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### Messages - Brian Bi

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16
##### Ch 3 / Re: Problem of the week 4b
« on: February 01, 2013, 12:13:33 AM »
Add and subtract the first and second equations to obtain:
\begin{align}
(y+z)'' + K(y+z) &= 0 \label{added} \\
(y-z)'' + (K+2L)(y-z) &= 0 \label{subtracted}
\end{align}
Since $K, L > 0$, both equations are of the form $u'' + \omega^2 u = 0$, with general solution $u = A \cos (\omega t) + B \sin (\omega t)$. So the general solution to $(\ref{added})$ is

y+z = A \cos (\sqrt{K} t) + B \sin (\sqrt{K} t)

and the general solution to $(\ref{subtracted})$ is

y - z = C \cos (\sqrt{K+2L} t) + D \sin(\sqrt{K+2L}t)

Using the identities $y = \frac{1}{2}((y+z)+(y-z))$ and $z = \frac{1}{2}((y+z)-(y-z))$ we obtain the general solution to $(\ref{eq-1})$:
\begin{align}
y &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) + C' \cos(\omega_2 t) + D' \sin(\omega_2 t) \\
z &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) - C' \cos(\omega_2 t) - D' \sin(\omega_2 t)
\end{align}
where $A' = A/2$ and so on, and the frequencies are $\omega_1 = \sqrt{K}$, and $\omega_2 = \sqrt{K+2L}$.

17
##### Ch 3 / Re: Problem of the week 4a
« on: January 31, 2013, 07:40:19 PM »
The amplitude $\sqrt{A^2 + B^2}$ is then:
\begin{align}
\sqrt{A^2+ B^2} &= \left[\left(\frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2}\right)^2 + \left(\frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}\right)^2\right]^{1/2} \\
&= \left[\frac{(\omega^2 - \beta^2)^2 + (\alpha\beta)^2}{((\omega^2-\beta^2)^2+(\alpha\beta)^2)^2}\right]^{1/2} \\
&= ((\omega^2 - \beta^2)^2 + (\alpha \beta)^2)^{-1/2}
\end{align}
This will be maximized if we minimize the expression $(\omega^2 - \beta^2)^2 + (\alpha \beta)^2$, which we expand to give
$\omega^4 + (\alpha^2 - 2\omega^2) \beta^2 + \beta^4$. This is a quadratic in $\beta^2$ with positive leading coefficient, so it has
a global minimum at $\beta^2 = -\frac{1}{2}(\alpha^2 - 2\omega^2)$, or $\beta = \sqrt{\omega^2 - \alpha^2/2}$. This value of $\beta$, the resonance frequency, is real and nonzero only if $\alpha < \sqrt{2} \omega$, so resonance occurs only in underdamped systems (and only those that are sufficiently underdamped).

18
##### 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)

19
##### Ch 3 / Re: Problem of the week 4a
« on: January 31, 2013, 06:19:20 PM »
The characteristic equation for (2) is as follows

r^2 + \alpha r + \omega^2 = 0

which has the solutions

r_{1,2} = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\omega^2}}{2}

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)

20
##### Quiz 2 / Re: Day Section, Question 1
« on: January 31, 2013, 04:26:03 PM »
The characteristic equation

r^2 - (2\alpha-1)r + \alpha(\alpha-1) = 0

factors as $(r - \alpha)(r - (\alpha - 1))$, so the general solution to the ODE is given by

y = A e^{\alpha t} + B e^{(\alpha-1)t}

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) $\alpha > 1$.

21
##### 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: $$f(x) = \begin{cases} 1 & \text{if |x| \leq 1} \\ 0 & \text{if |x| > 1} \end{cases}$$

22
##### 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.)

23
##### 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 $y = \cosh \omega x$.

24
##### Ch 3 / Re: Bonus problem for week 3b
« on: January 24, 2013, 11:19:31 PM »

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

separate variables

\frac{1}{\sqrt{C_1 - \omega^2 y^2}} \, dy = \, dx

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

\frac{1}{\omega} \arcsin\left(\frac{\omega}{\sqrt{C_1}}y\right) + C_2 = x

Isolate $y$ in terms of $x$:

y = \frac{\sqrt{C_1}}{\omega} \sin(\omega x - C_2)

Expand the sine term:

y = \frac{\sqrt{C_1}}{\omega} (\sin(\omega x) \cos C_2 - \sin(C_2) \cos(\omega x))

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 $\{y = A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$

25
##### 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?

26
##### MAT 244 Misc / Re: Comparison of 9th and 10th textbook editions
« on: January 22, 2013, 04:18:33 PM »
Yanyuan and I compared problem numbers in chapters 1 through 4 today. We found that:
• Most questions and question numbers are identical.
• A few question numbers are different. Yanyuan will post these shortly.
• There are some minor stylistic differences, such as writing equations in the form $$M(x,y) + N(x,y)y' = 0$$ instead of $$M(x,y)\, dx + N(x,y)\, dy = 0$$
There is one problem whose text was actually changed between the two editions. In section 2.3, problem 12 in the 10th edition is similar to problem 11 in the 9th edition. The 9th edition reads:
Quote from: 9th edition
11. A recent college graduate borrows $100,000 at an interest rate of 9% to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800(1 + t/120), where t is the number of months since the loan was made. (a) Assuming that this payment schedule can be maintained, when will the loan be fully paid? (b) Assuming the same payment schedule, how large a loan could be paid off in exactly 20 years? The 10th reads: Quote from: 10th edition 12. A recent college graduate borrows$150,000 at an interest rate of 6% to purchase a condominium.
Anticipating steady salary increases, the buyer expects to make payments at a
monthly rate of 800 + 10t, where t is the number of months since the loan was made.
(a) Assuming that this payment schedule can be maintained, when will the loan be fully
paid?
(b) Assuming the same payment schedule, how large a loan could be paid off in exactly
20 years?
A comparison of chapters 7, 9, 5, and 6 will be posted next week.

27
##### 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.

28
##### 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?

29
##### MAT 244 Misc / Re: Bonus problem of the week 1:
« on: January 15, 2013, 01:29:51 AM »
Submitting (first) the correct solution of the "Problem of the Week" (I will try to post them each week) you get karma which translates to bonus marks.
Is there any predictability as to when in the week you intend to post them, or should we just sit here and refresh the forum constantly?

30
##### MAT 244 Misc / Re: Comparison of 9th and 10th textbook editions
« on: January 07, 2013, 10:16:05 AM »
I wouldn't mind helping, though I haven't bought the book yet. Also, I don't think it'll need to be a weekly thing. I think all the assigned questions have already been posted, and there aren't any evaluated problem sets, just suggested practice problems.

I'll private message you to set up meeting times?

Yan
Yes, do that.

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