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### Messages - Wang Jingyao

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1
##### Quiz 3 / lec5101 Quiz3
« on: October 09, 2020, 02:42:03 PM »
Let $\gamma_1$ be the semicircle from $1$ to $-1$ through $i$ and $\gamma_2$ the semicirlce from $1$ to $-1$ through $-i$. Compute $\int_{\gamma_1}z^2dz$ and $\int_{\gamma_2}z^2dz$. Can you account for the fact that they are equal?

Compute $\int_{\gamma_1}z^2dz$
\begin{align}
\gamma_1(t)=e^{it}\ for\ -\pi \leq t \leq 0\\
\notag \\
\gamma_1'(t)=ie^{it}\\
\notag \\
\int_{\gamma_1}z^2dz = \int^{\pi}_{0}(e^{it})^2 \cdot ie^{it}dt =  \dfrac{1}{3}e^{3it} \Big|^{\pi}_{0}=-\dfrac{2}{3}\\

\end{align}

Compute $\int_{\gamma_2}z^2dz$
\begin{align}
\gamma_2(t)=e^{it}\ for\ 0 \leq t \leq \pi\\
\notag \\
\gamma_2'(t)=ie^{it}\\
\notag \\
\int_{\gamma_2}z^2dz = \int^{\pi}_{0}(e^{it})^2 \cdot ie^{it}dt =  \dfrac{1}{3}e^{3it} \Big|^{\pi}_{0}=-\dfrac{2}{3}\\

\end{align}

\begin{align}
\gamma_1 + (-\gamma_2) = \gamma\\
\notag \\
\int_{\gamma}z^2dz = \int_{\gamma_1}z^2dz - \int_{\gamma_2}z^2dz =0\\
\notag \\
\therefore\ equal.
\end{align}

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##### Quiz 2 / Quiz 2 LEC5101
« on: October 01, 2020, 07:23:22 PM »
\begin{align}
Find\ the\ limit\ of\ each\ function\ at\ the\ given\ point,\ or\ explain\ &why\ it\ does\ not\ exist. \\
\notag \\
f(z) = (z-2)log|z-2|\ at\ z_o = 2\\

\end{align}

\begin{align}
\lim_{z \to 2}f(z) &= \lim_{z \to 2}(z-2)log|z-2|\ at\ z_o = 2\\
\notag \\
\because z_0 &= 2\\
\notag \\
\therefore z-2 &\to 0\\
\notag \\
Let\ z' &= z-2\\
\notag \\
\therefore z' &\to 0\\
\end{align}

\begin{align}
\therefore \lim_{z' \to 0}f(z) &= \lim_{z' \to 0} z'log|z'|\\
\notag \\
\lim_{z' \to 0}|f(z)| &= \lim_{z' \to 0} |z'log|z'|| \\
\notag \\
&= \lim_{z' \to 0} \dfrac{log|z'|}{\dfrac{1}{|z'|}} \\
\notag \\
Take\ the\ derviative\ &both\ on\ numerator\ and\ denominator\\
&= \lim_{z' \to 0} \dfrac{\dfrac{1}{|z'|}}{-\dfrac{1}{|z'|^2}}\\
\notag \\
&= 0
\end{align}

3
##### Quiz-6 / Quiz6 LEC5101
« on: December 03, 2019, 10:53:51 PM »
a. Express the general solution of the given system of equations in terms of real-valued functions.
b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of
the solutions as $t \rightarrow \infty$.

$x’=$$\left ( \begin{matrix} 1 & 2 \\ -5 & -1 \end{matrix} \right )x det(A-\lambda I)=$$det \left [ \begin{matrix} 1-\lambda & 2 \\ -5 & -1-\lambda \end{matrix} \right ]=(\lambda-1)^2+10$

Solve for

$(\lambda-1)^2+10=0$

$\lambda_1=3i \;,\; \lambda_2=-3i$

Consider $\lambda=3i$

$x’=$$\left [ \begin{matrix} 1-3i & 2 \\ -5 & -1-3i \end{matrix} \right ] \left [ \begin{matrix} x_1 \\ x_2 \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 0 \end{matrix} \right ] let x_2=t, \left [ \begin{matrix} x_1 \\ x_2 \end{matrix} \right ] = \left [ \begin{matrix} -2 \\ 1-3i \end{matrix} \right ]t Consider e^{it}\left [ \begin{matrix} -2 \\ 1-3i \end{matrix} \right ] = (\cos3t+i\sin3t) \left [ \begin{matrix} -2 \\ 1-3i \end{matrix} \right ] = \left[ \begin{matrix} -2\cos3t\\ \cos3t+3\sin3t \end{matrix} \right] +i\left[ \begin{matrix} -2\sin3t\\ \sin3t-3\cos3t \end{matrix} \right] Therefore, x(t)=$$c_1 \left[ \begin{matrix} -2\cos3t\\ \cos3t+3\sin3t \end{matrix} \right] +c_2\left[ \begin{matrix} -2\sin3t\\ \sin3t-3\cos3t \end{matrix} \right]$

4
##### Quiz-6 / Re: Lec5101
« on: December 03, 2019, 06:37:22 PM »
I think the general solution should be $x(t)=$$c_1 \left [ \begin{matrix} 3 \\ 4 \end{matrix} \right ] +$$c_2 e^{-2t} \left [ \begin{matrix} 1 \\ 2 \end{matrix} \right ]$

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##### Quiz-7 / Quiz 7 LEC0101
« on: December 03, 2019, 06:14:59 PM »
(a) Determine all critical points of the given system of equations.

(b) Find the corresponding linear system near each critical point.

(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?

(d)  Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.

$\left\{\begin{array}{l}{\dfrac{dx}{dt}=-2x-y-x(x^2+y^2)} \\ {\dfrac{dy}{dt}=x-y+y(x^2+y^2)}\end{array}\right.$

To find critical points, let

$\dfrac{dx}{dt}=0\;,\; \dfrac{dy}{dt}=0$

Therefore,

$\left\{\begin{array}{l}{-2x-y-x(x^2+y^2)=0} \\ {x-y+y(x^2+y^2)=0}\end{array}\right.$

Then we have critical points

$(0,0),\quad \bigg(-\sqrt{\dfrac{4}{\sqrt{13}}-1}\;,\ (\dfrac{1}{2}+\dfrac{3}{2\sqrt{13}})\sqrt{4\sqrt{13}-13}\bigg),\quad \bigg(\sqrt{\dfrac{4}{\sqrt{13}}-1} \;,\ (-\dfrac{1}{2}-\dfrac{3}{2\sqrt{13}})\sqrt{4\sqrt{13}-13}\bigg)$

Which is

$(0,0),\quad (-0.330757,1.09242),\quad (0.330757,-1.09242)$

let

$\left\{\begin{array}{l}{f(t)=-2x-y-x(x^2+y^2)} \\ {g(t)=x-y+y(x^2+y^2)}\end{array}\right.$

$J[f(t),g(t)]=$$\left [ \begin{matrix} f_x & f_y \\ g_x & g_y \end{matrix} \right ]$$ =$$\left [ \begin{matrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{matrix} \right ] Plug in the critical points to find eigenvalues of each linear system. For (0,0), J(0,0)=$$ \left [ \begin{matrix} -2 & -1 \\ 1 & -1 \end{matrix} \right ]$

$det(A-\lambda I)=$$det \left [ \begin{matrix} -2-\lambda & -1 \\ 1 & -1-\lambda \end{matrix} \right ]=\lambda^2+3\lambda+3 Solve for \lambda^2+3\lambda+3=0 \lambda_1=\dfrac{-3+\sqrt{-3}i}{2}\;,\; \lambda_2=\dfrac{-3-\sqrt{-3}i}{2} Therefore, the system is a spiral and stable at (0,0). for (-0.330757,1.09242), J(-0.330757,1.09242)=$$ \left [ \begin{matrix} -3.521576422 & -0.2773500983 \\ 0.2773500983 & 2.689526127 \end{matrix} \right ]$

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##### Quiz-3 / TUT0602 Quiz3
« on: October 17, 2019, 12:11:10 AM »
Find the differiential equation whose general solution is y=c1e^2t+c2e^-3t

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##### Quiz-2 / Re: TUT0602 Quiz2
« on: October 04, 2019, 03:31:11 PM »
Find an integrating factor and solve the given equation

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##### Quiz-2 / TUT0602 Quiz2
« on: October 04, 2019, 03:27:45 PM »
FInd an integrating factor and solve the given equation

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##### Quiz-1 / TUT0602 Quiz1
« on: September 27, 2019, 06:07:44 PM »
Question:Find the general solution y'+3y=t+e^(-2t)

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