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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
$\mathbf x = c_1 \left(\begin{array}{c} -3 \\ 4 \\ 2 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) e^{2t} + c_3 \left( \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) t e^{2t} + \left(\begin{array}{c} -1 \\ -1 \\ 0 \end{array}\right) e^{2t}\right)$

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##### 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

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

$$r^2 = K - L \pm L$$

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)

$$y'' - w^2 y = 0 \\ z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\ \frac{dz}{dy} z = w^2 y \\ \frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\$$

\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1} \\

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 »
$$y'' + w^2 y = 0 \\ z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\ \frac{dz}{dy} z = -w^2 y \\ \frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\$$

\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\

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.

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##### Ch 1--2 / Re: Bonus problem for week 2
« on: January 18, 2013, 02:28:56 AM »