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Messages - Yuan Bian

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1
Quiz 5 / Re: Quiz 5
« on: December 02, 2014, 03:36:27 PM »
C)
nonlinear: near (0,0), saddle point, unstable;
                 near (2,1), node, AS;
                 near (-2,1), center or spiral point, indeterminate
                 near (-2, -4), spiral point, unstable

2
TT2 / Re: TT2 #2
« on: November 20, 2014, 08:58:13 PM »
I agree with you that typed version is better. As math students, I think many students wants to get bonus by answering online question and using typed version. but situation now, it's few of us can type math formula using latex(? or something else....),
As a math course, I think we should focus on math idea and correctness.
and I think copy my answer above to get bonus is not fair, and even not sure I get bonus or not.   

3
TT2 / Re: TT2 #1
« on: November 19, 2014, 10:53:37 PM »
homo:  r(r-1)(r-2)-3r(r-1)+6r-6=0
           (r-1)(r-2)(r-3)=0
r1=1,r2=2,r3=3
l(r)=r3-6r2+11r-6
let t=lnx
l(t)=t'''-6t''+11t'-6t=6t-24e-t
let yp1=At+B, yp2=Ce-t
so Ce-t(l(-1))=-24e-t, (l(0)(At+B)+l'(0)A)=6t
C=1, B=-11/6, A=-1
y=c1x+c2x2+c33+1/x-lnx-11/6
y(1)=−5/6 , y′(1)=−2 , y′′(1)=2
s0 c1+c2+c3+1-11/6=-5/6
    c1+2c2+3c3-2=-2
    2c2+6c3+3=2
get c1=-1/2, c2=1, c3=-1/2
y=-x3/2+x2-x/2+1/x-lnx-11/6

4
TT2 / Re: TT2 # 3
« on: November 19, 2014, 10:33:18 PM »
x′=(−1 −4)(x) .
y'=(1    -1) (y)

find eigenvalues

det(A−rI)=r2+2r+5=0
r1=-1+2i
r2=-1-2i
then, find eigenvectors

(-2i -4)   v1=(2i)
(1  -2i)         (1)



(2i -4)   v2=(2i)
(1  2i)         (-1)
x(t)=C1e(−1+2i)t(2i)+C2e−(-1-2i)t(2i)
                          (1)                      (-1)
 
stable spiral point
-4<0, counterclockwise

5
TT2 / Re: TT2 #2
« on: November 19, 2014, 10:31:33 PM »
different c1 and c2... ::)
and c2 is still in your final answer.

6
TT2 / Re: TT2 #2
« on: November 19, 2014, 10:19:36 PM »
x'= (0  1)
y'= (2  1)
r2-r-2=0
(r-2)(r+1)=0
r1=2, r2=-1
r1>0> r2
unstable saddle
(x)=c1e2t(1)+c2e-t(1)
(y)          (2)          (-1)
2b) c1+c2=2
      2c1-c2=1
so c1=1,c2=1
(x)=e2t(1)+e-t(1)
(y)       (2)     (-1)

7
MT / Re: MT problem 3
« on: November 18, 2014, 04:47:25 PM »
hi, prof
In the official soln of q3, I think there is a tiny mistake:
First, we set yp = Ax2 ln x + B
but after finding A and B, yp =(1/17)x2 ln x - (25/289)x2.... :o

8
Quiz 4 / Re: Q4 problem 2
« on: November 13, 2014, 09:13:45 PM »
from prof Victor Ivrii : "Why you just don't learn a very simple criteria for the center or focus: if the top-right element of 2×2 matrix is >0 the orientation is clock-wise; otherwise it is counter-clock-wise."

9
Quiz 4 / Re: Q4 problem 1
« on: November 13, 2014, 08:55:11 PM »
no, two distinct real eigenvalues both >0, it's unstable node; two distinct real eigenvalues both <0, then it's stable node.

10
Quiz 4 / Q4 problem 2
« on: November 12, 2014, 09:03:38 PM »
Q2: 7.6 #6
r2+9=0
r1=3i, r2=-3i
pure imaginary eigenvalue, so it's stable center
also whole soln is in the q1 fisrt post.

11
Quiz 4 / Re: Q4
« on: November 12, 2014, 08:56:52 PM »
Q1: 7.5 #5
r2+4r+3=0
(r+3)(r+1)=0
r1=-3, r2=-1
b/c r1<r2<0, stable node
sorry I don't know how to upload picture...I tried, but I failed
now I share the link of picture in the first post


12
Quiz 4 / Q4 problem 1
« on: November 12, 2014, 08:56:10 PM »
7.5 p 407 # 5

Draw a full phase portrait and describe completely the type of the stationary point  indicating if it is stable or unstable and in the case of the center and focus indicate orientation (clockwise or counter-clockwise)
\begin{equation*}
\textbf{x}'=\begin{pmatrix}
-2 & \hphantom{-}1\\  \hphantom{-}1 &-2
\end{pmatrix}\textbf{x}\ .
\end{equation*}




sorry two picture can't see normally, there are two links of picture of q1 and q2
http://www.imagebam.com/image/f84041363999992
http://www.imagebam.com/image/64d11b364003377

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