Messages

16

Quiz-6 / Re: Q6 TUT 5101

« on: November 17, 2018, 04:07:45 PM »

a) Finding the eigenvalues:

Set the determinant = 0

\begin{align}
(2 - \lambda)(-2 - \lambda) - (-5)(\alpha) &= 0\\
\lambda^2 - 4 + 5\alpha &= 0\\
\lambda &= \pm \sqrt{4 - 5\alpha}
\end{align}

b)

Case 1: Eigenvalues real and opposite sign
when: $\alpha$ < $\frac{4}{5}$ (unstable saddle)

Case 2: Eigenvalues complex and opposite sign
when: $\alpha$ > $\frac{4}{5}$ (stable centre)

c) will post below:

17

Quiz-5 / Re: Q5 TUT 5102

« on: November 10, 2018, 01:42:01 PM »

As mentioned before in the forum, $g(t)$ should be $\coth(t)$ not $\cosh(t)$

I recalculated:

$$
W(t) = 2e^{2t}\\

W_1(t) = -2e^{2t}\\

W_2(t) = e^t\\

W_3(t) = e^t

$$

This gives

$$
\begin{align}

y_p(t) &= \int{\frac{\coth(t)(-2e^{2t})dt}{2e^{2t}}} + e^t\int{\frac{\coth(t)e^t}{2e^{2t}}dt} + e^t\int{\frac{\coth(t)e^t}{2e^{2t}}dt}\\

y_p(t) &= -\int{\coth(t)dt} + e^t\int{\coth(t)e^{-t}dt}\\

y_p(t) &= -\ln|\sinh(t)| + e^t(e^{-t} + \ln|1-e^{-t}| - \ln|e^{-t} + 1|)\\

y(t) &= c_1+c_2e^t+c_3e^{-t} -\ln|\sinh(t)| + e^t(e^{-t} + \ln|1-e^{-t}| - \ln|e^{-t} + 1|)
\end{align}


$$

18

Quiz-5 / Re: Q5 TUT 0201

« on: November 04, 2018, 02:31:19 AM »

oh right! Totally my bad! time to relearn trig..   

19

Quiz-5 / Re: Q5 TUT 5102

« on: November 02, 2018, 04:36:21 PM »

Tianfangtong Zhang I think you made a small typo,

$W_1(t) = -2$

not 2.

I think you fixed it in your final answer though.

20

Quiz-5 / Re: Q5 TUT 0201

« on: November 02, 2018, 04:25:48 PM »

Hi Guanyao Liang,

Your answer is very close, but I think you messed up a calculation. The integral of $\tan(t)$ is $-\ln|\cos(t)|$, not $\ln|\sec(t)|$.

Therefore the solution should be:

$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$

21

Quiz-4 / Re: Q4 TUT 03014

« on: October 27, 2018, 09:07:57 PM »

Right, I think it should be:

$Y(t) = Y_c + Y_p$

$Y(t) = c_1 \cos(3t) + c_2 \sin(3t)+ \sin(3t) \ln| \sec(3t)+ \tan(3t)|-1 + c_3$, where $c_1, c_2, c_3 \in {\rm I\!R}$

Since $c_3$ can absorb the "-1", it becomes:

$Y(t) = c_1 \cos(3t) + c_2 \sin(3t)+ \sin(3t) \ln| \sec(3t)+ \tan(3t)| + c_3$, where $c_1, c_2, c_3 \in {\rm I\!R}$

22

Quiz-4 / Re: Q4 TUT 03014

« on: October 27, 2018, 03:57:36 PM »

So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{W(s)}\,ds=-1+sin(3t)ln$||sec(3t)+tan(3t)||

Small correction, likely due to typesetting error, but I think you mean the following:

So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{g(s)W_{2}(s)}{W(s)}\,ds=-1+sin(3t)ln|sec(3t)+tan(3t)|$

23

Quiz-3 / Re: Q3 TUT 0701

« on: October 13, 2018, 10:12:57 AM »

Thomas, your solution seems mostly correct, but I think you are missing something small. If I remember correctly, Abel's theorem has a negative factor you forgot.

Instead, the solution should be:

$$W(t)=e^{∫-p(t)dt}=e^{∫-\tan(t)dt}=e^{\ln(\cos(t))+c}=c(\cos(t))$$, for some constant c.