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Messages - Tim Mengzhe Geng

Pages: 1 [2]
16
Final Exam / Re: FE-P1
« on: April 11, 2018, 11:15:27 PM »
Actually Professor I remember on the test the last term is -5y^4 instead of -5y^3. I'm not sure whether I'm correct.

It is $-5y^3$.
Yes you are right and the solution is corresponding to the case $-5y^3$.

17
Final Exam / Re: FE-P2
« on: April 11, 2018, 11:14:33 PM »
First we find the solution for the homogeneous system
\begin{equation}
    y^{(3)}-2y^{(2)}+4y^{(1)}-8y=0
\end{equation}
The corresponding characteristic equation is
\begin{equation}
    r^3-2r^2+4r-8=0
\end{equation}
Three roots are
\begin{equation}
    r_1=2
\end{equation}
\begin{equation}
    r_2=2i
\end{equation}
\begin{equation}
    r_3=-2i
\end{equation}
Then the solution for the homogeneous system is

\begin{equation}
    y_c(t)=c_1e^{2t}+c_2\cos(2t)+c_3\sin(2t)
\end{equation}
Then we follow to find a particular solution $Y(t)$
We should have
\begin{equation}
    Y(t)=Y_1(t)+Y_2(t)
\end{equation}
where
\begin{equation}
    Y_1(t)=Ate^{2t}
\end{equation}
and
\begin{equation}
    Y_2(t)=M\sin(t)+N\cos(t)
\end{equation}
By plugging in to the equation, we can find that $A=2$, $M=2$ and $N=-4$
In this way we get the required general solution
\begin{equation}
    y(t)=c_1e^{2t}+c_2\cos(2t)+c_3\sin(2t)+2te^{2t}+2\sin(t)-4\cos(t)
\end{equation}

18
Final Exam / Re: FE-P1
« on: April 11, 2018, 10:51:53 PM »
Actually Professor I remember on the test the last term is -5y^4 instead of -5y^3. I'm not sure whether I'm correct.

19
Final Exam / Re: FE-P1
« on: April 11, 2018, 10:50:49 PM »
First we find the integrating factor.
Note that
\begin{equation}
    M_y=2x\cdot\cos(y) - 2xy\cdot\sin(y) - 2y\cdot\cos(x)
\end{equation}
\begin{equation}
    N_x=4x\cdot\cos(y) - 2xy\cdot\sin(y) - 3y\cdot\cos(x)
\end{equation}
\begin{equation}
    N_x - M_y=2x\cdot\cos(y) -  y\cdot\cos(x)
\end{equation}   
\begin{equation}
   (N_x - M_y)/M=1/y
\end{equation}
Therefore, the integrating factor is only dependent on y.
\begin{equation}
    \ln u= \ln y
\end{equation}
\begin{equation}
    u(y)= y
\end{equation}
Multiply u(y) on both sides of the equation. Then we have
\begin{equation}
    \phi_x=2xy^2\cos(y) - y^3\cos(x)
\end{equation}
\begin{equation}
    \phi=x^2y^2\cos(y)  -y^3\sin(x) +h(y)
\end{equation}
\begin{equation}
    \phi_y=2x^2y\cos(y) -x^2y^2\sin(y) - 3y^2\sin(x) +h^\prime(y)
\end{equation}
By comparison, we get
\begin{equation}
    h^\prime(y)=-5y^4
\end{equation}
\begin{equation}
    h^\prime(y)=-y^5
\end{equation}
Then we have
\begin{equation}
    \phi=x^2y^2\cos(y)-y^3\sin(x) -y^5
\end{equation}
The general solution is
\begin{equation}
\phi=x^2y^2\cos(y)-y^3\sin(x) -y^5=c
\end{equation}

20
MAT244--Misc / Mark for quiz 5
« on: March 15, 2018, 12:12:32 PM »
I wonder whether mark for quiz 5 has been posted on the Blackboard?
I ask this because I take this quiz in a different tutorial session and haven't seen my mark for it on Blackboard yet.

21
MAT244--Misc / Term Test 1 Grading
« on: March 06, 2018, 02:26:00 AM »
I have some doubt on the grading of Problem 2.
I guess the form of the solution, prove it works and get the answer right, but finally get no mark for part (a) and half mark for part (b).
May I know which TA I can turn to to discuss my solution?
Thanks in advance!

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