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### Messages - Brian Bi

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16
##### Ch 3 / Re: Problem of the week 4a
« on: January 31, 2013, 07:25:00 PM »
Note that the greater of the two roots in case 1 is always greater than the root $-\alpha/2 = -\omega$ in case 2. We call case 1 overdamped because $\alpha$ is large and the greater root (i.e., less negative) leads to a slower exponential decay. In contrast, case 2 is called critically damped because solutions decay with time constant $\alpha/2 = \omega$, which is maximal. Case 3 is called underdamped because $\alpha$ is small and the damping force is insufficient to prevent oscillation. In this case the decay will also be slow since the time constant is $\alpha/2 < \omega$.

The solution to equation (3) is given by the sum of the general solution to the homogeneous equation (1) and a particular solution to (3) which we will solve by the method of undetermined coefficients. As hinted, let $y_p = A \cos (\beta t) + B \sin(\beta t)$. Then

\begin{align}
y'' + \alpha y' + \omega^2 y
&= (A \cos(\beta t) + B \sin(\beta t))'' + \alpha (A \cos(\beta t) + B \sin(\beta t))' + \omega^2 (A \cos (\beta t) + B \sin(\beta t)) \\
&= -A\beta^2 \cos(\beta t) - B\beta^2 \sin(\beta t) + -A\alpha\beta\sin(\beta t) + B\alpha\beta\cos(\beta t) + A\omega^2 \cos(\beta t) + B\omega^2 \sin(\beta t) \\
&= (-A\beta^2 + B\alpha\beta + A\omega^2)\cos(\beta t) + (-B\beta^2 -A\alpha\beta + B\omega^2)\sin(\beta t)
\end{align}

By matching coefficients we find
\begin{align}
(-\beta^2 + \omega^2) A + \alpha\beta B &= 1 \\
(-\alpha\beta) A + (-\beta^2 + \omega^2) B &= 0
\end{align}
This is a linear system in two equations and two unknowns, which is therefore easy to solve. The solution is
\begin{align}
A &= \frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2} \\
B &= \frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}
\end{align}
so the particular solution to (3) is $y_p = A \cos(\beta t) + B\sin(\beta t)$ with $A, B$ given above and the general solution to (3) is given by adding $y_p$ and the general solution to (1) as found above.

(continued)

17
##### Ch 3 / Re: Problem of the week 4a
« on: January 31, 2013, 06:19:20 PM »
The characteristic equation for (2) is as follows
\begin{equation}
r^2 + \alpha r + \omega^2 = 0
\end{equation}
which has the solutions
\begin{equation}
r_{1,2} = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\omega^2}}{2}
\end{equation}
We consider three cases:
• $\alpha^2 > 4\omega^2$, i.e., $\alpha > 2\omega$ (assuming $\omega$ is always positive). In this case $r_1, r_2 \in \mathbb{R}$ and the general solution to (2) is given by $y = A e^{r_1 t} + B e^{r_2 t}$. Indeed, both roots are negative so all solutions decay monotonically to zero as $t \to \infty$ (unless $\omega = 0$ in which case we have constant solutions).
• $\alpha^2 = 4\omega^2$, i.e., $\alpha = 2\omega$. Here $r_1 = r_2 = -\alpha/2$ so the general solution is given by $y = (A + Bt)e^{-\alpha t/2} = (A+Bt)e^{-\omega t}$. Again, all solutions decay to zero as $t \to \infty$.
• $\alpha^2 < 4\omega^2$, i.e., $\alpha < 2\omega$. In this case we have two complex roots, $r_1 = -\alpha/2 + i \sqrt{4\omega^2 - \alpha^2}/2$ and $r_2 = -\alpha/2 - i \sqrt{4 \omega^2 - \alpha^2}/2$. The solutions therefore have both a time-decaying exponential factor $e^{-\alpha t/2}$ and a sinusoidal factor with angular frequency $\sqrt{4 \omega^2 - \alpha^2}/2$. The general solution is $y = e^{-\alpha t/2} (A \cos (\sqrt{4 \omega^2 - \alpha^2}t/2) + B \sin (\sqrt{4\omega^2 - \alpha^2}t/2))$.

(continued)

18
##### Quiz 2 / Re: Day Section, Question 1
« on: January 31, 2013, 04:26:03 PM »
The characteristic equation
\begin{equation}
r^2 - (2\alpha-1)r + \alpha(\alpha-1) = 0
\end{equation}
factors as $(r - \alpha)(r - (\alpha - 1))$, so the general solution to the ODE is given by
\begin{equation}
y = A e^{\alpha t} + B e^{(\alpha-1)t}
\end{equation}
where $A, B \in \mathbb{R}$.

We consider the following cases:
• $\alpha < 0$: Both exponentials will be decaying, so each solution tends to zero as $t \to \infty$.
• $\alpha = 0$ or $\alpha = 1$: Each $y = c$ for constant $c$ is a solution, so there exist solutions that neither tend to zero nor become unbounded as $t \to \infty$.
• $0 < \alpha < 1$: One exponential is growing and the other decaying, so there exist nonzero solutions that tend to zero as well as solutions that tend to infinity.
• $\alpha > 1$: Both exponentials will be growing. The larger of the two, $Ae^{\alpha t}$, dominates as $t \to \infty$, so $y$ is unbounded unless $A = 0$. If $A$ vanishes identically, then all nonzero solutions $Be^{(\alpha-1)t}$ again become unbounded.
We conclude that the answer is: (i) $\alpha < 0$, and (ii) $\alpha > 1$.

19
##### Technical Questions / Re: Typesetting piecewise functions
« on: January 25, 2013, 12:59:16 AM »
Sorry for necroposting but I find it curious that nobody here mentioned the cases environment, which is designed specifically to typeset piecewise functions and similar things.

Code: [Select]
f(x) = \begin{cases} 1 & \text{if $|x| \leq 1$} \\ 0 & \text{if $|x| > 1$} \end{cases}
Output: $$f(x) = \begin{cases} 1 & \text{if |x| \leq 1} \\ 0 & \text{if |x| > 1} \end{cases}$$

20
##### Ch 3 / Re: Bonus problem for week 3b
« on: January 25, 2013, 12:54:10 AM »
I'm not sure whether that image got corrupted on upload or whether you're just pointing out that I erred in putting $y$ outside the braces instead of inside where it should be. (I fixed that now, btw.)

21
##### Ch 3 / Re: Bonus problem for week 3a
« on: January 25, 2013, 12:50:45 AM »
$$y = \left( e^{2\left(wx+i\frac{\pi}{2} \right)} + 1\right) e^{-\left(wx+i\frac{\pi}{2} \right)} \frac{1}{2}$$
Pretty sure this isn't right; the correct solution is $y = \frac{1}{2}\left( e^{\omega x} + e^{-\omega x}\right)$. How exactly did you calculate $C_1$ and $C_2$?

BTW, this problem can be solved analogously to the 3b problem by using a hyperbolic trig substitution instead of a circular trig substitution. The result is, unsurprisingly, $y = A \cosh \omega x + B \sinh \omega x$. This form is more convenient for the initial conditions given because here $y(0) = A$ and $y'(0) = \omega B$, whence $A = 1, B = 0$ so indeed $y = \cosh \omega x$.

22
##### Ch 3 / Re: Bonus problem for week 3b
« on: January 24, 2013, 11:19:31 PM »
\begin{equation}
\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\
\end{equation}
separate variables
\begin{equation}
\frac{1}{\sqrt{C_1 - \omega^2 y^2}} \, dy = \, dx
\end{equation}
integrate LS by substituting $y = \frac{\sqrt{C_1}}{\omega}u$
\begin{align}
\int \frac{1}{\sqrt{C_1 - \omega^2 y^2}} \, dy
&= \int \frac{1}{\sqrt{C_1 - \omega^2 ((\sqrt{C_1}/\omega)u)^2}} \frac{\sqrt{C_1}}{\omega} \, du \\
&= \int \frac{1}{\sqrt{C_1(1 - u^2)}} \frac{\sqrt{C_1}}{\omega} \, du \\
&= \frac{1}{\omega} \arcsin u + C_2 \\
&= \frac{1}{\omega} \arcsin\left(\frac{\omega}{\sqrt{C_1}}y\right) + C_2
\end{align}
Integrate also the RS to obtain
\begin{equation}
\frac{1}{\omega} \arcsin\left(\frac{\omega}{\sqrt{C_1}}y\right) + C_2 = x
\end{equation}
Isolate $y$ in terms of $x$:
\begin{equation}
y = \frac{\sqrt{C_1}}{\omega} \sin(\omega x - C_2)
\end{equation}
Expand the sine term:
\begin{equation}
y = \frac{\sqrt{C_1}}{\omega} (\sin(\omega x) \cos C_2 - \sin(C_2) \cos(\omega x))
\end{equation}
By choosing $C_1 = \omega^2$ and either $C_2 = 0$ or $C_2 = 3\pi/2$ respectively we obtain particular solutions $y_1 = \sin(\omega x)$ and $y_2 = \cos(\omega x)$. The Wronskian is
\begin{align}
W &= \left|\begin{array}{cc} \sin(\omega x) & \cos(\omega x) \\ \omega \cos(\omega x) & -\omega \sin( \omega x) \end{array}\right| \\
&= -\omega \sin^2(\omega x) - \omega \cos^2(\omega x) \\
&= -\omega \neq 0
\end{align}
so we conclude that the general solution to the linear homogeneous second-order ODE is $\{y = A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$

23
##### Ch 3 / Re: Bonus problem for week 3a
« on: January 24, 2013, 11:46:09 AM »
What's the difference between the 3a and 3b questions?

24
##### Ch 1--2 / Re: Bonus problem for week 2
« on: January 21, 2013, 01:13:56 AM »
I see. Is this covered in class or in the textbook? I can't find it in the textbook.

25
##### Ch 1--2 / Re: Bonus problem for week 2
« on: January 20, 2013, 08:59:46 PM »
let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$
What's the motivation for this? Inspired guess?

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