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Messages - Tzu-Ching Yen

Pages: [1] 2
1
Final Exam / Re: FE-P6
« on: December 16, 2018, 08:15:54 PM »
Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.
No idea tbh. Joyce provided exactly the same answer as your corrected version.

2
Final Exam / Re: FE-P6
« on: December 16, 2018, 07:40:28 PM »
Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C

$+$ missing

3
Final Exam / Re: FE-P6
« on: December 16, 2018, 02:22:20 PM »
The term in $H(x, y)$ should be $\frac{1}{2}x^4$ as well. Perhaps an extra constant. Otherwise I couldn't find anything else.

4
Final Exam / Re: FE-P2
« on: December 16, 2018, 11:07:35 AM »
I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's
You mean $2\cos(t) + 6\sin(t)$ right?

5
Final Exam / Re: FE-P6
« on: December 16, 2018, 10:50:13 AM »
$G_y = -4xy$

6
Final Exam / Re: FE-P2
« on: December 14, 2018, 11:47:35 AM »
I also think one of the particular solution should be $10te^{t}$ $L'(1) = 1$, by $AL'(1) = 10$ $A = 10$. In your solution the $e^{t}$ part had an extra $-2A$.

7
Quiz-7 / Re: Q7 TUT 0701
« on: November 30, 2018, 04:28:06 PM »
$ 0 = y - x^2, 0 = x - y^2$
$ x = x^4, (x,y) = (1, 1), (0, 0)$

$F = x - y^2, F_x = 1, F_y = -2y$
$G = y - x^2, G_x = -2x, G_y = 1$
at $(x, y) = (1, 1)$
$
\left[ {\begin{array}{cc}
    1 & -2  \\
    -2 & 1  \\
\end{array} } \right]
$
gives characteristic equation
$r^2 - 2r - 3 = 0, r = 3, -1$
Solutions are real but opposite sign so near (1, 1) it's a saddle point

at $(x, y) = (0, 0)$
$
\left[ {\begin{array}{cc}
    1 &  0  \\
    0 &  1  \\
\end{array} } \right]
$

$r = 1$
Solutions are real and positive. Near (0, 0) is a unstable node.

8
Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 02:47:25 PM »
I agree with Jingze's solution. It is obvious that Boyu's answer is not (0, 0) at $t =0$

9
Term Test 2 / Re: TT2-P2
« on: November 20, 2018, 07:45:15 AM »
Blair, I don't think there should be any relationship between b) and c). Since Wronskian is only dependent on solutions to homogeneous equation while particular solution is dependent on g(t) ($8e^t$ in this case). Not sure thou.

10
Term Test 2 / Re: TT2-P2
« on: November 20, 2018, 07:02:49 AM »
a)
$\frac{dW}{W} = -1 dt$
$W = e^{\int -1 dt} = ce^{-t}$

b) Characteristic equation reads
$r^3 + r^2 - r - 1 = (r^2 - 1)(r+1) = (r+1)^2(r-1)$
$r = 1, -1$ where $-1$ is a repeated eigenvalue, hence the solutions are
$y_1 = e^{-t}, y_2 = te^{-t}, y_3 = e^t$
After some row operations,
$ W =
e^{-t}det \bigl(\left[ {\begin{array}{ccc}
    1 & t & 1 \\
    0 & 1 & 2 \\
    0 & -2 & 0 \\
\end{array} } \right]\bigr) = 4e^{-t}
$
This agree with part a) where $c = 4$

c) Since $e^{-t}, te^{-t}$ are solutions to homogeneous equation, the form of particular solution is $At^2e^{-t}$, where
$L''(-1) = -4$
$AL''(-1) = 8, A = -2$
Hence the solution is
$y = c_1e^{-t} + c_2te^{-t} + c_3e^t - 2t^2e^{-t}$

11
Quiz-6 / Re: Q6 TUT 0701
« on: November 17, 2018, 04:59:10 PM »
det($M - rI$) gives
$ (3-r)(r^2 - 3r - 4) - 2(-2r -2) + 4(4 + 4r) = (r+1)((3-r)(r-4)-4 + 16) = -(r+1)(r^2-7r -8) = -(r+1)^2(r-8) $
Set this to be zero, $r = -1, 8$

Let $r = 8$, matrix is
$
    M=
   \left[ {\begin{array}{ccc}
    -5 & 2 & 4 \\
    2 & -8 & 2 \\
    4 & 2 & -5 \\
   \end{array} } \right]
$
By inspection solution is $x_1 = [2, 1, 2]$

Let $r = -1$
$
    M=
   \left[ {\begin{array}{ccc}
    4 & 2 & 4 \\
    2 & 1 & 2 \\
    4 & 2 & 4 \\
   \end{array} } \right]
 $

gives equation $r = -2s - 2t$ where solution is $[r, s, t]$. Hence two solutions are $x_2 = [1, -2, 0]$ and $x_3 = [0, -2, 1]$

General solution is therefore
$x = c_1e^{8t}x_1 + e^{-t}(c_2x_2 + c_3x_3)$

12
Quiz-5 / Re: Q5 TUT 0701
« on: November 02, 2018, 03:17:35 PM »
characteristic equation is
$$r^3 - r = 0\\
r(r+1)(r-1) = 0\\
r = 0, 1, \text{  or   } -1$$
The solution to homogeneous equations are $c_1e^t, c_2e^{-t}, c_3$
Let $y_1 = e^t, y_2 = e^{-t}, y_3 = 1$
$$W(y_1, y_2, y_3) = y_1'y_2'' - y_2'y_1'' = 1 + 1 = 2$$
Let $W_i$ be the determinant of wronskian with ith column substituted with $(0, 0, 1)$
$$W_1(t) = -e^{-t}\\
$$W_2(t) = e^{t}\\
$$W_3(t) = -2$$
given that $g(t) = \tanh(t)$
$$Y_1 = y_1 \int \dfrac{W_1(s)  g(s)}{W(s)}ds = \dfrac{-e^t}{2}(\int \tanh(s)e^{-t}ds) = -\dfrac{1}{2} + e^t\arctan(e^{-x})\\
Y_2 = y_2 \int \dfrac{W_2(s)  g(s)}{W(s)}ds = \dfrac{e^{-t}}{2}(\int \tanh(s)e^{t}ds) = \dfrac{1}{2} - e^{-t}\arctan(e^{x})\\
Y_3 = \int -tanh(s) ds =  -\ln(\cosh(x))$$
The general solution is therefore
$$y = c_1e^{-t} + c_2e^t + c_3 + e^t\arctan(e^{-x}) - e^{-t}\arctan(e^{x}) - \ln(\cosh(x))$$

13
Term Test 1 / Re: TT1 Problem 2 (main)
« on: October 16, 2018, 07:29:31 AM »
(a) reformat equation into
$$y'' + p(x)y' + q(x)$$
where
$$p(x) = -\frac{x}{x-1}$$
by equation of wronskian
\begin{gather*}
W(x) = \exp(-\int p(x) dx)\\
W(x) = \exp(\int \frac{x}{x-1} dx) = \exp(\int \frac{u+1}{u}du) = c_0ue^u = c_0(xe^{x-1} - e^{x-1})
\end{gather*}
choose $c_0 = e$
$$W(x) = xe^x - e^x$$
(b) Plug in $y = x, y' = 1, y'' = 0$, equation becomes
$$-x + x = 0$$
therefore $y=x$ is a solution, let second solution be $y_2$ and let $y_1 = x$ since
$$W(x) = y_1y_2' - y_1'y_2$$
Plug in $y_1 = x, y_1' = 1$
$$xe^x - e^x = xy_2' - y_2$$
By inspection
$$y_2 = e^x$$
Hence, general solution is
$$y(x) = c_1x + c_2e^x$$
c) Set $y(0) = 1$
$$c_2 = 1$$

Set $y'(0) = 0, y'(0) = c_1 + c_2e^x$
$$c_1 + c_2 = 0, c_1 = -c_2 = -1$$
Hence the solution is
$$y(x) = -x + e^x$$

14
Quiz-3 / Re: Q3 TUT 0701
« on: October 12, 2018, 06:09:14 PM »
Reformat equation into form $y'' + p(t)y' + q(t) = 0$
We can see that
$p(t) = \frac{\sin(t)}{\cos(t)}$
By the equation for wronskian
$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$
where $c_0$ depends on choice of $y_2$ and $y_2$

15
Thanksgiving Bonus / Re: Thanksgiving bonus 3
« on: October 05, 2018, 06:35:26 PM »
$y_1 = x^3, y_1' = 3x^2, y_1'' = 6x$
$y_2 = y_2' = y_2'' = e^x$
Expand W(y, y_1, y_2)
$$y(y_1'y_2'' - y_2'y_1'') - y'(y_1y_2'' - y_2y_1'') + y''(y_1y_2' - y_2y_1') = 0.$$
Plug in $y_1$, $y_2$ and their derivatives, equation becomes
$$(x^3 - 3x^2)e^xy'' + (6x - x^3)e^xy' + (3x^2 - 6x)e^xy' = 0$$
or
$$ (x^2 - 3x)y'' + (6 - x^2)y' + (3x - 6)y' = 0$$

Btw, I think typo is still there. Two $y_2''$.  FIXED

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