### 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 - Boyu Zheng

Pages: [1]
1
##### Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 07:39:51 AM »
here is my answer

2
##### Term Test 2 / Re: TT2B-P2
« on: November 20, 2018, 07:39:28 AM »
Here is my answer

3
##### Term Test 2 / Re: TT2-P2
« on: November 20, 2018, 07:38:42 AM »
$$a)w=ce^{-\int p(t) \mathrm{d}t}=ce^{-\int 1\mathrm{d}t}=ce^{-t}$$
$$b)\gamma^3+\gamma^2-\gamma-1=0$$
\begin{align*}
(\gamma-1)(\gamma^2+2\gamma+1)=0\\
\gamma_1=1,\gamma_2=\gamma_3=-1\\
yc(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t
\end{align*}
\begin{align*}
w&=      \begin{vmatrix}
e^t   &      e^{-t}       &te^{-t}\\
e^t   &      -e^{-t}       &e^{-t}-te^{-t}\\
e^t   &      e^{-t}       &-2e^{-t}+te^{-t}
\end{vmatrix}
=e^t(-e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(e^{-t}-te^{-t}))\\
&-e^t(e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(te^{-t}))+e^t(e^{-t}(e^{-t}-te^{-t})+e^{-t}(te^{-t}))\\
&=4e^{-t}
\end{align*}
\centerline{Compared with(a),c=4}
\begin{align*}
c) yp(t)&=At^2e^{-t}\\
y'p(t)&=2Ate^{-t}-At^2e^{-t}\\
y''p(t)&=A(-4e^{-t}t+2e^{-t}+e^{-t}t^2)\\
y'''p(t)&=A(6e^{-t}t-6e^{-t}-e^{-t}t^2)\\
y'''+y''&-y'-y=8e^{-t}
\end{align*}
$$6Ae^{-t}t-6Ae^{-t}-Ae^{-t}t^2-4Ae^{-t}+2Ae^{-t}+Ae^{-t}t^2-2Ate^{-t}+At^2e^{-t}-At^2e^{-t}=8e^{-t}$$
$$-4A=8$$
$$A=-2$$
$$y(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t-2e^{-t}$$

4
##### Quiz-6 / Re: Q6 TUT 0501
« on: November 17, 2018, 04:21:46 PM »
\begin{equation*}
det
\begin{pmatrix}
-3\lambda    &0          &2 \\
1         & -1\lambda    &0 \\
-2         & -1            & -\lambda
\end{pmatrix}
=-\lambda^3-4\lambda^2-7\lambda-6=-(\lambda+2)(\lambda+2\lambda+3)=0
\end{equation*}
$$\lambda=-2,\lambda=\sqrt{2}\qquad i-1,\lambda=-\sqrt{2}\qquad i-1$$

when $\lambda$=-2
\begin{equation*}
\begin{pmatrix}
-1          &0           &2 \\
1         & 1             &0 \\
-2         & -1            & 2
\end{pmatrix}

\begin{pmatrix}
x_1            \\       x_2       \\       x_3
\end{pmatrix}=0
\end{equation*}
$$\text{let } x_3=t,x_1=2t,x_2=-2t, x= \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}t$$

when $\lambda=\sqrt{2}\qquad i-1$

\begin{equation*}
\begin{pmatrix}
-2-\sqrt{2}\qquad i           &0           &2 \\
1         & \sqrt{2}\qquad i            &0 \\
-2         & -1            & -\sqrt{2}\qquad i+1
\end{pmatrix}
\begin{pmatrix}
x_1    \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$x=\begin{pmatrix} \frac{2}{3}-\frac{i\sqrt{2}\qquad}{3}\\\frac{-1}{3}-\frac{i\sqrt{2}\qquad}{3} \\1 \end{pmatrix}t$$
\begin{equation*}
x(t)=c_1e^{-2t}
\begin{pmatrix}
2\\-2\\1
\end{pmatrix}
+c_2e^{-t}
\begin{pmatrix}
\frac{2}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
-\frac{1}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
\cos\sqrt{2}\theta
\end{pmatrix}
+c_3e^{-t}i
\begin{pmatrix}
\frac{2}{3}\sin \sqrt{2}\theta-\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
-\frac{1}{3}\sin \sqrt{2}\theta+\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
\sin\sqrt{2}\theta
\end{pmatrix}
\end{equation*}

5
##### Quiz-6 / Re: Q6 TOT 0301
« on: November 17, 2018, 04:10:39 PM »
\begin{equation*}
det
\begin{pmatrix}
1-\lambda    &1           &2 \\
1         & 2-\lambda    &1 \\
2         & 1            & 1-\lambda
\end{pmatrix}
=-\lambda^3+4\lambda^2+\lambda-4=-(\lambda-1)(\lambda-4)(\lambda+1)=0
\end{equation*}
$$\lambda=1,\lambda=4,\lambda=-1$$

when $\lambda$=1
\begin{equation*}
\begin{pmatrix}
0           &1           &2 \\
1         & 1             &1 \\
2         & 1            & 0
\end{pmatrix}
\sim
\begin{pmatrix}
2           &1           &0 \\
0         & 1             &2 \\
1         & 1            & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2           &0           &-2 \\
-1         & 0             &1 \\
1         & 1            & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2           &0           &-2 \\
0         & 0             &0 \\
1         & 1            & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2           &0           &-2 \\
1         & 1             &1 \\
0         & 0            & 0
\end{pmatrix}
\sim
\begin{pmatrix}
x_1            \\       x_2       \\       x_3
\end{pmatrix}=0
\end{equation*}
$$\text{let } x_3=t,x_1=t,x_2=-2t x= \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}t$$

when $\lambda$=4
\begin{equation*}
\begin{pmatrix}
-3           &1           &2 \\
1         & -2             &1 \\
2         & 1            & -3
\end{pmatrix}
\sim
\begin{pmatrix}
-3           &1           &2 \\
0         & -5             &5 \\
2         & 1            & -3
\end{pmatrix}
\begin{pmatrix}
x_1    \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$x=\begin{pmatrix} 1 \\1 \\1 \end{pmatrix}t$$
when $\lambda$=-1
\begin{equation*}
\begin{pmatrix}
2           &1           &2 \\
1         & 3             &1 \\
2         & 1            & 2
\end{pmatrix}
\sim
\begin{pmatrix}
2           &1           &2 \\
0         & 5             &0 \\
0         & 0            & 0
\end{pmatrix}
\begin{pmatrix}
x_1    \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$x=\begin{pmatrix}-1\\0\\1\end{pmatrix}$$
$$x(t)=c_1e^4t\begin{pmatrix}1\\1\\1\end{pmatrix}+c_2e^{-t}\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3e^t\begin{pmatrix}1\\-2\\1\end{pmatrix}$$

6
##### MAT244--Misc / concern about the quiz6 grade
« on: November 11, 2018, 08:47:30 PM »
I wrote the quiz  during the tutorial and i receive a zero but I don't know what's going on with it. And I didn't get the feedback from the tutorial. I just wonder whether I can see the result and figure it out the problem.

7
##### MAT244--Misc / Re: MAT244 study group
« on: October 10, 2018, 04:12:11 PM »
Lets do this together

8
##### MAT244--Misc / Ask about the test
« on: October 10, 2018, 04:06:55 PM »
Do we have to know Eulerâ€˜s method for this test ?

9
##### Quiz-1 / Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« on: September 30, 2018, 12:10:17 AM »
Solution for TUT5101
Question: Find the solution of the given initial value problem y'-2y = e^2t , y(0)=2
let p(t)=-2 and set u=e^(integral p(t)dt) then you get u = e^(-2t)
then you multiply u on each side of the standard form equation and you get
e^(-2t)y'-2e^(-2t)y = 1
then you can find the LHS is equal to (e^(-2t)y)' = 1
integral each side you get (e^-2t)y = t +C
rearrange the equation you get y=e^2t(t+C)
y(t)=(t+C)e^2t
plug in y(0)=2 and you get C=2
y can be written as y=e^2t(t+2)

10
##### MAT244--Lectures & Home Assignments / Re: Question about how to deal with non exact equation
« on: September 20, 2018, 12:29:07 AM »
I just reread your notes from the lecture but I still don't understand for try3 why do you use -(dM/dy - dN/dx)/Mx-Ny = f(xy) to get M+Ny'=0?

11
##### MAT244--Lectures & Home Assignments / Question about how to deal with non exact equation
« on: September 19, 2018, 11:01:15 PM »
Hi professor Ivrii, You ask everybody, not just me. V.I.
When I do this question (3x+6/y) +(x^2/y+3y/x)dy/dx = 0 , I tried to get My and Nx and after take the partial derivative. I found My is not equal to Nx and its not exact.
I just wonder how to find the intergation factor in order to make the equation exact?

12
##### MAT244--Lectures & Home Assignments / Last question from todays lecture
« on: September 13, 2018, 08:08:58 PM »
Hi,
I have a question from today's lecture. The initial condition equation is $y'= (x-y)/(x+y-2)$ and after a couple steps we get
$$dw/dt = (t-w+a-b)/(t+w+a-b-2) = t-w/t+w.$$
And I am confused about why you say that $a-b=0$ and $a-b-2=0$?

Pages: [1]