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### Messages - RunboZhang

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1
##### Quiz 7 / LEC5101-Quiz7-FIVE-E
« on: December 08, 2020, 07:55:16 PM »
$\textbf{Problem: } \\$
$\text{Using argument principle along line on the picture, calculate the number of zeroes of the following function in the upper half-plane.}$ $$2z^4-2iz^3+z^2+2iz-1$$
$\text{(The contour is attached in the picture below)}$

$\textbf{Solution: }\\$
$\text{Denoting the contour } c_1 ,\ c_2 \text{ as indicated in the picture below.}$

$\text{Firstly, discuss contour }c_1 \text{. On }c_1 \text{, we have: }$ $$z=x ,\ x:-R \rightarrow R$$

$\text{And f(z) becomes}$ $$f(z) = 2x^4-2ix^3+x^2+2ix-1=(2x^4+x^2-1)+i(-2x^3+2x)$$

$\text{When } x \in [-R,1]$ $$Re(f(z))>0 ,\ Im(f(z))<0 ,\ \text{and } \frac{d}{dx}{Re(f(z))}>\frac{d}{dx}{Im(f(z))}$$
$\text{Hence, as Real part is growing much faster than Imaginary part in interval } x \in [-R,1] ,\ arg(f(z)) \rightarrow 0 \text{ in the first quadrant.}$

$\text{Moving forward, as x increases from -1 }$
$$\text{when } x \in (-1, -0.707) ,\ Re>0 ,\ Im<0 \Rightarrow 4^{th} quadrant \\ \text{when } x \in (-0.707,0),\ Re<0 ,\ Im<0 \Rightarrow 3^{rd} quadrant\\ \text{when } x \in (0,0.707),\ Re<0 ,\ Im>0 \Rightarrow 2^{nd} quadrant\\ \text{when } x \in (0.707,1) ,\ Re>0 ,\ Im>0 \Rightarrow 1^{st} quadrant$$

$\text{The change of Real part and Imaginary part makes f(z) travels from forth to third to second to first quadrant.}$

$\text{When } x \in (1, R),\ Re>0 ,\ Im<0 \text{, thus f(z) goes to the } 4^{th} \text{ quadrant.}$

$\text{As } x \rightarrow R ,\ \frac{d}{dx}{Re}>\frac{d}{dx}{Im} \Longrightarrow arg(f(z)) \rightarrow 0 \text{ from the forth quadrant.}$

$\text{Therefore, in contour } c_1 \text{, the total change of argument is } -2\pi \text{ as f(z) rotates clockwise.}$

$\text{Then observe the half-arc contour }c_2 \text{, we have }$ $$z=Re^{it} ,\ t \in (0, \pi)$$

$\text{And f(z) becomes }$
\begin{gather} \begin{aligned} f(z) &= 2R^{4}e^{4it}-2iR^{3}e^{3it}+R^{2}e^{2it}-1 \\ &=2R^{4}e^{4it}\cdot(1-\frac{ie^{-it}}{R} +\frac{e^{-2it}}{2R^2}-\frac{e^{-4it}}{2R^4}) \\ &\rightarrow 2R^{4}e^{4it} \ \ \text{as } R\rightarrow \infty \end{aligned} \end{gather}

$\text{Thus as t changes from 0 to }\pi ,\ 2R^{4}e^{4it} \text{ changes from 0 to } 4\pi \text{. And the change of argument in contour } c_2 \text{ is }4\pi$

$\text{Therefore, by the Argument Principle, the number of zeroes equals to }$ $$\frac{1}{2\pi} (-2\pi +4\pi) = 1$$
$\text{only 1 zero.}$

2
##### Test 4 / LEC0201-TT4-ALF-F-Q2
« on: December 03, 2020, 02:20:22 PM »
$\textbf{Problem1:} \\$
$\textbf{(a)} \text{ Find the general solution of }$ $$x'=\begin{bmatrix} 13 & -9\\ 6 & -8 \end{bmatrix}x$$
$\text{classify fixed point (0, 0) and sketch trajectories.} \\$
$\textbf{(b)} \text{ Find the general solution}$ $$x'=\begin{bmatrix} 13 & -9\\ 6 & -8 \end{bmatrix}x + \begin{bmatrix} 0\\ \frac{150e^{25t}}{e^{30t}+1} \end{bmatrix}x$$

$\textbf{Solution:} \\$
$\textbf{(a)}$

$\text{Let }$ $$A=\begin{bmatrix} 13 & -9\\ 6 & -8 \end{bmatrix}$$

$\text{Then }$ $$det(A-\lambda I) = (13-\lambda)(-8-\lambda)-6\cdot(-9)=0$$

$\text{Solve for } \lambda:$ $$\lambda_1 = 10 , \lambda_2 = -5$$

$\text{When }\lambda = 10,$ $$A-\lambda I = \begin{bmatrix} 3 & -9\\ 6 & -18 \end{bmatrix} \xrightarrow{\text{ref}}\begin{bmatrix} 1 & -3\\ 0 & 0 \end{bmatrix}$$
$\text{Hence the eigenvector is }$ $$\vec{v_1}=\begin{bmatrix} 3\\ 1 \end{bmatrix}$$

$\text{When }\lambda = -5,$ $$A-\lambda I = \begin{bmatrix} 18 & -9\\ 6 & -3 \end{bmatrix} \xrightarrow{\text{ref}}\begin{bmatrix} 2 & -1\\ 0 & 0 \end{bmatrix}$$
$\text{Hence the eigenvector is }$ $$\vec{v_2}=\begin{bmatrix} 1\\ 2 \end{bmatrix}$$

$\text{Therefore we have }$ $$x=c_1e^{10t}\begin{bmatrix} 3\\ 1 \end{bmatrix} + c_2e^{-5t}\begin{bmatrix} 1\\ 2 \end{bmatrix}$$

$\text{(graph is attached below)}$

$\textbf{(b)}$

$\text{Let }$ $$\varphi (t)=\begin{bmatrix} 3e^{10t} & e^{-5t}\\ e^{10t} & 2e^{-5t} \end{bmatrix}, u'(t) = \begin{bmatrix} u_1'(t)\\ u_2'(t) \end{bmatrix}$$

$\text{Thus we have }$ $$\left[ \begin{array}{cc|c} 3e^{10t} & e^{-5t} & 0 \\ e^{10t} & 2e^{-5t} & \frac{150e^{25t}}{e^{30t}+1} \end{array}\right] \xrightarrow{R_2=3R_2-R_1} \left[ \begin{array}{cc|c} 3e^{10t} & e^{-5t} & 0 \\ 0 & 5e^{-5t} & \frac{450e^{25t}}{e^{30t}+1} \end{array}\right]$$

$\text{Observe the second row, we can get }$
\begin{gather} \begin{aligned} 5e^{-5t}\cdot u_2'(t) &= \frac{450e^{25t}}{e^{30t}+1} \\ \\ u_2'(t) &= \frac{90e^{30t}}{e^{30t}+1}\\ \\ u_2(t) &=\int{\frac{90e^{30t}}{e^{30t}+1}} \,dt \\ \\ &= 90\int{\frac{e^{30t}}{e^{30t}+1}} \,dt \ \ \ \text{(integral by substitution)}\\ \\ &= 3ln(e^{30t}+1)+c_2 \end{aligned} \end{gather}

$\text{Then compute the first row}$
\begin{gather} \begin{aligned} 3e^{10t}u_1'(t)+e^{-5t}u_2'(t) &= 0 \\ \\ 3e^{10t}u_1'(t)+e^{-5t}\frac{90e^{30t}}{e^{30t}+1} &= 0 \\ \\ u_1 &= -30\int{\frac{e^{15t}}{e^{30t}+1}} \,dt \ \ \ \text{(integral by substitution)}\\ \\ &=-30\int{\frac{\frac{1}{15}}{u^2+1}} \,du \\ \\ &= -2arctan(e^{15t})+c_1 \end{aligned} \end{gather}

$\text{Therefore }$
\begin{gather} \begin{aligned} x&=\varphi{(t)}\cdot u(t) \\ \\ &= \begin{bmatrix} 3e^{10t} & e^{-5t}\\ e^{10t} & 2e^{-5t} \end{bmatrix} \cdot \begin{bmatrix} -2arctan(e^{15t})+c_1\\ 3ln(e^{30t}+1)+c_2 \end{bmatrix} \\ \\ &= c_1\begin{bmatrix} 3e^{10t}\\ e^{10t} \end{bmatrix} +c_2\begin{bmatrix} e^{-5t}\\ 2e^{-5t} \end{bmatrix} + \begin{bmatrix} -6e^{10t}arctan(e^{15t})+3e^{-5t}ln(e^{30t}+1)\\ -2e^{10t}arctan(e^{15t})+6e^{-5t}ln(e^{30t}+1) \end{bmatrix} \end{aligned} \end{gather}

3
##### Test 4 / LEC0201-TT4-ALF-F-Q2
« on: December 03, 2020, 01:24:53 PM »
$\textbf{Problem2:}$

$\text{Find the general solution of }$ $$x' = \begin{bmatrix} -3 & 25\\ -9 & 3 \end{bmatrix} x$$
$\text{classify fixed point (0,0) (type, is stable or not, orientation if applicable) and sketch trajectories.}$

$\textbf{Solution:}$

$\text{Let }$ $$A= \begin{bmatrix} -3 & 25\\ -9 & 3 \end{bmatrix}$$

$\text{Then we have}$ $$det(A-\lambda I) = (-3-\lambda)(3-\lambda)-25\cdot(-9)=0$$
$\text{Solve for } \lambda ,$  $$\lambda_1 = 6 \sqrt{6}i ,\ \lambda_2=-6\sqrt{6}i$$

$\text{Take } \lambda = 6\sqrt{6}i \text{, then }$
$$A-\lambda I = A-6\sqrt{6}iI=\begin{bmatrix} -3-6\sqrt{6}i & 25\\ -9 & 3-6\sqrt{6}i \end{bmatrix} \xrightarrow{\text{ref}} \begin{bmatrix} -3-6\sqrt{6}i & 25\\ 0 & 0 \end{bmatrix}$$

$\text{Hence we have }$ $$null(A-\lambda I) = \begin{bmatrix} 25 \\ 3+6\sqrt{6}i \end{bmatrix}$$

$\text{and}$ $$x=e^{6\sqrt{6}it}\begin{bmatrix} 25 \\ 3+6\sqrt{6}i \end{bmatrix}$$

$\text{Further expand it and we get }$

\begin{gather} \begin{aligned} x &= [cos(6\sqrt{6}t)+isin(6\sqrt{6}t)]\begin{bmatrix} 25 \\ 3+6\sqrt{6}i \end{bmatrix} \\\\ &=\begin{bmatrix} 25cos(6\sqrt{6}t)+i25sin(6\sqrt{6}t) \\ 3cos(6\sqrt{6}t)+i3sin(6\sqrt{6}t)+i6\sqrt{6}cos(6\sqrt{6}t)-6\sqrt{6}sin(6\sqrt{6}t) \end{bmatrix} \\\\ &=c_1 \begin{bmatrix} 25cos(6\sqrt{6}t)\\ 3cos(6\sqrt{6}t)-6\sqrt{6}sin(6\sqrt{6}t) \end{bmatrix} + c_2 \begin{bmatrix} 25sin(6\sqrt{6}t) \\ 3sin(6\sqrt{6}t)+6\sqrt{6}cos(6\sqrt{6}t) \end{bmatrix} \end{aligned} \end{gather}

$\text{(graph is attached in the pic below)}$

4
##### Test 3 / LEC5101-TT3-ALT-F-Q2
« on: November 27, 2020, 08:03:20 PM »
$\textbf{Problem2:} \\ \\ \text{Find the power series expansion at z=0 of}$
$$f(z) = \frac{1}{z} \int_{0}^{z}{\frac{1}{z}\int_{0}^{z}{cos(w^2)\,dw}\,dz}$$
$\text{What is the radius of convergence?}$

$\textbf{Solution:} \\ \\ \text{Firstly, we have power series,}$
$$cos(w) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{2n}}{(2n)!}} \\$$
$\text{with a radius of convergence of } \infty \\$

$\text{Then substitute } w \text{ by } w^2$
$$cos(w^2) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{4n}}{(2n)!}}$$

$\text{Now substitute the series in the integral and calculate the integral}$

\begin{gather} \begin{aligned} f(z) &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \int_{0}^{z}{cos(w^2)\,dw}\,dz \\\\ &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}(\int_{0}^{z} {\sum_{n=0}^{\infty}{\frac{(-1)^{n}w^{4n}}{(2n)!}\,}}\,dw) \,dz \\\\ &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}\sum_{n=0}^{\infty}{ \int_{0}^{z}{(-1)^{n} \cdot \frac{w^{4n}}{(2n)!}}\,dw}\,dz \\\\ &=\frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \sum_{n=0}^{\infty}{(-1)^{n}\cdot \frac{z^{4n+1}}{(2n)! \cdot (4n+1)}}\,dz \\\\ &= \frac{1}{z} \int_{0}^{z}{\sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}} \,dz\\\\ &= \frac{1}{z} \sum_{n=0}^{\infty}{\int_{0}^{z}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}}\,dz \\\\ &= \frac{1}{z} \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n+1}}{(2n)! \cdot (4n+1)^{2}}} \\\\ &= \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}} \end{aligned} \end{gather}

$\text{Hence the power series expansion is }$
$$f(z) = \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}$$

$\text{Since the radius of convergence of } cos(z) \text{ is } \infty \text{. Thus } f(z) \text{ has the same radius of convergence of } R=\infty.$

5
##### Test 3 / LEC5101-TT3-ALT-F-Q1
« on: November 27, 2020, 07:40:12 PM »
$\textbf{Problem 1:} \\\\ \text{Using Cauchy's Integral Formula or Residue Theorem to calculate}$
$$I:= \int_{\gamma}{\frac{z-1}{z^2-4z+5}} \,dz$$
$\text{where } \gamma \text{ is a contour on the picture with the shown direction; poles are shown like colored dots.}$

$\textbf{Solution:} \\\\ \text{Since contour } \gamma \text{ is not a simple closed curve, then denote the contour } c_1 ,\ c_2 \text{ as following (pic is attached below)}$

$\text{Let }$ $$f(z) = z^2-4z+5 = 0$$

$\text{Then }$$z=\frac{4 \pm \sqrt{4^2-4\cdot5}}{2} = 2 \pm i$$$\text{For clockwise, simple closed contour } c_2 \text{, there is no pole. By Cauchy Theorem: }$$$\int_{c_2}{\frac{z-1}{z^2-4z+5}} \,dz = 0$$$\text{For clockwise, simple closed contour } c_1 \text{, there is two poles} z_1 = 2 + i ,\ z_2 = 2-i \text{. Then by Residual Theorem, we have:}
\begin{gather}
\begin{aligned}

\int_{c_1}{\frac{z-1}{z^2-4z+5}} \,dz &= -2 \pi i \sum{Res} \\\\
&= -2 \pi i \cdot (\frac{z-1}{z-2+i}|_{z=2+i} + \frac{z-1}{z-2-i}|_{z=2-i}) \\\\
&= -2 \pi i \cdot (\frac{1+i}{2i} + \frac{1-i}{-2i}) \\\\
&= -2 \pi i

\end{aligned}
\end{gather}
\text{Therefore we have}\begin{gather}
\begin{aligned}
\int_{\gamma}{\frac{z-1}{z^2-4z+5}} \,dz &=\int_{c_1}{\frac{z-1}{z^2-4z+5}} \,dz + \int_{c_2}{\frac{z-1}{z^2-4z+5}} \,dz \\\\ &= -2\pi i
\end{aligned}
\end{gather}
$6 ##### Quiz 6 / LEC5101-Quiz6-C « on: November 22, 2020, 10:49:48 AM »$\textbf{Problem:} \\\\
\text{Compute the following integral}$$$\int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^2}} \,dx$$$\text{with } b>0.\textbf{Solution:} \\\\\text{Let }$$$f(z) = \frac{e^{iz}}{(z+a)^{2}+b^{2}}$$$\text{Then we have two poles: } z_1 = -a-ib ,\ z_2 = -a+ib\text{Since we are looking for upper-half plane, then we have pole at } z=-a+ib \text{ with order 1.}\text{(Denote the path as the pic attached below)}\text{By Residual Theorem we have: }
\begin{gather}
\begin{aligned}

\int_{\gamma}{f(z)} \,dz &= 2 \pi i \cdot Res{f; -a+ib} \\\\
&=2 \pi i \cdot \frac{e^{iz}}{z+a+ib}|_{z=-a+ib} \\\\
&= \frac{\pi e^{-b} e^{-ia}}{b}

\end{aligned}
\end{gather}
\text{For path } c_2:
\begin{gather}
\begin{aligned}

|\int_{c_2}{f(z)} \, dz| &= |\int_{0}^{\pi}{f(Re^{i\theta})Re^{i\theta}} \,d\theta| \\\\
&= |\int_{0}^{pi} {\frac{e^{iRe^{i\theta}}}{(Re^{i\theta}+a)^{2}+b^{2}}} \,d\theta| \\\\
&\leq \pi R \cdot |max(f(z))| \\\\
&= \pi R \cdot max|\frac{e^{iRe^{i\theta}}}{(Re^{i\theta}+a)^{2}+b^{2}}| \\\\
&= \pi R \cdot max\frac{e^{|iR(cos(\theta)+isin(\theta))}|}{|(Re^{i\theta}+a)^{2}+b^{2}|}\\\\
&\leq \pi R \cdot \frac{e^{-Rsin(\theta)}}{|(R+a)^{2} - b^{2}|} \ \ \ \ \ \text{(since triangle inequality)} \\\\
&\leq \pi R \cdot \frac{e^{0}}{(R+a)^{2}+b^{2}} \\\\
&= 0  \ \ \text{as R} \rightarrow \infty \\\\
\end{aligned}
\end{gather}
$$$\text{(By comparing the power of R on both numerator and denominator.)}$$$\text{Now solve for path } c_1:
\begin{gather}
\begin{aligned}
\int_{c_1}{f(z)}\,dz &= \int_{-\infty}^{\infty}{\frac{e^{ix}}{(x+a)^{2}+b^{2}}}\,dx \\\\
&= \int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^{2}}}\,dx + i\int_{-\infty}^{\infty}{\frac{sin(x)}{(x+a)^{2}+b^{2}}}\,dx \\\\
\end{aligned}
\end{gather}
\text{Since the imaginary part of the integral is an odd function, thus its integral is 0.}\text{Thus we have:}$$$\int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^{2}}}\,dx = \frac{\pi e^{-b} e^{-ia}}{b}$$ 7 ##### Quiz 5 / LEC5101-Quiz6-C « on: November 19, 2020, 07:36:37 PM » Sorry this post has been moved to the new section, Quiz6 8 ##### Test 2 / TT3-LEC0201-ALT-E-Q2 « on: November 19, 2020, 01:12:05 PM »$
\textbf{Problem2:} \\\\

\text{Find the general solution of }$$$y'''+y' = \frac{2}{sin(2t)}$$$\text{Hint: Use method of variations.} \int{sec(t)}\,dt = ln(tan(t)+sec(t)) \text{ and } \int{csc(t)}\,dt = -ln(cot(t)+csc(t))\textbf{Solution: }\text{Consider the characteristic polynomial:}$$$\lambda^3 + \lambda = 0 \\\\ \lambda(\lambda^2+1) = 0 \\\\ \lambda_1 =0, \lambda_2=i, \lambda_3=-i$$$\text{Hence we have }$$$y_1 = c_1e^0 = c_1 \\\\ y_2 = c_2cos(t) \\\\ y_3 = c_3sin(t)$$$\text{Take} c_1 = c_2 = c_3 = 1 \text{, we have}$$$w_1 =\begin{bmatrix} cos(t) & sin(t) \\ -sin(t) & cos(t) \\ \end{bmatrix} = 1$$ $$w_2 = -\begin{bmatrix} 1 & sin(t) \\ 0 & cos(t) \\ \end{bmatrix} = -cos(t)$$ $$w_3 = \begin{bmatrix} 1 & cos(t) \\ 0 & -sin(t) \\ \end{bmatrix} = -sin(t)$$$\text{Then we have Wronskian of }y_1 ,\ y_2 ,\ y_3
\begin{gather}
\begin{aligned}
W&= y_1''w_1 + y_2''w_2+y_3''w_3 \\\\
&= 0+ (cos(t))''(-cos(t)) + (sin(t))''(-sin(t)) \\\\
&= cos^2(t)+sin^2(t) \\\\
& =1
\end{aligned}
\end{gather}
\text{To compute }u_1 ,\ u_2,\ u_3 \text{, we need to use the hind in integral computation} 
\begin{gather}
\begin{aligned}

u_1 &= \int{\frac{1\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\
&= 2 \int{csc(2t)} \,dt \ \ \ \text{ (by the hint)} \\\\
&= -2 \cdot ln(cot(2t)+csc(2t)) + c_1

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

u_2 &=\int{\frac{(-cos(t))\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\
& = -2\int{\frac{cos(t)}{sin(2t)}} \,dt \\\\
&= -2\int{\frac{cos(t)}{2sin(t)cos(t)}} \,dt \\\\
&= - \int{csc(t)} \,dt \ \ \ \text{ (by the hint)} \\\\
&= ln(cot(t) + csc(t)) + c_2

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

u_3 &=\int{\frac{(-sin(t))\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\
& = -2\int{\frac{sin(t)}{2sin(t)cos(t)}} \,dt \\\\
& = - \int{sec(t)} \,dt \ \ \ \text{ (by the hint)} \\\\
&= -ln(tan(t) + sec(t)) +c_3

\end{aligned}
\end{gather}
\text{Therefore, the general solution is}
\begin{gather}
\begin{aligned}

y &= u_1y_1 + u_2y_2 + u_3y_3 \\\\
&= -2 \cdot ln(cot(2t)+csc(2t)) + c_1 + cos(t)\cdot(ln(cot(t) + csc(t)) + c_2) + sin(t) \cdot (-ln(tan(t) + sec(t)) +c_3) \ \ \ \ \text{(by rearranging, we have)}\\\\
& =c_1 + c_2cos(t) + c_3sin(t) - 2 \cdot ln(cot(2t)+csc(2t)) + cos(t) \cdot ln(cot(t) + csc(t)) - sin(t) \cdot ln(tan(t) + sec(t)

\end{aligned}
\end{gather}
$9 ##### Test 2 / TT3-LEC0201-ALT-E-Q1 « on: November 19, 2020, 12:45:28 PM »$
\textbf{Problem 1: } \\\\
\text{Consider the equation}$$$y''' - 9y'' + 27y' -27y = 24e^{3t}$$$
\textbf{a:}
\text { Write a differential equation for Wronskian of } y_1 ,\ y_2 ,\ y_3 ,\ \text{ which are solutions for homogeneous equation and solve it.}\\\\ \\\\
\textbf{b:}
\text { Find fundamental system } {y_1, y_2, y_3} \text{of solutions for homogeneous equation, and find their Wronskian. Compare with part (a).} \\\\ \\\\
\textbf{c:}
\text { Find the general solution of the equation.} \\\\

\textbf{Solution for part (a):} \\\\

\begin{gather}
\begin{aligned}
W &= ce^{- \int{-9} \,dt} \\\\
&= ce^{\int{9}\, dt} \\\\
&= ce^{9t}
\end{aligned}
\end{gather}

\textbf{Solution for part (b):} \\\\
\text{We have characteristic polynomial: }$$$\lambda ^{3} - 9\lambda ^{2} + 27 \lambda -27 = 0$$$\text{Then solve it we have: }$$$\lambda_1 =\lambda_2 =\lambda_3 = 3$$$\text{Hence: }$$$y_1 = c_1e^{3t} \\\\ y_2 = c_2 t e^{3t} \\\\ y_3 = c_3 t^{2}e^{3t}$$$\text{Take } c_1 =c_2 =c_3 = 1 \text{, we have:}
\begin{gather}
\begin{aligned}

w_1 &= \begin{bmatrix}
te^{3t} & t^{2}e^{3t}  \\
(3t+1)e^{3t} & (3t^{2}+2t)e^{3t} \\
\end{bmatrix} \\\\
&= (3t^{3}+2t^{2})e^{6t} - (3t^{3}+t^{2})e^{6t} \\\\
&= t^{2}e^{6t}

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

w_2 &= - \begin{bmatrix}
e^{3t}& t^{2}e^{3t}  \\
3e^{3t} & (3t^{2}+2t)e^{3t} \\
\end{bmatrix} \\\\
&= -[(3t^{2} + 2t)e^{6t} -3t^{2}e^{6t}] \\\\
&= -2te^{6t}

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

w_3 &= \begin{bmatrix}
e^{3t}& te^{3t}  \\
3e^{3t} & (3t+1)e^{3t} \\
\end{bmatrix} \\\\
&= (3t+1)e^{6t}-3te{6t} \\\\
&= e^{6t}

\end{aligned}
\end{gather}
\text{Therefore we have: }
\begin{gather}
\begin{aligned}

W &= y_1''w_1 +y_2''w_2 + y_3''w_3 \\\\
&= (e^{3t})'' t^{2}e^{6t} + (te^{3t})''(-2te^{6t}) + (t^{2}e^{3t})''e^{6t} \\\\
&= 9t^{2}e^{9t}-18t^{2}e^{9t} -12te^{9t} +9t^{2}e^{9t} +12te^{9t} + 2e{9t} \\\\
&= 2e^{9t}

\end{aligned}
\end{gather}
\text{Compared to part(a), Wronskian in part (b) satisfies that of in part (a) and c = 2.}
\textbf{Solution for part (c):} \\\\
\text{using the method of variation, we have}

\begin{gather}
\begin{aligned}

u_1 & = \int{\frac{24e^{3t}\cdot t^{2}e^{6t}}{2e^{9t}}} \,dt \\\\
&= 12 \int{\frac{ t^{2}e^{9t}}{e^{9t}}} \,dt \\\\
&= 4t^{3} + c_1

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

u_2 & = \int{\frac{24e^{3t}\cdot (-2)te^{6t}}{2e^{9t}}} \,dt \\\\
&= -24 \int{t} \,dt \\\\
&= -12t^{2} + c_2

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

u_3 & =  \int{\frac{24e^{3t}\cdot e^{6t}}{2e^{9t}}} \,dt \\\\
& = 12 \int{1} \,dt \\\\
&= 12t + c_3

\end{aligned}
\end{gather}
\text{Therefore the general solution is: }
\begin{gather}
\begin{aligned}
y &= u_1y_1 + u_2y_2 + u_3y_3 \\\\
&= (4t^{3} + c_1)e^{3t} + (-12t^{2} + c_2)te^{3t} + (12t + c_3)t^{2}e^{3t} \\\\
&= c_1e^{3t} + c_2te^{3t} +c_3t^{2}e^{3t}+4t^{3}e^{3t}
\end{aligned}
\end{gather}
$10 ##### Test 1 / Re: Test 1 Question 2 « on: November 16, 2020, 01:47:37 PM »$
\textbf{For part (a):} \\\\
\text{By ratio test, we have : }\\\\

\begin{gather}
\begin{aligned}

\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty}|\frac{(1+ i\sqrt{3})^{n+1}}{4^{n+1}(n+1)log^{3}(n+1)} \cdot \frac{n 4^{n} log^{3}(n)}{(1+i\sqrt{3})^{n}}| \\\\
&= \lim_{n \to \infty} |\frac{1+i\sqrt{3}}{4} \cdot \frac{nlog^{3}(n)}{(n+1)log^3(n+1)}| \\\\
&= \frac{1}{2} \text{(by further expanding this absolute value)}\\\\
&= \frac{1}{R}

\end{aligned}
\end{gather}
\text{Therefore the radius of convergence is 2.}
\textbf{For part (b):} \\\\
\text{We can follow the similar method. By ratio test, we have : }\\\\

\begin{gather}
\begin{aligned}

\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty} |\frac{(i+\sqrt{3})^{n+1} \cdot \frac{e^{2(n)+1}-1}{e^{2(n+1)}+1}}{4^{n+1}} \cdot \frac{4^n}{(i+\sqrt{3})^{n}\cdot \frac{e^{2n}-1}{e^{2n}+1}}| \\\\
&= \lim_{n \to \infty} |\frac{i+\sqrt{3}}{4} \cdot \frac{e^{2n+2}-1}{e^{2n+2}+1} \cdot \frac{e^{2n}+1}{e^{2n}-1}| \\\\
&= \frac{1}{2} \text{(by further expanding this absolute value, and most terms can be canceled out.)}\\\\
&= \frac{1}{R}

\end{aligned}
\end{gather}
\text{Therefore the radius of convergence is 2.} \\\\\text{Above is how I got my answers. Correct me if I made any mistakes. Feel free to comment below }$11 ##### Quiz 5 / LEC5101-Quiz5-E « on: November 05, 2020, 07:27:03 PM »$
\textbf{Problem: }\\\\
\text{Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity.} \\\\ \text{If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole}\\\\

\textbf{Solution:}

\text{Let  } \\\\
\begin{gather}
\begin{aligned}

f(z) &= \frac{e^{z}-1}{e^{2z}-1} \\\\
&= \frac{e^{z}-1}{(e^{z}-1)(e^{z}+1)}

\end{aligned}
\end{gather}

\text{Then it shows two singularities: }

\begin{aligned}

e^{z} = -1 ,\ e^{z} = 1

\end{aligned}

\text{Therefore we have} \\\\

\begin{gather}
\begin{aligned}

z_1&=log(1)\\\\
&=ln|1|+iarg(1)\\\\
&= 2k\pi i,  k\in \mathbb{Z}\\\\

z_2&=log(-1)\\\\
&=ln|-1|+iarg(-1)\\\\
&=(2k+1)\pi i, k\in \mathbb{Z}
\end{aligned}
\end{gather}

\text{When } z = z_1 \\\\

\begin{gather}
\begin{aligned}

\lim_{z \to z_1} \frac{e^{z}-1}{e^{2z}-1} &= \lim_{z \to 2k\pi i} \frac{e^{z}-1}{(e^{z}-1)(e^{z}+1)} \\\\
&=  \lim_{z \to 2k\pi i} \frac{1}{e^{z}+1}\\\\
&= \frac{1}{2}

\end{aligned}
\end{gather}
\text{Therefore } z_1 \text{ is a removable singularity with value } \frac{1}{2}
\text{When } z = z_2 \\\\

\begin{gather}
\begin{aligned}

\lim_{z \to z_2} \frac{e^{z}-1}{e^{2z}-1} &= \lim_{z \to (2k+1)\pi i} \frac{e^{z}-1}{(e^{z}-1)(e^{z}+1)} \\\\
&=  \lim_{z \to (2k+1)\pi i} \frac{1}{e^{z}+1}\\\\
&= \infty

\end{aligned}
\end{gather}
\text{Hence } z_2 \text{ is a pole.}\text{Since the numerator has an order of 0 and the denominator has an order of 1}\text{Therefore } z_2 \text{ is a pole of order 1.}$12 ##### Test 2 / 2020F Test2-ALT-K « on: November 04, 2020, 07:45:49 PM »$
\textbf{Problem:} \\\\
\text{(a) Find the general solution for equation: } \ y'' -4y' + 5y = 30 + 40cos(3t) \\\\
\text{(b) Find solution that satisfies:} \ y(0) = 0, y'(0) = 0

\textbf{Solution for part (a): } \\\\
\text{Solve for homogenous solution:} \\\\

\begin{gather}
\begin{aligned}

r^{2} - 4r + 5 = 0 \Longrightarrow r = 2 \pm i \\\\

\end{aligned}
\end{gather}

\text{Therefore, } \\\\
\begin{gather}
\begin{aligned}

y_h = c_1 e^{2t}cos(t) + c_2 e^{2t}sin(t)

\end{aligned}
\end{gather}

\text{Let } y_p = At+B+Csin(3t)+Dcos(3t) \ \text{ , then we have} \\\\

\begin{gather}
\begin{aligned}

y'_p &= A+3Ccos(3t) -3Dsin(3t) \\\\
y''_p &= -9Csin(3t)-9Dcos(3t)

\end{aligned}
\end{gather}

\text{Plug them in the original equation, we have: } \\\\

\begin{gather}
\begin{aligned}

y'' -4y' + 5y &= -9Csin(3t)-9Dcos(3t) -4(A+3Ccos(3t) -3Dsin(3t)) + 5(At+B+Csin(3t)+Dcos(3t)) \\\\
&= 30 + 40cos(3t)

\end{aligned}
\end{gather}

\text{Solve for parameter A, B, C, and D, we get } A=0 ,\ B=6,\ C=-3,\ D=-1\\\\
\text{Thus we have: } \\\\ \\\\

\begin{gather}
\begin{aligned}

Y &= y_h + y_p \\\\
&= c_1 e^{2t}cos(t) + c_2 e^{2t}sin(t) -3sin(3t) -cos(3t) + 6

\end{aligned}
\end{gather}

\textbf{Solution for part (b): } \\\\

\text{Plug in } y(0)=0 \text{, we get } \ c_1 = -5 \text{, thus we have } \\\\
\begin{gather}
\begin{aligned}
Y = -5e^{2t}cos(t) + c_2 e^{2t}sin(t) -3sin(3t) -cos(3t) + 6
\end{aligned}
\end{gather}

\text{Now consider } y'(0)=0 \text{, then we have:}\\\\
\begin{gather}
\begin{aligned}
0 = -5(2e^{2t}cos(t) - e^{2t}sin(t)) + c_2(2e^{2t}sin(t)+e^{2t}cos(t)) - 9cos(3t) + 3sin(3t) \\\\
\Longrightarrow c_2 = 19
\end{aligned}
\end{gather}

\text{Thus the solution is:}\\\\

\begin{gather}
\begin{aligned}
Y = -5e^{2t}cos(t) + 19 e^{2t}sin(t) -3sin(3t) -cos(3t) + 6
\end{aligned}
\end{gather}
$13 ##### Test 2 / 2020F Test2-ALT-F Q2 « on: November 04, 2020, 07:03:43 PM »$
\textbf{Problem 2:} \\\\
\text{Calculate directly (that means, without Cauchy's or Green's theorem the integral)} \\\\

\begin{gather}
\begin{aligned}
\int_{L}{Re(z) \,dz}
\end{aligned}
\end{gather}

\text{where L is a path, consisting of two acrs of the radii 2, from 1 to -1 and to 1 on the figure (fig is attached below).}

\textbf{Solution: } \\\\
\text{Denote the path on upper-half plane } L_1 \text{ and the lower-half plane } L_2 \ \text{, we have:} \\\\

\begin{gather}
\begin{aligned}
L_1 = -i\sqrt{3}+2e^{it_1}, \ t_1 \in [\pi/3, 2\pi/3] \\\\
L_2 = i\sqrt{3}+2e^{it_2}, \ t_2 \in [4\pi/3, 5\pi/3]
\end{aligned}
\end{gather}

\text{Now we calculate } L_1 \ \text{ and } \ L_2 \text{respectively: }\\\\

\begin{gather}
\begin{aligned}

\int_{L_1}{Re(-i\sqrt{3}+2e^{it}) \cdot d(-i\sqrt{3}+2e^{it})} \,dt &=  \int_{\pi/3}^{2\pi/3}{2cos(t)\cdot i2e^{it}}\,dt\\\\
&= i\cdot 4 \int_{\pi/3}^{2\pi/3}{cos(t)e^{it}} \,dt\\\\
&= i\cdot 4 \int_{\pi/3}^{2\pi/3}{\frac{1}{2}\cdot (e^{it}+e^{-it})\cdot e^{it}} \,dt\\\\
&= i\cdot 2 \int_{\pi/3}^{2\pi/3}{(e^{2it}+1)} \,dt\\\\
&= i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2})

\end{aligned}
\end{gather}

\begin{gather}
\begin{aligned}

\int_{L_2}{Re(i\sqrt{3}+2e^{it}) \cdot d(i\sqrt{3}+2e^{it})} \,dt &=  \int_{4\pi/3}^{5\pi/3}{2cos(t)\cdot i2e^{it}}\,dt\\\\
&= i\cdot 4 \int_{4\pi/3}^{5\pi/3}{cos(t)e^{it}} \,dt \\\\
&= i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2})

\end{aligned}
\end{gather}

\text{Lastly,} \\\\ \\\\

\begin{gather}
\begin{aligned}

\int_{L}{Re(z)} \,dz &= \int_{L_1}{Re(z)} \,dz + \int_{L_2}{Re(z)} \,dz\\\\
&= i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2}) + i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2})\\\\
&= i\frac{4}{3}\pi - i2\sqrt{3}

\end{aligned}
\end{gather}
$14 ##### Test 2 / 2020F Test2-ALT-F Q1 « on: November 04, 2020, 06:36:41 PM »$
\textbf{Problem 1:} \\\\
\text{(a) Show that  } \ 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x  \text{ is a harmonic function;} \\\\
\text{(b) Find the harmonic conjugate function } v(x,y) \\\\
\text{(c) Consider } u(x,y) + iv(x,y) \ \text{ and write it as a function } f(z) \text{ of } z=x+iy

\textbf{Solution for part (a): } \\\\

\text{Since we have:} \\\\

\begin{gather}
\begin{aligned}

U(x,y) = 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x \\\\

\end{aligned}
\end{gather}

\text{then we have the following:} \\\\

\begin{gather}
\begin{aligned}

u_x &= 12x^{3} - 36xy^{2} - 3 \\\\
u_{xx} &= 36x^{2} - 36y^{2}\\\\
u_y &= -36x^{2}y + 12y^{3} \\\\
u_{yy} &= -36x^{2} + 36y{2}

\end{aligned}
\end{gather}
\text{Thus we have: } \\\\
\begin{gather}
\begin{aligned}

\Delta u &= u_{xx} + u_{yy} \\\\
&= 36x^{2} - 36y^{2} - 36x^{2} + 36y^{2} \\\\
&= 0

\end{aligned}
\end{gather}
\text{Therefore the function is harmonic. } \\\\
\\\\
\\\\
\textbf{Solution for part (b): }
\\\\

\text{We have:} \\\\
\begin{gather}
\begin{aligned}

v_x &= -u_y = 36x^{2}y - 12y^{3} \\\\
v_y &= u_x = 12x^{3} -36xy^{2} -3

\end{aligned}
\end{gather}

\text{Then} \\\\

\begin{gather}
\begin{aligned}

v_{x,y} &= \int{v_x} \,dx + \phi(y) \\\\
&= 12x^{3}y-12xy^{3}+\phi(y)

\end{aligned}
\end{gather}

\text{Now we calculate  } \phi(y) \text{ by following:} \\\\

\begin{gather}
\begin{aligned}

v_y &= (v_{x,y})' \\\\
&= 12x^{3} -36xy^{2}-3

\end{aligned}
\end{gather}

\text{thus, } \\\\

\begin{gather}
\begin{aligned}

\phi'(y) &= -3 \\\\
\phi(y) &= \int{-3}\,dy \\\\
&= -3y + c

\end{aligned}
\end{gather}

\text{Therefore} \\\\

\begin{gather}
\begin{aligned}

v_{x,y} = 12x^{3}y-12xy^{3}-3y+c
\end{aligned}
\end{gather}

\\\\
\\\\
\textbf{Solution for part (c): }
\\\\
\text{Let } \ z=x+iy \\\\

\begin{gather}
\begin{aligned}

u_{x,y} + iv_{x,y} &= 3x^{4} -12x^{2}y^{2} + 3y^{4} -3x +i(12x^{3}y-12xy^{3}-3y)+iC \\\\
&= 3(3x^{4} -12x^{2}y^{2} + 3y^{4} + i12x^{3}y-i12xy^{3}) -3(x+iy) +iC\\\\
&= 3(x+iy)^{4}-3(x+iy)+iC\\\\
&= 3z^{4} -3z + iC

\end{aligned}
\end{gather}
$15 ##### Test 2 / Re: Question 3c for 2020S Night Sitting « on: October 27, 2020, 07:22:04 PM »$\text{Since we have: } \ cosh(x) = cos(ix) \ \text{ and } \ sinh(x) = -i \cdot sin(ix) \\
\text{By substituting and rearranging, we have the following: }
\begin{gather}
\begin{aligned}

f(x,y) &= cosh(x) \cdot sin(y) + ( - sinh(x) \cdot cos(y)) \cdot i + C \cdot i \\\\
&= cos(ix) \cdot sin(y) - sin(ix) \cdot cos(y) + C \cdot i \\\\
&= sin(y - ix) + C \cdot i \\\\
&= sin(-i \cdot (x+iy)) + C \cdot i \\\\
&= - sin(i(x+iy)) + C \cdot i \\\\
&= -i \cdot sinh(x + iy) + C \cdot i \\\\
&= -i \cdot sinh(z) + C \cdot i \\\\

\end{aligned}
\end{gather}
\text{I think the answer has a typo, I got} \ -i \cdot sinh(z) + C \cdot i \ \text{ instead of } cosh(z) + C \cdot i\$

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