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### Messages - Yiheng Bian

Pages:  2
1
##### Quiz 3 / TUT5101 QUIZ3
« on: February 06, 2020, 12:34:14 AM »
$$\int_{\Upsilon}e^z dz$$
$$\text{line from 0 to } z_{0}$$
Therefore
$$r(t)= tz_{0}$$
$$r'(t) = z_{0} (0\leq t \leq 1)$$
And since
$$f(z)=e^z$$
$$f(r(t))= e^{z_{0}}$$
So
$$\int_{\Upsilon}e^z dz = \int_{0}^{1}f(r(t))r'(t)dt = \int_{0}^{1}e^{tz_{0}}z_{0} dt = z_{0}(\frac{1}{z_{0}}e^{z_{0}}- \frac{1}{z_{0}})= e^{z_{0}} - 1$$

2
##### Quiz 2 / Re: TUT5101 QUIZ2
« on: January 31, 2020, 11:31:11 PM »
Ok, I got it. Thank you sir

3
##### Quiz 2 / Re: TUT5101 QUIZ2
« on: January 30, 2020, 09:25:16 PM »
Sorry sir, , could you point it out?

4
##### Quiz 2 / TUT5101 QUIZ2
« on: January 30, 2020, 12:42:13 AM »
Question:
$$h(z)=\frac{Log(z)}{z}$$
We know when
$$z \rightarrow \infty$$
$$Log(z) \rightarrow \infty$$
So we can use L'Hopital rules and get
$$\frac{1}{z}$$
So
$$\text{limit is zero}$$

5
##### Quiz 1 / Re: TUT5101
« on: January 26, 2020, 04:25:25 PM »
I have modified it sir.

6
##### Quiz 1 / TUT5101
« on: January 26, 2020, 01:14:20 AM »
Find all solutions of the given equation:
$$z^8 =1$$
We know
$$z=r(\cos\theta +I\sin\theta)$$
$$z^8=r^8(\cos8\theta +I\sin8\theta)$$
Then
$$1=1+0i=1(\cos2k\pi + I\sin2k\pi)$$
So we have
$$r^8(\cos8\theta +I\sin8\theta) = 1(\cos2k\pi + I\sin2k\pi)$$
Therefore we get
$$r=1$$
$$8\theta=2k\pi$$
$$\theta= \frac{k\pi}{4}$$
So when
$$k=0, \theta=0,z=1(\cos0+i\sin0)=1$$
$$k=1, \theta=\frac{\pi}{4},z=1(\cos(\frac{\pi}{4})+I\sin{\frac{\pi}{4}})$$
$$k=2, \theta=\frac{\pi}{2},z=1(\cos(\frac{\pi}{2})+i\sin{\frac{\pi}{2}})$$
$$k=3, \theta=\frac{3\pi}{4},z=1(\cos(\frac{3\pi}{4})+i\sin{\frac{3\pi}{4}})$$
$$k=4, \theta=\pi,z=1(\cos\pi+i\sin\pi)$$
$$k=5, \theta=\frac{5\pi}{4},z=1(\cos(\frac{5\pi}{4})+i\sin{\frac{5\pi}{4}})$$
$$k=6, \theta=\frac{3\pi}{2},z=1(\cos(\frac{3\pi}{2})+i\sin{\frac{3\pi}{2}})$$
$$k=7, \theta=\frac{7\pi}{4},z=1(\cos(\frac{7\pi}{4})+i\sin{\frac{7\pi}{4}})$$

7
##### Term Test 2 / Re: Problem 1 (noon)
« on: November 19, 2019, 04:33:17 AM »
No double-dipping

(a):
We can solve homo firstly:
$$r^2-3r+2=0\\ (r-2)(r-1)=0\\ r_1=2,r_2=1$$
Therefore:
$$y=c_1e^{2t}+c_2e^t$$
So we can get:
$$W=\begin{vmatrix} e^{2t} & e^t \\ 2e^{2t} & e^t \end{vmatrix}=-e^{3t}\\ W_1=\begin{vmatrix} 0 & e^{t} \\ 1 & e^{t} \end{vmatrix}=-e^{t}\\ W_2=\begin{vmatrix} e^{2t} & 0 \\ 2e^{2t} & 1 \end{vmatrix}=e^{2t}$$
So we can get:
$$Y(t)=e^{2t}\int{\frac{-e^{s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds + e^{t}\int{\frac{e^{2s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds\\ Y(t)=e^{2t}\int{\frac{e^{s}}{e^{2s}+1}}ds - e^{t}\int{\frac{e^{2s}}{e^{2s}+1}}ds\\ Y(t)=e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)$$
Finally:
$$y(t)=c_1e^{2t}+c_2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)$$

(b):
So we can get y'(t):
$$y'=2c_1e^{2t}+c_2e^t+2e^{2t}arctan(e^t)+e^{2t}*\frac{e^t}{e^{2t}+1}-0.5e^t*ln(e^{2t}+1)-0.5e^t*\frac{e^{2t}}{e^{2t}+1}$$
We take y(0)=y'(0)=0,so we can get:
$$2c_1+2c_2+0.5\pi-ln2=0\\ 2c_1+c_2+0.5\pi-0.5ln2=0$$
So
$$c_1=-0.25\pi,c_2=0.5ln2$$
Therefore:
$$y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)$$

8
##### Term Test 2 / Re: Problem 4 (main sitting)
« on: November 19, 2019, 04:30:28 AM »
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} 3-\lambda & 3 \\ -2 & -1-\lambda \end{vmatrix}={\lambda -1}^2+2=0$$
WRONG So
$$\lambda_1=1 +\sqrt{2}i\\ \lambda_2=1-\sqrt{2}i$$
$$\text{when } \lambda=1 +\sqrt{2}i\\ \begin{vmatrix} 2-\sqrt{2}i & 3 \\ -2 & -2-\sqrt{2}i \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \end{vmatrix}$$
RREF:
$$\begin{pmatrix} 2-\sqrt{2}i & 3 \\ 0 & 0 \end{pmatrix} \quad = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$

Let x_2=t
So we can get:
$${(2-\sqrt{2}i})x_1=-3x_2=-3t\\ x_1=\frac{-3t}{2-\sqrt{2}i}$$
So
$$t*\begin{pmatrix} -1-\frac{\sqrt{2}i}{2} \\ 1 \end{pmatrix} \quad$$
Therefore:
$$e^{1+i\sqrt2t}\begin{pmatrix} -1-\frac{\sqrt{2}i}{2} \\ 1 \end{pmatrix} \quad = e^t[\begin{pmatrix} -cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\ cos(\sqrt{2}t) \end{pmatrix} \quad + i\begin{pmatrix} -sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\ sin(\sqrt{2}t) \end{pmatrix} \quad]$$
So, general solution:
$$y=c_1e^t\begin{pmatrix} -cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\ cos(\sqrt{2}t) \end{pmatrix} \quad + c_2e^t\begin{pmatrix} -sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\ sin(\sqrt{2}t) \end{pmatrix} \quad$$

9
##### Term Test 2 / Re: Problem 3 (main sitting)
« on: November 19, 2019, 04:29:59 AM »
(a):
We solve homo firstly:
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} 1-\lambda & 1 \\ -2 & 4-\lambda \end{vmatrix}=-5\lambda+{\lambda}^2+6=0\\ (\lambda-2)(\lambda-3)=0\\ \lambda_1=2,\lambda_2=3$$
Then:
$$(A-{\lambda}I)x=0$$
$$When \lambda=2$$
$$\begin{pmatrix} -1 & 1 \\ -2 & 2 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
RREF:
$$\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
$$\text{Let x_2=t, so }x_1=x_2=t\\ t*\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad$$
$$When \lambda=3$$
$$\begin{pmatrix} -2 & 1 \\ -2 & 1 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
RREF:
$$\begin{pmatrix} 2 & -1 \\ 0 & 0 \end{pmatrix} \quad= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
$$\text{Let x_2=t, so }x_1=0.5t, x_2=t\\ t*\begin{pmatrix} 1 \\ 2 \end{pmatrix} \quad$$
So, the general solution is :
$$y= c_1e^{2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad +c_2e^{3t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} \quad$$

(b):
$$\phi = \begin{pmatrix} e^{2t} & e^{3t} \\ e^{2t} & 2e^{3t} \end{pmatrix} \quad$$
$$\phi * u' = g(t)$$
$$\begin{pmatrix} e^{2t} & e^{3t} \\ e^{2t} & 2e^{3t} \end{pmatrix} \quad *{\begin{pmatrix} u_1' \\ u_2' \end{pmatrix} \quad}=\begin{pmatrix} \frac{e^{4t}}{e^{2t} + 1 } \\ 0 \end{pmatrix} \quad$$
Simplify we can get:
$$u_1'=-\frac{-e^t}{e^{2t} + 1}\\ u_2'=\frac{2e^{2t}}{e^{2t}+1}$$
Therefore:
$$u_1=ln(e^{2t}+1)+c_1\\ u_2=-arctane^t+c_2$$
Finally:
$$x=\phi * u=(ln(e^{2t}+1)+c_1)*{\begin{pmatrix} e^{2t} \\ e^{2t} \end{pmatrix} \quad} +(-arctane^t+c_2)*\begin{pmatrix} e^{3t} \\ 2e^{3t} \end{pmatrix} \quad$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

10
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 04:29:25 AM »
(a):
$$W=ce^{-\int{p(t)}}dt\\ W=ce^{-\int{4}dt}\\ W=ce^{-4t}$$

(b):
We can get:
$$r^3+4r^2+r-6=0\\ (r-1)(r+2)(r+3)=0\\ r_1=1,r_2=-2,r_3=-3$$
Therefore:
$$y=c_1e^t+c_2e^{-2t}+c_3e^{-3t}$$
$$\begin{vmatrix} e^t & e^{-2t} & e^{-3t} \\ e^t & -2e^{-2t} & -3e^{-3t} \\ e^t & 4e^{-2t} & 9e^{-3t} \end{vmatrix}=-12e^{-4t}$$
compare with (a) we get
$$c=-12$$

(c):
Let:
$$y_p(t)=Ate^t\\ y'=A(e^t+te^t)\\ y''=A(2e^t+te^t)\\ y'''=A(3e^t+te^t)$$
We take these into equation and get:
$$A(3e^t+te^t)+4A(2e^t+te^t)+A(e^t+te^t)-6Ate^t=24e^t\\ 12Ae^t=24e^t\\ A=2$$
So:
$$y_p(t)=2te^t$$
Therefore:
$$y=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$$

OK. V.I.

11
##### Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 04:29:03 AM »
(a):
$$r^2+4=0\\ r_1=2i,r_2=-2i$$
So we can get:
$$y_c(t)=c_1cos2t+c_2sin2t$$
$$W=\begin{vmatrix} cos2t & sin2t \\ -2sin2t & 2cos2t \end{vmatrix}=2$$
$$W_1=\begin{vmatrix} 0 & sin2t \\ 1 & 2cos2t \end{vmatrix}=-sin2t$$
$$W_2=\begin{vmatrix} cos2t & 0 \\ -2sin2t & 1 \end{vmatrix}=cos2t$$
Therefore:
$$Y(t)=cos2t\int{\frac{-sin2s*\frac{1}{(cos(s))^2}}{2}}ds+sin2t\int{\frac{cos2s*\frac{1}{(cos(s))^2}}{2}}ds\\ Y(t)=cos2t\int{-\frac{sins}{coss}}ds+sin2t\int{\frac{2(cos(s))^2-1}{2(cos(s)^2)}}ds\\ Y(t)=cost2t*lncost+sin2t*(t-0.5tant)$$
So general solution is :
$$y(t)=c_1cos2t+c_2sin2t+cost2t*lncost+sin2t*(t-0.5tant)$$

(b):
$$y'(t)=-2c_1sin2t+2c_2cos2t+cos2t*\frac{-sint}{cost} - 2sin2t*lncost+sin2t(1-0.5(sect)^2)+2cos2t(t-0.5tant)$$
$$\text{Take } y(0)=y'(0)=0\\ \text{We can get that: }c_1=c_2=0$$
So finally:
$$y=cost2t*lncost+sin2t*(t-0.5tant)$$

OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
$$\boxed{ y= \ln (\cos(t)) \cos(2t) + \Bigl(t-\frac{1}{2}\tan (t)\Bigr)\sin(2t). }$$

12
##### Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 04:28:39 AM »
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} 2-\lambda & -3 \\ 4 & -2-\lambda \end{vmatrix}={\lambda}^2+8=0$$
So
$$\lambda_1=\sqrt{8}i\\ \lambda_2=-\sqrt{8}i$$
$$\text{when } \lambda=\sqrt{8}I\\ \begin{vmatrix} 2-\sqrt{8}i & -3 \\ 4 & -2-\sqrt{8}i \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \end{vmatrix}$$
RREF:
$$\begin{pmatrix} 2-\sqrt{8}i & -3 \\ 0 & 0 \end{pmatrix} \quad = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
Let x_2=t
So we can get:
$${(2-\sqrt{8}i})x_1=3x_2=3t\\ x_1=\frac{3t}{2-\sqrt{8}i}$$
So
$$t*\begin{pmatrix} \frac{3}{2-\sqrt8i} \\ 1 \end{pmatrix} \quad$$
Therefore:
$$e^{i\sqrt8t}\begin{pmatrix} \frac{3}{2-\sqrt8i} \\ 1 \end{pmatrix} \quad = (cos(\sqrt8t)+isin(\sqrt8t))\begin{pmatrix} \frac{3}{2-\sqrt8i} \\ 1 \end{pmatrix} \quad=\begin{pmatrix} \frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} +\frac{isin(\sqrt8t)+i\sqrt2cos(\sqrt8t)}{2}\\ cos(\sqrt8t )+ isin(\sqrt8t)) \end{pmatrix} \quad$$
So
$$y=c_1\begin{pmatrix} \frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} \\ cos(\sqrt8t) \end{pmatrix} \quad +c_2\begin{pmatrix} \frac{sin(\sqrt8t)+\sqrt2cos(\sqrt8t)}{2}\\ sin(\sqrt8t)) \end{pmatrix} \quad$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

13
##### Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 04:28:19 AM »
(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:
$$x=\phi * u=(-0.5ln(t^2+1)+c_1)*{\begin{pmatrix} e^{-t} \\ e^{-t} \end{pmatrix} \quad} +(arctant+c_2)*\begin{pmatrix} te^{-t} \\ e^{-t}(t+1) \end{pmatrix} \quad$$

14
##### Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 04:27:56 AM »
No double-dipping
(a):
$$W=ce^{-\int{p(t)}}dt\\ W=ce^{-\int{-2}dt}\\ W=ce^{2t}$$

(b):
We can get:
$$r^3-2r^2+4r-8=0\\ (r-2)(r^2+4)=0\\ r_1=2,r_2=2i,r_3=-2i$$
Therefore:
$$y=c_1e^2t+c_2sin2t+c_3cos2t$$
$$\begin{vmatrix} e^2t & sin2t & cos2t \\ 2e^2t & 2cos2t & -2sin2t \\ 4e^2t & -4sin2t & -4cos2t \end{vmatrix}=e^{2t}({-8(cos2t)^2}-8(sin2t)^2)-sin2t(-8e^{2t}cos2t+8e^{2t}sin2t)+cos2t(-8e^{2t}sin2t-8e^{2t}cos2t)=-8e^{2t}-8e^{2t}=-16e^{2t}$$
compare with (a) we get
$$c=-16$$

(c):
Let
$$y=Acost+Bsint$$
Therefore:
$$y'=-Asint+Bcost\\ y''=-Acost-Bsint\\ y'''=Asint-Bcost$$
Next We take these into equation and get:
$$(-3A-6B)sint+(3B-6A)cost=15cost\\ -3A-6B=0:3B-6A=15\\ A=-2,B=1$$
Finally:
$$y=c_1e^2t+c_2sin2t+c_3cos2t-2cost+sint$$

15
##### Term Test 2 / Re: Problem 1 (morning)
« on: November 19, 2019, 04:27:31 AM »
You should not double-dip. V.I.
The first question
we solve homo firstly, so we get:
$$r^2 - 1=0\\ r^2 =1\\ r_1=1\\ r_2=-1\\ \text{Therefore: }y=c_1e^t+c_2e^{-t}$$
So we can get
$$W = \begin{vmatrix} e^t & e^{-t} \\ e^t & -e^{-t} \end{vmatrix}=-2\\ W_1=\begin{vmatrix} 0 & e^{-t} \\ 1 & -e^{-t} \end{vmatrix}=-e^{-t}\\ W_2=\begin{vmatrix} e^t & 0 \\ e^t& 1 \end{vmatrix}=e^t$$
So we can get
$$Y(t)=e^t\int{\frac{-e^{-s}*\frac{12}{e^s+1}}{-2}}ds + e^{-t}\int{\frac{e^{s}*\frac{12}{e^s+1}}{-2}}ds\\ Y(t)=6e^t\int{\frac{e^{-s}}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{1}{e^s*(e^s+1)}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{e^s+1-e^s}{e^s*(e^s+1)}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{1}{e^s}-\frac{1}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{1}{e^s}-\frac{e^s+1-e^s}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\$$
Finally:
$$Y(t)=6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)\\ y(t)=c_1e^t+c_2e^{-t}+6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)$$

The second question:
since we have y(t), so we can get y'(t)
$$y'(t)=c_1e^t-c_2e^{-t}+6e^t[-e^{-t}-t+ln(e^t+1)]+6e^t[e^{-t}-1+\frac{e^t}{e^t+1}]+6e^{-t}ln(e^t+1)-6e^{-t}*\frac{e*t}{e^t+1}$$
Then we take y(0)=y'(0)=0 into y and y'
Therefore:
$$c_1+c_2-6=0\\ c_1-c_2-6+12ln2=0\\ c_1=6-6ln2, c_2=6ln2$$
So
$$y(t)=(6-6ln2)e^t+6ln2e^{-t}6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)$$

Pages:  2