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### Messages - Yulin WANG

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1
##### Quiz-7 / Re: Q7 TUT 0201
« on: November 30, 2018, 11:13:57 PM »
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer?

2
##### Quiz-7 / Re: Q7 TUT 0801
« on: November 30, 2018, 04:40:26 PM »
(a)

\left\{
\begin{array}{**lr**}
y+x-x^{3}-xy^{2}=0 &  \\
-x+y-x^{2}y-y^{3}=0\\
\end{array}
\right.

\left\{
\begin{array}{**lr**}
x^{2}+y^{2}=0
\end{array}
\right.

\left\{
\begin{array}{**lr**}
x=0 &  \\
y=0\\
\end{array}
\right.

Therefore, the only critical point is (0,0)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
1-3x^{2}-y^{2} & 1-2xy \\
-1-2xy & 1-x^{2}-3y^{2}
\end{bmatrix}\\
~\\
J(0,0) &= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
\end{align*}
(c)
\begin{align*}
For (0,0), let A&= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
-1 & 1-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda-1)^{2}+1=0\\
~\\
\lambda &= 1 \pm i \\
~\\
Then \ the \ system \ has \ a \ clockwise \ spiral \ outwards \ at \ (0,0) \\
\end{align*}
(d) In the attachment.

3
##### Quiz-7 / Re: Q7 TUT 0201
« on: November 30, 2018, 04:39:52 PM »
(a)

\left\{
\begin{array}{**lr**}
1-xy=0 &  \\
x-y^{3}=0\\
\end{array}
\right.

\left\{
\begin{array}{**lr**}
xy=1 &  \\
x=y^{3}\\
\end{array}
\right.

\left\{
\begin{array}{**lr**}
x=1 &  \\
y=1\\
\end{array}
\right.
or
\left\{
\begin{array}{**lr**}
x=-1 &  \\
y=-1\\
\end{array}
\right.

Therefore, the critical points are (1,1) and (-1,-1)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
-y & -x \\
1 & -3y^{2}
\end{bmatrix}\\
~\\
J(1,1) &= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
J(-1,-1) &= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}
\end{align*}
(c)
\begin{align*}
For (1,1), let A&= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
-1-\lambda & -1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda+1)+1=0\\
~\\
\lambda_{1} &= \lambda_{2} = -2 \\
~\\
Then \ the \ system \ has \ a \ stable \ improper \ node \ at \ (1,1) \\
~\\
For (-1,-1), let A&= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda-1)-1=0\\
~\\
\lambda = -1 \pm \sqrt{5} \\
~\\
Then \ the \ system \ has \ a \ unstable \ saddle \ point \ at \ (1,1) \\
\end{align*}
(d) In the attachment.

4
##### Term Test 2 / Re: TT2B-P3
« on: November 20, 2018, 07:55:59 PM »
My answer is different from Xiaoyuan.
(a)
\begin{align*}
Let A &= \begin{bmatrix}
-1 & 1 \\
-1 & -3
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-1 - \lambda & 1 \\
-1 & -3 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (\lambda + 3)(\lambda +1) + 1\\
~\\
&= \lambda^{2} + 3\lambda + \lambda + 3 +1\\
~\\
&= (\lambda + 2)^{2} = 0\\
~\\
\lambda_{1} = \lambda_{2} &= -2\\
~\\
For \ \lambda = -2, A - \lambda I &= \begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}\\
~\\
Since,\ null(\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}) &= span\{\begin{bmatrix}
1\\
-1 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}\\
~\\
want\ to\ find\ a\ vector\ w, \ where\ &(A - \lambda I)w = v\\
~\\
Then,\ \begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}w &= \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}\\
~\\
Then,\ pick\ w &= \begin{bmatrix}
1\\
0 \\
\end{bmatrix}\\
~\\
Thus,\ x(t) &= c_{1}e^{\lambda t}v + c_{2}e^{\lambda t}(w + tv)\\
&= c_{1}e^{-2t}\begin{bmatrix}
1\\
-1 \\
\end{bmatrix} + c_{2}e^{-2t}(\begin{bmatrix}
1\\
0 \\
\end{bmatrix} + t \begin{bmatrix}
1\\
-1 \\
\end{bmatrix})\\
\end{align*}
(b) In the attachment.

(c)
\begin{align*}
Since\ W(x_{1}, x_{2})(t) &= det \begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix}\\
~\\
&= -te^{-4t} + e^{-4t} + te^{-4t}\\
~\\
&= e^{-4t} \neq 0\\
~\\
So\ x_{1}(t) \ and \ x_{2}(t) \ form \ a \ fundamental &\ set\ of \ solutions\\
~\\
Then\ the\ fundamenta l\ matrix\ \varphi (t) &= \begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix}\\
~\\
Guess\ that\ x(t) &= \varphi (t)\mu(t)\\
~\\
Then\ solve\ that\ \varphi (t)\mu'(t) &= g(t)\\
~\\
\begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix} \begin{bmatrix}
\mu_{1}'(t)\\
\mu_{2}'(t) \\
\end{bmatrix} &= \begin{bmatrix}
\frac{e^{-2t}}{t^{2} + 1}\\
0\\
\end{bmatrix}\\
~\\
use\ row\ reduction\ of\ the\ matrix,\ we\ have:\\
~\\
\begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
0 & e^{-2t}
\end{bmatrix} \begin{bmatrix}
\mu_{1}'(t)\\
\mu_{2}'(t) \\
\end{bmatrix} &= \begin{bmatrix}
\frac{e^{-2t}}{t^{2} + 1}\\
\frac{e^{-2t}}{t^{2} + 1}\\
\end{bmatrix}\\
~\\
\mu_{1}'(t) &= \frac{-t}{t^{2} + 1}\\
\mu_{2}'(t) &= \frac{1}{t^{2} + 1}\\
~\\
Then,\ we\ have&:\\
~\\
\mu_{1}(t) &= -\frac{1}{2}\ln(t^{2} + 1) + c_{1}\\
\mu_{2}(t) &= arctan(t) + c_{2}\\
~\\
Thus,\ x(t) &= \varphi (t)\mu(t)\\
~\\
&= \begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix} \begin{bmatrix}
-\frac{1}{2}\ln(t^{2} + 1) + c_{1} \\
arctan(t) + c_{2}
\end{bmatrix}\\
~\\
&= \begin{bmatrix}
-\frac{1}{2}e^{-2t}\ln(t^{2} + 1) + (e^{-2t} +te^{-2t})arctan(t)\\
\frac{1}{2}e^{-2t}\ln(t^{2} + 1) - te^{-2t}arctan(t)
\end{bmatrix} + c_{1}e^{-2t}\begin{bmatrix}
1\\
-1 \\
\end{bmatrix} + c_{2}e^{-2t}(\begin{bmatrix}
1\\
0 \\
\end{bmatrix} + t \begin{bmatrix}
1\\
-1 \\
\end{bmatrix})\\
~\\
&= e^{-2t}\ln(t^{2} + 1) \begin{bmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
\end{bmatrix} + te^{-2t}arctan(t) \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}) + e^{-2t}arctan(t) \begin{bmatrix}
1\\
0 \\
\end{bmatrix} + c_{1}e^{-2t}\begin{bmatrix}
1\\
-1 \\
\end{bmatrix} + c_{2}e^{-2t}(\begin{bmatrix}
1\\
0 \\
\end{bmatrix} + t \begin{bmatrix}
1\\
-1 \\
\end{bmatrix})\\
~\\
Sine\ x(0) &= \begin{bmatrix}
0\\
0\\
\end{bmatrix}\\
~\\
Then,\ c_{1} + c_{2} &= 0 \ and\ -c_{1} = 0\\
~\\
Thus,\ c_{1} &= c_{2} = 0\\
~\\
Therefore,\ x(t) &= e^{-2t}\ln(t^{2} + 1) \begin{bmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
\end{bmatrix} + te^{-2t}arctan(t) \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}) + e^{-2t}arctan(t) \begin{bmatrix}
1\\
0 \\
\end{bmatrix}\\
\end{align*}

5
##### Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 04:03:36 PM »
here is my answer
Hi, my answer is same like yours except I got $C_1=-2ln2, C_2=0$
I am not confident about my answer, feel free to correct me
My answer is the same as yours.

6
##### Term Test 2 / Re: TT2B-P4
« on: November 20, 2018, 03:50:55 PM »
\begin{align*}
Let A &= \begin{bmatrix}
-3 & -2 \\
5 & -5
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-3 - \lambda & -2 \\
5 & -5 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (5 + \lambda)(3 + \lambda) + 10\\
~\\
&= \lambda^{2} + 8\lambda +25\\
~\\
&= (\lambda +4)^{2} + 9 = 0\\
~\\
\lambda &= -4 \pm 3i \\
~\\
For \lambda = -4 + 3i, A - \lambda I &= \begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}\\
~\\
Since,\ nul(\begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}) &= span\{\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
Since \ e^{\lambda t}v &= e^{-4 + 3i}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}e^{3it}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
&= e^{-4t}(cos3t + isin3t)\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}\begin{bmatrix}
3icos3t - 3sin3t + cos3t + isin3t\\
5cos3t +5isin3t \\
\end{bmatrix}\\
~\\
So, \ \phi_{1}(t) &= e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix}\\
~\\
\phi_{2}(t) &= e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
~\\
Thus, \ x(t) &= c_{1}\phi_{1}(t) + c_{2}\phi_{2}(t)\\
~\\
&= c_{1}e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix} + c_{2}e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
\end{align*}
(b) In the attachment.
Hi Yulin, I think you made a mistake in your graph, since c=5>0, your graph should be counterclockwise instead of clockwise
Thanks for your help.

7
##### Term Test 2 / Re: TT2B-P4
« on: November 20, 2018, 03:10:39 PM »
(a)
\begin{align*}
Let A &= \begin{bmatrix}
-3 & -2 \\
5 & -5
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-3 - \lambda & -2 \\
5 & -5 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (5 + \lambda)(3 + \lambda) + 10\\
~\\
&= \lambda^{2} + 8\lambda +25\\
~\\
&= (\lambda +4)^{2} + 9 = 0\\
~\\
\lambda &= -4 \pm 3i \\
~\\
For \ \lambda = -4 + 3i, A - \lambda I &= \begin{bmatrix}
1 - 3i & -2 \\
5 & -1 - 3i
\end{bmatrix}\\
~\\
Since,\ null(\begin{bmatrix}
1 - 3i & -2 \\
5 & -1 - 3i
\end{bmatrix}) &= span\{\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
Since \ e^{\lambda t}v &= e^{-4 + 3i}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}e^{3it}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
&= e^{-4t}(cos3t + isin3t)\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}\begin{bmatrix}
3icos3t - 3sin3t + cos3t + isin3t\\
5cos3t +5isin3t \\
\end{bmatrix}\\
~\\
So, \ \phi_{1}(t) &= e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix}\\
~\\
\phi_{2}(t) &= e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
~\\
Thus, \ x(t) &= c_{1}\phi_{1}(t) + c_{2}\phi_{2}(t)\\
~\\
&= c_{1}e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix} + c_{2}e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
\end{align*}
(b) In the attachment.

8
##### Quiz-5 / Re: Q5 TUT 0601
« on: November 02, 2018, 03:22:26 PM »
(1)
Rewrite the first equation: $x_{2} = \frac{1}{2}x_{1}' + \frac{1}{4}x_{1}$
Then, $x_{2}' = \frac{1}{2}x_{1}'' + \frac{1}{4}x_{1}'$
Plug into the second equation: $\frac{1}{2}x_{1}'' + \frac{1}{4}x_{1}' = -2x_{1} -\frac{1}{2}(\frac{1}{2}x_{1}' + \frac{1}{4}x_{1})$
Then, we get: $x_{1}'' + x_{1}' + \frac{17}{4}x_{1} = 0$ which is  a second order ODE of $𝑥_{1}$
(2)
\begin{align*}
Let A &=
\begin{bmatrix}
-\frac{1}{2} & 2 \\
-2 & -\frac{1}{2}
\end{bmatrix}\\
A-\lambda I &=
\begin{bmatrix}
-\frac{1}{2} -\lambda & 2 \\
-2 & -\frac{1}{2} -\lambda
\end{bmatrix}\\
det(A-\lambda I) &= (\lambda + \frac{1}{2})^{2} + 4 = 0\\
\lambda + \frac{1}{2} &= \pm 2i\\
\lambda &= -\frac{1}{2} \pm 2i\\
\end{align*}
For $\lambda = -\frac{1}{2} + 2i$:
\begin{align*}
A-\lambda I &= \begin{bmatrix}
-2i & 2 \\
-2 & -2i
\end{bmatrix}\\
null\begin{bmatrix} -2i & 2 \\ -2 & -2i \end{bmatrix}
&= span{\begin{bmatrix} 1\\ i \end{bmatrix}\\}\\
so, \ the \ eigenvector \  V &= \begin{bmatrix} 1\\ i \end{bmatrix}\\
Then, \  e^{\lambda t}V &= e^{(-\frac{1}{2} + 2i)t}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= e^{-\frac{1}{2}t}e^{2it}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&=  e^{-\frac{1}{2}t}(cos2t + isin2t)\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= \begin{bmatrix} cos2t + isin2t\\ icos2t - sin2t \end{bmatrix}\\
Thus, \  \phi_{1}(t) &= e^{-\frac{1}{2}t}\begin{bmatrix} cos2t\\- sin2t \end{bmatrix}\\
\phi_{2}(t) &= e^{-\frac{1}{2}t}\begin{bmatrix} sin2t\\cos2t \end{bmatrix}\\
Therefore, \ x_{1} &= c_{1}e^{-\frac{1}{2}t}cos2t + c_{2}e^{-\frac{1}{2}t}sin2t\\
x_{2} &= -c_{1}e^{-\frac{1}{2}t}sin2t + c_{2}e^{-\frac{1}{2}t}cos2t\\
Since, \ x_{1}(0) &= -2 \ and \  x_{2}(0) = 2\\
So, \ c_{1} &= -2 \ and \ c_{2} = 2\\
\end{align*}
Therefore,
\begin{align*}
x_{1} &= -2e^{-\frac{1}{2}t}cos2t + 2e^{-\frac{1}{2}t}sin2t\\
x_{2} &= 2e^{-\frac{1}{2}t}sin2t + 2e^{-\frac{1}{2}t}cos2t\\
\end{align*}

9
##### Quiz-5 / Re: Q5 TUT 5101
« on: November 02, 2018, 03:21:48 PM »
(1)
Rewrite the first equation: $x_{2} = \frac{1}{2}x_{1}'$
Then: $x_{2}'' = \frac{1}{2}x_{1}''$
Plug into the second equation: $\frac{1}{2}x_{1}'' = -2x_{1}$
Therefore, we get $x_{1}'' + 4x_{1} = 0$ which is a second order ODE of $𝑥_{1}$
(2)
\begin{align*}
Let A &=
\begin{bmatrix}
0 & 2 \\
-2 & 0
\end{bmatrix}\\
A-\lambda I &=
\begin{bmatrix}
-\lambda & 2 \\
-2 & -\lambda
\end{bmatrix}\\
det(A-\lambda I) &= \lambda^{2} + 4 = 0\\
\lambda &= \pm 2i\\
\end{align*}
For $\lambda = 2i$:
\begin{align*}
A-\lambda I &=
\begin{bmatrix}
-2i & 2 \\
-2 & -2i
\end{bmatrix}\\
null\begin{bmatrix} -2i & 2 \\ -2 & -2i \end{bmatrix}
&= span{\begin{bmatrix} 1\\ i \end{bmatrix}\\}\\
so, \ the \ eigenvector \  V &= \begin{bmatrix} 1\\ i \end{bmatrix}\\
Then, \  e^{\lambda t}V &= e^{2it}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= (cos2t + isin2t)\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= \begin{bmatrix} cos2t + isin2t\\ icos2t - sin2t \end{bmatrix}\\
Thus, \  \phi_{1}(t) &= \begin{bmatrix} cos2t\\- sin2t \end{bmatrix}\\
\phi_{2}(t) &= \begin{bmatrix} sin2t\\cos2t \end{bmatrix}\\
Therefore, \ x_{1} &= c_{1}cos2t + c_{2}sin2t\\
x_{2} &= -c_{1}sin2t + c_{2}cos2t\\
Since, \ x_{1}(0) &= 3 \ and \  x_{2}(0) = 4\\
So, \ c_{1} &= 3 \ and \ c_{2} = 4\\
\end{align*}
Therefore,
\begin{align*}
x_{1} &= 3cos2t + 4sin2t\\
x_{2} &= -3sin2t + 4cos2t\\
\end{align*}

10
##### Quiz-4 / Re: Q4 TUT 5101
« on: October 26, 2018, 06:03:13 PM »
Step1: Find the homogeneous solution $y_{h}$
\begin{align*}
4y'' + y &= 0 \\
4r^{2} + 1 &= 0\\
4r^{2} &= -1\\
r^{2} &= -\frac{1}{4}\\
r &= \pm\frac{1}{2}i\\
\end{align*}
So, $y_{1} = cos(\frac{t}{2})$ and $y_{2} = sin(\frac{t}{2})$

Then, $y_{h} = c_{1}cos(\frac{t}{2}) + c_{2}sin(\frac{t}{2})$

Step2: Find the particular solution $y_{p}$
\begin{align*}
W(y_{1}, y_{2}) = det
\begin{bmatrix}
y_{1} & y_{2} \\
y_{1}' & y_{2}'
\end{bmatrix}
&= y_{1}y_{2}' - y_{2}y_{1}'\\
&= \frac{1}{2}cos(\frac{t}{2})cos(\frac{t}{2}) + \frac{1}{2}sin(\frac{t}{2})sin(\frac{t}{2})\\
&= \frac{1}{2}cos^{2}(\frac{t}{2}) + \frac{1}{2}sin^{2}(\frac{t}{2})\\
&= \frac{1}{2}
\end{align*}
Rewrite the equation: $y'' + \frac{1}{4}y = \frac{1}{2}sec(\frac{t}{2})$

Then g(t) = $\frac{1}{2}sec(\frac{t}{2})$

Since $y_{1} = cos(\frac{t}{2})$, $y_{2} = sin(\frac{t}{2})$, $W(y_{1}, y_{2}) = \frac{1}{2}$, g(t) = $\frac{1}{2}sec(\frac{t}{2})$

So, we have:
\begin{align*}
u_{1} &= -\int \frac{g(t)y_{2}}{W(y_{1}, y_{2})}\\
&= -\int\frac{\frac{1}{2}sec(\frac{t}{2})sin(\frac{t}{2})}{\frac{1}{2}}dt\\
&= -\int tan(\frac{t}{2})dt\\
&= -2ln\lvert sec(\frac{t}{2})\rvert\\
~\\
u_{2} &= \int \frac{g(t)y_{1}}{W(y_{1}, y_{2})}\\
&= \int\frac{\frac{1}{2}sec(\frac{t}{2})cos(\frac{t}{2})}{\frac{1}{2}}dt\\
&= \int1dt\\
&= t
\end{align*}
So, $y_{p} = y_{1}u_{1} + y_{2}u_{2} = -2cos(\frac{t}{2})ln\lvert sec(\frac{t}{2})\rvert + tsin(\frac{t}{2})$

Therefore, $y = y_{h} + y_{p} = c_{1}cos(\frac{t}{2}) + c_{2}sin(\frac{t}{2}) - 2cos(\frac{t}{2})ln\lvert sec(\frac{t}{2})\rvert + tsin(\frac{t}{2})$

11
##### Term Test 1 / Re: TT1 Problem 2 (afternoon)
« on: October 18, 2018, 05:52:51 PM »
I calculated three times again, but I didn't find my mistake.

12
##### Term Test 1 / Re: TT1 Problem 2 (afternoon)
« on: October 18, 2018, 12:08:52 PM »
I have cleaned my post. Is that OK now? Thanks.

13
##### Term Test 1 / Re: TT1 Problem 2 (afternoon)
« on: October 16, 2018, 10:23:49 AM »
(a) Rewrite the equation: $$y'' - \frac{2+\cos^{2}x}{\sin x\cos x}y' + \frac{3}{sin^{2}x}y = 0.$$
Then $$p(x) = - \frac{2+\cos^{2}x}{\sin x\cos x} = -\frac{2}{\sin x\ cos x} - \frac{\cos x}{\sin x}.$$
By Abel's Theorem, we have:
$$W(y_1,y_2)(x) =c\exp \bigl(\int -p(x)dx\bigr) = c\exp\bigl(\int(\frac{2}{\sin x \cos x} + \frac{\cos x}{sin x}) dx\bigr)=\\ C\exp \bigl(\int \frac{2}{\sin x\cos x} dx + \int \frac{\cos x}{\sin x} dx\bigr)$$

Since $\int( \frac{2}{sinxcosx} )dx = 2\int(\frac{1}{sinxcosx})dx = 2\int(\frac{sex}{sinx})dx$

$= 2\int\frac{sec^{2}x}{sinxsecx}dx = 2\int\frac{du}{u} =2lnu = 2ln(tanx)$

(by substitute: $u = tanx, du = sec^{2}xdx)$

and $\int(\frac{cosx}{sinx})dx = \int\frac{du}{u} = lnu = ln(sinx)$

(by substitute: $u = sinx, du = cosxdx$)

So, $$W(y_1,y_2)(x) = ce^{2\ln(\tan x) +\ln(\sin x)}= \tan^{2}x\sin x$$

(b) Since $y_1(x) = sinx$, so $y_1'(x) = cosx, y_1''(x) = -sinx$

Plug in: $-sinxsin^{2}x - tanx(2 + cos^{2}x)cosx + 3sinx$
= $-sin^{3}x -sinx(2 + 1 - sin^{2}x) + 3sinx$
= $-sin^{3}x -3sinx +sin^{3}x + 3sinx$
= 0

So, $y_1(x) = \sinx$ is a solution.

Take c = 1, then $W(y_1,y_2)(x) = tan^{2}xsinx$

By Reduction of Order, we have:

$y_2 = y_1\int(\frac{(e^{\int-p(x)dx})}{y_1^{2}})dx$ = $sinx\int(\frac{tan^{2}xsinx}{sin^{2}x})dx$

= $sinx\int(\frac{sinx}{cos^{2}x})dx = -sinx\int\frac{du}{u^{2}} = -sinx(-\frac{1}{u}) = \frac{sinx}{cosx} = \tan x$

(By substitute: u = cosx, du= -sinxdx)

(c) By (b), we have:

$$y = c_1\sin x + c_2\tan x$$ is the general solution.

then $y' = c_1cosx + c_2sec^{2}x$

Since $y(\frac{\pi}{3})=0, y'(\frac{\pi}{3})=7$

So, $sin(\frac{\pi}{3})c_1 + tan(\frac{\pi}{3})c_2 = 0$ and $cos(\frac{\pi}{3})c_1 + sec^2(\frac{\pi}{3})c_2 = 7$

Then $(\frac{\sqrt{3}}{2})c_1 + \sqrt{3}c_2 = 0$ and $\frac{c_1}{2} + 4c_2 = 7$

Thus, $c_1= -\frac{14}{3}$ and $c_2 = \frac{7}{3}$

Therefore, $y = -\frac{14}{3}sinx + \frac{7}{3}tanx$ is a solution to the IVP.

14
##### Term Test 1 / Re: TT1 Problem 2 (noon)
« on: October 16, 2018, 09:46:21 AM »
(a) Rewrite the equation:

$y'' -(2x/x^{2} +1)y' + 2/(x^{2} + 1)y = 0$

Then p(x) = $-2x/(x^{2} +1)$

By Abel's Thereom, we have:

$W(y_1,y_2)(x) = ce^{\lmoustache-p(x)dx} = ce^{\lmoustache(2x/(x^{2} +1))dx} = c(x^2 + 1)$

(b) Since $y_1(x) = x$, so $y'_1(x) = 1 , y''_1(x) = 0$

Then plug in:

$0 - 2x/(x^{2} +1) + 2x/(x^{2} +1) = 0$

Thus, $y_1(x)$ is a solution.

Take c = 1, then $W(y_1,y_2)(x) = x^{2} + 1$

By Reduction of Oder, we can have:

$y_2 = y_1\lmoustache(e^{\lmoustache-p(x)dx}/y_1^{2})dx = x\lmoustache(1 + 1/x^{2})dx = x(x - 1/x) x^{2} -1$

Thus, $y_2(x) = x^{2} -1$

(c) By (b), we know $y = c_1x + c_2(x^{2} -1)$

Since y(0) = 1, y'(0) = 1

So $-c_2 = 1, c_1 = 1, then c_1 = 1, c_2 = -1$

Therefore, y = x- x^2 +1 is the solution to the IVP.

15
##### Term Test 1 / Re: TT1 Problem 1 (morning)
« on: October 16, 2018, 09:21:08 AM »
Let $M(x,y) = 4x^{2}ylny + 3xy$, $N(x,y) = x^{3}lny + x^{3} +x^{2}$

Then $My = 4x^{2}lny + 4x^{2} + 3x$,  $Nx = 3x^{2}lny +3x^{2} +2x$

Since, My $\neq$ Nx, so the equation is not exact.

Since $R = (My - Nx)/N = [(4x^{2}lny + 4x^{2} + 3x) - (3x^{2}lny +3x^{2} +2x)] / (x^{3}lny + x^{3} +x^{2}) = 1/x$

So the integrating factor is $u(x) = e^{\lmoustache Rdx} = e^{\lmoustache(1/x)dx} = e^{lnx} = x$

Then multiply u(x) = x on both sides, then the equation becomes exact.

Let $M' = 4x^{3}ylny + 3x^{2}y, N' = x^{4}lny + x^{4} +x^{3}$

Since $\lmoustache M'dx = x^{4}ylny + x^{3}y + h(y)$,and $\lmoustache N'dy = x^{4}ylny + x^{3}y +g(x)$

So $x^{4}ylny + x^{3}y = c$ is the general soluttion.

Since y(1) = 1, so ln1 + 1 = c, then c = 1.

Therefore, $x^{4}ylny + x^{3}y = 1$ is a solution to the IVP.

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