THE SOLUTION FOR SKETCHING.

Ruojing Chen:
a)$$x'= \left( \begin{matrix} -2 & 1 \\ -1 & 0 \\ \end{matrix} \right )x+\left( \begin{matrix} 0 \\ \frac{e^{-t}}{t^2+1} \\ \end{matrix} \right )$$

$$det(A-\lambda I)=0$$
$$(-2-\lambda)(-\lambda)+1=0$$
$$\lambda^2+2\lambda+1=0$$
$$\lambda_1=\lambda_2=-1$$

$$when \lambda =-1$$
$$(A-\lambda I)x=0$$
$$\left( \begin{matrix} -1 & 1 \\ -1 & 1 \\ \end{matrix} \right )\left( \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right )=\left( \begin{matrix} 0 \\ 0 \\ \end{matrix} \right )$$
$$x_1=x_2 \Rightarrow x=t\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )$$

$$\left( \begin{matrix} -1 & 1 \\ -1 & 1 \\ \end{matrix} \right )\left( \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right )=\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )$$
$$x_1+x_2=1 \Rightarrow x=t\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right )$$

$$\therefore y=c_1e^{-t}\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )+c_2e^{-t}(\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )t+\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right ))$$

B)$$\phi u'=g(t)$$
$$\left( \begin{matrix} e^{-t} & e^{-t}t \\ e^{-t} & e^{-t}t+e^{-t}\\ \end{matrix} \right ) \left( \begin{matrix} u_1' \\ u_2' \\ \end{matrix} \right )=\left( \begin{matrix} 0 \\ \frac{e^{-t}}{t^2+1} \\ \end{matrix} \right )$$
$$\left \{ \begin{array}{lr} u_1'=-\frac{t}{t^2+1} &\\ u_2'=\frac{1}{t^2+1} \end{array} \right.$$
$$\Rightarrow \left \{ \begin{array}{lr} u_1=-\frac{1}{2} \ln(t^2+1)&\\ u_2=\arctan t \end{array} \right.$$
$$x=\phi u$$
$$\therefore x=\left( \begin{matrix} e^{-t} & e^{-t}t \\ e^{-t} & e^{-t}t+e^{-t}\\ \end{matrix} \right )\left( \begin{matrix} -\frac{1}{2}\ln(t^2+1)\\ \arctan t\\ \end{matrix} \right )$$

$$x=(-\frac{1}{2}\ln(t^2+1)+c_1)\left( \begin{matrix} e^{-t} \\ e^{-t} \\ \end{matrix} \right )+(c_2+\arctan t)\left( \begin{matrix} e^{-t}t \\ e^{-t}t+e^{-t} \\ \end{matrix} \right)$$

OK, but LaTeX sucks:
1) \det should be escaped as well
2) Text should not be  included in math formulae, or included  through  \tex{blah blah } to make it upright and properly spaced

3) Directions are opposite

Yiran Wang:
Det(A-rI)=0
(-2-r)(-r)+1=2r+r^2+1=0
r1=r2=-1
(A-rI)=0
we have eigenvector (1 1)
Let (A-rI)=(1 1)
We have another eignvector (-1 0)
We have x=C1e^-t *(1 1)+ C2e^-t *[ t(1 1)+(-1 0)]
e^-t u1+(te^-t – e^-t)u2=0   e^-t u1+te^-t u2=e^-t/(t^2+1)
we have u1=-t/(t^2+1)
u2=1/(t^2+1)
U1=-0.5ln(t^2+1)+C1
U2=arctant+C2
X=U1(e^-t  e^-t)+ U2(te^-t  te^-t +e^-t)