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### Messages - Yunqi(Yuki) Huang

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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

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

10
##### MAT244--Lectures & Home Assignments / Re: non-homogenous equation
« on: October 10, 2018, 10:25:45 AM »
I mean we usually solve the second order equation like y''+2y'+y= 2e^(-t), which right-hand side is a non-homogenous equation. We could assume right-hand side is Y(t)=Ae^(-t), then substitute y'' and y' and y in the ordinary equation. However, how can we solve the equation like y''+2y'+y=3? If it is right to assume Y(t)=A for the right-hand side?

11
##### MAT244--Lectures & Home Assignments / non-homogenous equation
« on: October 10, 2018, 01:36:33 AM »
I wonder that how can we solve the problem when the right-hand side of the non-homogenous equation is only a constant?

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