### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Shengying Yang

Pages: [1]
1
##### Term Test 2 / Re: TT2B Problem 5
« on: November 24, 2018, 08:34:22 AM »
$$f(z)=\frac{8}{(z-3)(z+5)}=\frac{1}{z-3}-\frac{1}{z+5}$$
(a)as$|z|<3$,
$$f(z)= -\frac{1}{3} \frac{1}{1-\frac{z}{3}}- \frac{1}{5} \frac{1}{1-(-\frac{z}{5})}=-\frac{1}{3}\sum_{n=0}^{\infty}(\frac{z}{3})^n-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n$$
(b)as$3<|z|<5$,
$$f(z)= \frac{1}{z} \frac{1}{1-\frac{3}{z}}- \frac{1}{5} \frac{1}{1-(-\frac{z}{5})}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n=\sum_{n=0}^{\infty}3^nz^{-n-1}-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n$$
(c)as$|z|>5$,
$$f(z)= \frac{1}{z} \frac{1}{1-\frac{3}{z}}- \frac{1}{z} \frac{1}{1-(-\frac{5}{z})}=\sum_{n=0}^{\infty}3^nz^{-n-1}-\sum_{n=0}^{\infty}(-5)^nz^{-n-1}=\sum_{n=0}^{\infty}z^{-n-1}(3^n-(-5)^n)$$

2
##### Quiz-6 / Re: Q6 TUT 5201
« on: November 17, 2018, 04:18:55 PM »
let $h(z)=1, g(z)=1-\cos z$
as $z_0=0$, $$h(0)=1\ne0$$
$$g(0)=0, g'(0)=0, g''(0)=1\ne0$$
$$\therefore2-0=2$$, order of pole = 2
$$\therefore\frac{1}{1-\cos z}=a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+\cdots$$
$$\frac{1}{1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots}=a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+\cdots$$ Errors in the line above; correct below
$$\therefore(\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots)(a_{-2}z^{-2}+a_{-1}z^{-1}+a_0+a_1z+a_2z^2+\cdots)=1$$
Since corresponding coefficients equal each other, we have
$$\frac{a_{-2}}{2!}=1\therefore a_{-2}=2$$
$$\frac{a_{-1}}{2!}=0\therefore a_{-1}=0 \therefore Res(f;0)=0$$
$$\frac{a_0}{2!}-\frac{a_{-2}}{4!}=0\therefore a_0=\frac{1}{6}$$
$$\frac{a_1}{2!}-\frac{a_{-1}}{4!}=0\therefore a_1=0$$
$$\therefore \frac{2}{z^2}+\frac{1}{6}+\cdots$$

3
##### Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P5
« on: November 03, 2018, 06:48:33 PM »
$$f(z)=\frac{16}{(z^2-16)(z^2+25)}$$
By doing partial fraction, we can get
$$f(z)=\frac{2}{41}\frac{1}{z-1}-\frac{2}{41}\frac{1}{z+4}-\frac{16}{41}\frac{1}{10i}(\frac{1}{z-5i}-\frac{1}{z+5i})$$
a) as $|z|<4,$
$$f(z)=\frac{2}{41}\frac{(-1)}{4}\frac{1}{1-\frac{z}{4}}-\frac{2}{41}\frac{1}{4}\frac{1}{1-\frac{(-z)}{4}}+\frac{16}{41}\frac{1}{10i}\frac{1}{5i}(\frac{1}{1-\frac{z}{5i}}+\frac{1}{1-\frac{-z}{5i}})$$
$$∴f(z)=\frac{-2}{164}(\sum_{n=0}^{\infty}(\frac{z}{4})^n+\sum_{n=0}^{\infty}(\frac{-z}{4})^n)-\frac{16}{2050}(\sum_{n=0}^{\infty}(\frac{z}{5i})^n+\sum_{n=0}^{\infty}(\frac{-z}{5i})^n)$$
Similarily,
b)as $<4|z|<5$,
$$f(z)=\frac{2}{41z}(\sum_{n=0}^{\infty}(\frac{4}{z})^n-\sum_{n=0}^{\infty}(\frac{-4}{z})^n)-\frac{16}{2050}(\sum_{n=0}^{\infty}(\frac{z}{5i})^n+\sum_{n=0}^{\infty}(\frac{-z}{5i})^n)$$
c)as $|z|>5$,
$$f(z)=\frac{2}{41z}(\sum_{n=0}^{\infty}(\frac{4}{z})^n-\sum_{n=0}^{\infty}(\frac{-4}{z})^n)-\frac{16}{410iz}(\sum_{n=0}^{\infty}(\frac{5i}{z})^n-\sum_{n=0}^{\infty}(\frac{-5i}{z})^n)$$

4
##### Term Test 1 / Re: TT1 Problem 3 (morning)
« on: October 16, 2018, 10:30:15 AM »
Zixuan, I think you made a mistake.
Plugging in $y(0)=0$ ,  you should get $0=C_1+C_2+2=0$. Therefore, $C_1=1, C_2=-2$
$$∴y=-2e^t+4te^t+2e^{-t}$$

5
##### Term Test 1 / Re: TT1 Problem 3 (noon)
« on: October 16, 2018, 08:05:30 AM »
There is a mistake in your answer. Plugging in $y(0)=0$ , you should get $0=C_1+C_2-\frac{1}{2}$ . Therefore, $C_1=\frac{1}{2}, C_2=0$
$$∴y(t)=\frac{1}{2}e^{-7t}-\frac{1}{2}e^t+4te^{-t}$$

6
##### Term Test 1 / Re: TT1 Problem 1 (noon)
« on: October 16, 2018, 07:53:51 AM »
$$M_y =4\ln (xy) +5 \ne N_x=\ln (xy)+2$$
∴not exact.
$$R=\frac{M_y-N_x}{N}= \frac{3(\ln (xy)+1)}{x(\ln (xy)+1)}=\frac{3}{x}$$
$$\mu=e^{\int \frac{3}{x} dx}=x^3$$
$$∴x^3y(4\ln (xy) +1)+x^4(\ln (xy)+1)y'=0$$
There exist a ​ $\psi(x,y)​$ such that ​$\psi_x=M, \psi_y=N​$ .
$$\psi=\int x^3y(4\ln (xy) +1)dx=x^4y\ln (xy)+h(y)$$
$$\psi_y=x^4\ln (xy)+x^4+h'(y)$$
$$∴h'(y)=0, h(y)=C$$
$$∴\psi(x,y)= x^4y\ln (xy)=C$$
as x=1, y=1, we get C=0
$$∴x^4y\ln (xy)=0$$

7
##### Term Test 1 / Re: TT1 Problem 4 (noon)
« on: October 16, 2018, 07:41:48 AM »
First, we consider $y''+6y'+10y=0​$
$$r^2+6r+10=0$$
$$∴r=-3\pm i​$$
$$∴y_c(t)=C_1e^{-3t}\cos t+C_2e^{-3t}\sin t$$
Second, we consider $y''+6y'+10y=5e^{-3t}​$, let $$y_{p1}(t)= Ae^{-3t}, y'_{p1}(t)=-3Ae^{-3t}, y''_{p1}(t)=9Ae^{-t}​$$
$$9Ae^{-3t} -18Ae^{-3t}+10Ae^{-3t} =5e^{-3t}$$
$$∴A=5$$
$$∴y_{p1}(t)= 5e^{-3t}$$
Third, we consider $y''+6y'+10y=13\cos t$ , let $$y_{p2}(t)= B\cos t+C\sin t, y'_{p2}(t)= -B\sin t+C\cos t,y''_{p2}(t)= -B\cos t-C\sin t​$$
$$∴-B\cos t-C\sin t+(-B\sin t+C\cos t)+10(B\cos t+C\sin t)=13\cos t$$
$$∴9C-6B=0, 9B+6C=13$$
$$∴ B=1, C=\frac{2}{3}$$
$$∴y_{p2}(t)= \cos t+\frac{2}{3}\sin t​$$
Therefore,
$$y(t)=y_c(t)+y_{p1}(t)+y_{p2}(t) =C_1e^{-3t}\cos t+C_2e^{-3t}\sin t+5e^{-3t}+\cos t+\frac{2}{3}\sin t$$

8
##### Term Test 1 / Re: TT1 Problem 4 (noon)
« on: October 16, 2018, 07:37:25 AM »
For the homogeneous part, I think $r=-3\pm i​$ . Therefore, $$y_c(t)=C_1e^{-3t}cost+C_2e^{-3t}sint$$
I will post my answer below.

9
##### Term Test 1 / Re: TT1 Problem 1 (noon)
« on: October 16, 2018, 07:22:50 AM »
$$M=y(4lnx + 4lny +1)=y(4ln(xy)+1),N=x(lnx+lny+1)=x(ln(xy)+1)$$
$$M_y =4ln(xy) +5, N_x=ln(xy)+2$$
Since $M_y\ne N_x​$​, it is not exact.
$$R=\frac{M_y-N_x}{N}=\frac{3ln(xy)+3}{x(ln(xy)+1)} = \frac{3(ln(xy)+1)}{x(ln(xy)+1)}=\frac{3}{x}$$
R is a function of x only, so there is an integrating factor of the form
$$\mu=e^{\int R dx}=e^{\int \frac{3}{x} dx}=e^{3lnx }=x^3$$
multiply the differential equation by $\mu​$
$$x^3y(4ln(xy) +1)+x^4(ln(xy)+1)y'=0$$
Since ​$M_y = N_x​$, It is exact now. There exist a ​ $\psi(x,y)​$ such that ​$\psi_x=M, \psi_y=N​$ .
$$\psi=\int M dx=\int x^3y(4ln(xy) +1)dx=x^4yln(xy)+h(y)$$
$$\psi_y=x^4ln(xy)+x^4+h'(y)=x^4(ln(xy)+1)=N$$
$$∴h'(y)=0, h(y)=C$$
$$Therefore, \psi(x,y)= x^4yln(xy)=C$$
as x=1, y=1, we get C=0
$$∴x^4yln(xy)=0$$

10
##### Quiz-3 / Re: Q3 TUT 0601
« on: October 12, 2018, 07:04:07 PM »
I agree with Nick's answer. Here is just a version without word explanation which is easier to see.
\begin{align*} ∵ r=2 , r=-3 \end{align*}
\begin{align*} ∴(r-2)(r+3)=0 \end{align*}
\begin{align*} ∴ r^2-2r+3r-6=0 \end{align*}
\begin{align*} ∴r^2+r-6=0 \end{align*}

\begin{align*} ∴ y''+y'-6y=0 \end{align*}

Pages: [1]