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 - Yunqi(Yuki) Huang

Pages: [1]
1
Quiz-5 / Re: Q5 TUT 0501
« on: November 19, 2018, 11:20:47 PM »
Isolate the first equation for$ x_2:x_2=0.5x′_1+0.25x_1 (1)$
Differentiating both sides we get$x′_2=0.5x″_1+0.25x′_1(2)$
Substitute (1) and (2) in the above into the second equation$x″_1+x_1+4.25x_1=0(3)$
Solving (3):r=−0.5±2i
So, we now know $x_1=e^{-0.5t}(c_1cos(2t)+c_2sin(2t))$ and $x2=e^{−0.5t}(-c_1sin(2𝑡)+c_2cos(2𝑡))$
By initial conditions:$c_1 = -2$ and $c_2=2$
Final solution:$x_1=e^{-0.5t}(−2cos(2𝑡)+2sin(2𝑡))$ and $x_2=e^{-0.5t}(2cos(2𝑡)+2sin(2𝑡))$

2
Quiz-4 / Re: Q4 TUT 03014
« on: October 28, 2018, 12:18:31 AM »
For the left-hand side, we could write as $r^2+9=0$.Thus $r=3i$ or $r=-3i.
$
Now  $Y_{c}=C_{1}cos3t+C_{2}sin3t
$
For the Right-hand side, $Y_{1}(t)=cos3t$$.  Y_{2}(t)+sin3t$$. g(t)=9sec^2(3t)$
 $$W(t)=\left[
 \begin{matrix}
   cos3t & sin3t \\
   -3sin3t& 3cos3t
  \end{matrix}
  \right] \tag{3}=3
$$
$$W_{1}(t)=\left[
 \begin{matrix}
   0 & sin3t \\
   1 & 3cos3t
  \end{matrix}
  \right] \tag{3}=-sin3t
$$
$$W_{2}(t)=\left[
 \begin{matrix}
   cos3t & 0 \\
   -3sin3t& 1
  \end{matrix}
  \right] \tag{3}=cos3t
$$
We could check the integral table that $\int\tan(t)\sec(t)dt=sec(t)$. $\int\sec(t)dt=ln|sec(t)+tan(t)|$
So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{g(s)W_{2}(s)}{W(s)}\,ds=-3\cos(3t)\int\tan(3s)\sec(3s)ds+3\sin(3t)\int\sec(3t)ds=-3\cos3t*\frac{1}{3}sec(3t)+3\sin(3t)ln|\sec3t+tan3t|*\frac{1}{3}$
Thus, the general solution is $Y(t)=C_{1}\cos(3t)+C_{2}\sin(3t)+\sin(3t)ln|\sec(3t)+\tan(3t)|-1$

3
Quiz-4 / Re: Q4 TUT 03014
« on: October 26, 2018, 05:59:28 PM »
For the left-hand side, we could write as $r^2+9=0$.Thus $r=3i$ or $r=-3i.
$
Now  $Y_{c}=C_{1}cos3t+C_{2}sin3t
$
For the Right-hand side, $Y_{1}(t)=cos3t$$.  Y_{2}(t)+sin3t$$. g(t)=9sec^2(3t)$
 $$W(t)=\left[
 \begin{matrix}
   cos3t & sin3t \\
   -3sin3t& 3cos3t
  \end{matrix}
  \right] \tag{3}=3
$$
$$W_{1}(t)=\left[
 \begin{matrix}
   0 & sin3t \\
   1 & 3cos3t
  \end{matrix}
  \right] \tag{3}=-sin3t
$$
$$W_{2}(t)=\left[
 \begin{matrix}
   cos3t & 0 \\
   -3sin3t& 1
  \end{matrix}
  \right] \tag{3}=cos3t
$$
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)||
Thus, the general solution is $Y(t)=C_{1}cos(3t)+C_{2}sin(3t)+sin(3t)ln$|sec(3t)+tan(3t)|-1

4
Quiz-4 / Re: Q4 TUT 03014
« on: October 26, 2018, 05:46:15 PM »
For the left-hand side, we could write as $r^2+9=0$.Thus $r=3i$ or $r=-3i.
$
Now  $Y_{c}=C_{1}cos3t+C_{2}sin3t
$
For the Right-hand side, $Y_{1}(t)=cos3t$, $Y_{2}(t)+sin3t$, $g(t)=9sec^2(3t)$
 $$W(t)=\left[
 \begin{matrix}
   cos3t & sin3t \\
   -3sin3t& 3cos3t
  \end{matrix}
  \right] \tag{3}=3
$$
$$W_{1}(t)=\left[
 \begin{matrix}
   0 & sin3t \\
   1 & 3cos3t
  \end{matrix}
  \right] \tag{3}=-sin3t
$$
$$W_{2}(t)=\left[
 \begin{matrix}
   cos3t & 0 \\
   -3sin3t& 1
  \end{matrix}
  \right] \tag{3}=cos3t
$$
So, the particular solution is $$Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{g(s)W_2(s)}{W(s)}\,ds=-1+\sin(3t)\ln |\sec(3t)+\tan(3t)|$$
Thus, the general solution is $Y(t)=C_{1}cos(3t)+C_{2}sin(3t)+sin(3t)ln$||sec(3t)+tan(3t)||-1

5
Quiz-3 / Re: Q3 TUT0401
« on: October 12, 2018, 07:16:55 PM »
$$
f(t)=t
$$
$$
So f'(t)=1
$$
$$
W=tg'(t)-g(t)=t^2e^t
$$
$$
g'(t)-\frac{1}{t}g(t)=te^t
$$
$$
p(t)=-\frac{1}{t}
$$
$$
u(t)=e^{\int1p(t)dt}
$$
$$
wherep(t)=-\frac{1}{t}
$$
$$
u(t)=\frac{1}{t}
$$
$$
\frac{1}{t}g(t)=\int e^tdt
$$
$$
g(t)=te^t+ct$$where let c=1
$$
$$
Thus g(t)=te^t+t

6
Quiz-3 / Re: Q3 TUT 0301
« on: October 12, 2018, 06:49:55 PM »
$$
r=-\frac{1}{2}or-2
$$
$$
(r+\frac{1}{2})(r+2) = 0
$$
$$
r^2+ 2r+\frac{1}2r +1 = 0
$$
$$
r^2+\frac{5}{2}r+1 = 0
$$
$$
2r^2 + 5r +2 = 0
$$
Therefore
$$
2y'' + 5y' + 2y = 0
$$

7
Quiz-3 / Re: Q3 TUT 0801
« on: October 12, 2018, 06:28:18 PM »
the new following attachment is right. sorry for my previous mistake to the answer

8
Quiz-3 / Re: Q3 TUT0401
« on: October 12, 2018, 06:20:48 PM »
in the attachment

9
Quiz-3 / Re: Q3 TUT 0301
« on: October 12, 2018, 06:16:38 PM »
in the attchement

Pages: [1]