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Messages - Changyu Li

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1
Quiz 4 / Q4--day section--problem 1
« on: March 22, 2013, 10:22:22 PM »
I think the first problem was from the textbook
$
\mathbf x' = \left(\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & -1 \\
0 & -1 & 1 \\
\end{array}\right) \mathbf x \\
$
The characteristic polynomial is
$
-4 + 3k^2 - k^3 = 0 \\
k = 2, 2, -1
$
The eigenvectors are
$
\lambda = -1,
\left(\begin{array}{c}
-3 \\
4 \\
2
\end{array}\right),
\lambda = 2,
\left(\begin{array}{c}
-0 \\
-1 \\
1
\end{array}\right)

$
Jordan decomposition yields the similarity transform of
$
\mathbf T = \left(\begin{array}{ccc}
-3 & 0 & -1 \\
4 & -1 & -1 \\
2 & 1 & 0
\end{array}\right)
$
Thus the solution is
$
\mathbf x = c_1 \left(\begin{array}{c} -3 \\ 4 \\ 2 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) e^{2t} + c_3 \left( \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) t e^{2t} + \left(\begin{array}{c} -1 \\ -1 \\ 0 \end{array}\right) e^{2t}\right)
$

2
Quiz 3 / Re: Day Section Problem 1
« on: February 28, 2013, 11:29:34 AM »
$$
(r-1)(r^2+1) = 0 \\
r = 1, \pm i \\
y_h = c_1 e^t + c_2 e^{it} + c_3 e^{-it} \\
y_p = A e^{-t} \sin t + B e^{-t} \cos t \\
y_p' = e^{-t} \left(\left(A-B\right) \cos t - \left(A+B\right) \sin t \right) \\
y_p'' = -2 e^{-t} \left(A \cos t - B \sin t\right) \\
y_p''' = 2 e^{-t}\left(\left(A-B\right) \sin t + \left(A+B\right) \cos t \right) \\
A = 0, B = -\frac{1}{5} \\
y = c_1 e^t + c_2 e^{it} + c_3 e^{-it} -\frac{1}{5}e^{-t} \cos t
$$

3
Quiz 3 / Re: Day Section Problem 2
« on: February 27, 2013, 09:17:36 PM »
I think the question was
$$
y'''-y= 2 \sin t
$$
solution:
$$
r^3 - 1 = 0 \\
(r+1)(r^2+r+1) = 0\\
r = -1, \frac{-1 \pm \sqrt{3} i}{2}\\
y_h = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} \\
y_p = A \sin t + B \cos t \\
y_p' = A \cos t - B \sin t \\
y_p'' = -A \sin t - B \cos t \\
y_p''' = - A \cos t + B \sin t \\
A = -1, B = 1\\
y  = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} + \cos t - \sin t
$$

4
Term Test 1 / Re: TT1--Problem 4
« on: February 14, 2013, 12:04:40 AM »
$$
r^4 + 8 r^2 + 16 = 0 \\
r = \pm 2 i, \pm 2 i\\
y = c_1 e^{2i x} + c_2 e^{-2i x} + c_3 x e^{2i x} + c_4 x e^{-2i x}\\
y' = 2 i c_1 e^{2 i x}-2 i c_2 e^{-2 i x}+c_3 e^{2 i x}+2 i c_3 e^{2 i x} x+c_4 e^{-2 i x}-2 i c_4 e^{-2 i x} x \\
y'' = -4 c_1 e^{2 i x}-4 c_2 e^{-2 i x}+c_3 \left(4 i e^{2 i x}-4 e^{2 i x} x\right)+c_4 \left(-4 i e^{-2 i x}-4 e^{-2 i x} x\right)\\
y''' = -8 i c_1 e^{2 i x}+8 i c_2 e^{-2 i x}+c_3 \left(-8 i e^{2 i x} x-12 e^{2 i x}\right)+c_4 \left(8 i e^{-2 i x} x-12 e^{-2 i x}\right)\\
c_1+c_2=1 \\
2 i c_1-2 i c_2+c_3+c_4=0 \\
-4 c_1-4 c_2+4 i c_3-4 i c_4=0\\
-8 i c_1+8 i c_2-12 c_3-12 c_4=0\\
c_1 = \frac{1}{2}, c_2 = \frac{1}{2}, c_3 = \frac{-i}{2}, c_4 = \frac{i}{2} \\
y = \frac{1}{2} e^{2 i x}+\frac{1}{2} e^{-2 i x} + \frac{1}{2} i x e^{2 i x} - \frac{1}{2} i x e^{-2 i x}
$$

5
Ch 3 / Re: Problem of the week 4b
« on: January 31, 2013, 10:45:31 PM »
2)
guess $ y = A e^{rt}$, $z = B e^{rt}$
$$
A r^2 + K A + L\left(A-B\right) = 0 \\
B r^2 + K B + L\left(B-A\right) = 0 \\

\left(
\begin{array}{cc}
r^2 + K + L & - L \\
-L & r^2 + K + L \\
\end{array} \right)
\left( \begin{array}{c}
A \\
B
\end{array}
\right) = 0

$$

nontrivial solution exists if there is no inverse to the ugly matrix, therefore its determinant is 0

$$
\left( r^2 + K + L \right)^2 + L^2 = 0
$$

expand and use quadratic equation
$$
r^2 = K - L \pm L
$$

therefore the frequencies are $\pm K$, $\pm\left(K-2L\right)$

6
Ch 3 / Re: Bonus problem for week 3a
« on: January 25, 2013, 01:43:42 AM »
To solve for $C_1$ and $C_2$, I tried substituting $dy/dx = 0,\;y = 1$ into $\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1}$ and $x = 0,\;y = 1$ into $\ln \left( \sqrt{ y^2 -1} + y \right) = wx + C_2$. I'm not sure why it gives a different result.

7
Ch 3 / Re: Bonus problem for week 3b
« on: January 24, 2013, 11:38:01 PM »
$y = \{A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$
let me plot that for you



8
Ch 3 / Re: Bonus problem for week 3a
« on: January 24, 2013, 11:28:07 PM »
fixed.

9
Ch 3 / Re: Bonus problem for week 3a
« on: January 24, 2013, 01:33:03 PM »
a)

$$
y'' - w^2 y = 0 \\
z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\
\frac{dz}{dy} z = w^2 y \\
\frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\
$$
\begin{equation}
\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1} \\
\end{equation}
factor out $w^2$ from $C_1$
$$
\frac{dy}{dx} = w \sqrt{y^2 + C_1} \\
\frac{dy}{ \sqrt{y^2 + C_1}} = w dx \\
$$
use trig substitution
$$
y = \sqrt{C_1} \tan u, \; dy=\sqrt{C_1} \sec^2 u\;du \\
\int \frac{\sqrt{C_1} \sec^2 u \; du}{\sqrt{C_1 \left( \tan^2 u + 1 \right)}} = \int wdx \\
\int \frac{\sqrt{C_1} \sec^2 u \; du}{\sqrt{C_1 \sec^2 u }} = \int wdx \\
\int \sec u \; du = \int wdx \\
\ln\left( \tan u + \sec\; u \right) = wx + C_2

$$
note $\sec \theta = \sqrt{1 + \tan^2 \theta}$ and $u = \arctan \frac{y}{\sqrt{C_1}}$
$$
\ln\left( \sqrt{y^2 + C_1 } + y \right) = wx + C_2 \\
\sqrt{y^2 + C_1 }  = e^{wx + C_2} - y \\
y^2 + C_1  = \left( e^{wx + C_2} - y\right)^2 \\
$$
expand and simplify
$$
y = \left( e^{2\left(wx+C_2\right)} - C_1 \right) e^{-\left(wx+C_2\right)} \frac{1}{2}
$$

b)
insert IC to (2) to get $C_1 = -w^2$
$$
\frac{dy}{dx} = w \sqrt{y^2-1} \\
\frac{1}{\sqrt{y^2-1}}  dy = w dx \\
\ln \left( \sqrt{ y^2 -1} + y \right) = wx + C_2 \\$$

insert IC to get $C_2 = \ln\left( i \right)$.
simplify using euler's identity
$$
e^{i\frac{\pi}{2}} = i \\
i\frac{\pi}{2} = \ln i \\
$$

$$
y = \left( e^{2\left(wx+i\frac{\pi}{2} \right)} + 1\right) e^{-\left(wx+i\frac{\pi}{2} \right)} \frac{1}{2}
$$

10
Ch 3 / Re: Bonus problem for week 3b
« on: January 24, 2013, 01:07:24 PM »
$$
y'' + w^2 y = 0 \\
z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\
\frac{dz}{dy} z = -w^2 y \\
\frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\
$$
\begin{equation}
\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\
\end{equation}
insert IC to get $C_1 = w^2$
$$
\frac{dy}{dx} = w \sqrt{1 - y^2} \\
\frac{1}{\sqrt{1-y^2}}  dy = w dx \\
wx = \arcsin y + C_2 \\
$$
insert IC to get $C_2 = -\pi / 2$
$$
y = \sin\left( wx + \pi/2 \right) \\
y = \cos\left( wx \right)
$$

general solution to $\frac{dy}{dx}= w \sqrt{C_1-y^2}$ using trig substitution
$$
\frac{dy}{\sqrt{C_1-y^2}}  = w dx \\
wx = \frac{\arcsin\left( y / \sqrt{C_1} \right)}{\sqrt{C_1}} + C_2 \\
y = \sqrt{C_1} \sin \left( \sqrt{C_1}\left( wx + C_2 \right) \right)
 
$$

11
Quiz 1 / Re: Day section 2.6 #25
« on: January 21, 2013, 11:03:52 PM »
Fixed, I think.

12
Quiz 1 / Re: Day section 2.6 #25
« on: January 21, 2013, 02:49:19 PM »
$$
M = 3x^2 y + 2xy + y^3 \\
N = x^2 + y^2 \\
\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{N} \mu = \frac{d\mu}{dt} \\
\frac{3x^2+3y^2}{x^2+y^2} \mu = \frac{d\mu}{dx} \\
$$
new equation is separable
$$
\mu = e^{3x} \\
\frac{\partial \psi}{\partial x} = e^{3x}\left( 3 x^2 y + 2 x y + y^3 \right) \\
\psi = e^{3x} y \left( x^2 + y^2/3 \right) + g(y) \\
\\
\frac{\partial \psi}{\partial y} = e^{3x} \left( x^2 + y^2 \right) \\
\psi = e^{3x} y \left( x^2 + y^2/3 \right) + \tilde{g}(x) \\
\psi = e^{3x} y \left( x^2 + y^2/3 \right) \\
e^{3x} y \left( x^2 + y^2/3 \right) = C
$$

13
Quiz 1 / Re: Day section 2.1 #18
« on: January 21, 2013, 12:27:01 PM »
$$
y'+\frac{2}{t} y = \frac{\sin t}{t} \\
\mu y' + \frac{2 \mu}{t} y = \mu \frac{\sin t}{t} \\
\mu = t^2 \\
\frac{d}{dt} t^2 y = t \sin t \\
t^2 y = \int t \sin t dt \\
y = \frac{\sin t - t \cos t + C}{t^2}
$$
use initial value
$$
1 = \frac{4\left(1 - 0 + C\right)}{\pi^2}\\
C= \frac{\pi^2-4}{4}\\
y = \frac{\sin t - t \cos t + \frac{\pi^2-4}{4}}{t^2}\\
y = \frac{4\sin t - 4t \cos t + \pi^2-4}{4t^2}
$$

14
Ch 1--2 / Re: Bonus problem for week 2
« on: January 21, 2013, 01:50:39 AM »
Problem 30 from chapter 2.2 in the tenth edition is a similar problem with a partial solution.

15
Ch 1--2 / Re: Bonus problem for week 2
« on: January 18, 2013, 02:28:56 AM »
Changes were made per your suggestions.

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