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### Messages - Devin Jeanpierre

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##### Term Test 1 / Re: TT1--Problem 3
« on: February 14, 2013, 07:34:58 AM »
for me, wolfram alpha spits out $y(t) = c_1 e^{-2 t} \sin(t)+c_2 e^{-2 t} \cos(t)+\frac{1}{2} e^{-2 t} (t (\sin(t)+2)+\cos(t))$. It's technically correct, because you can just change the constants to turn it into my solution, but...

PS Usage of double dollars in LaTeX is deprecated.
For  most everything what I really wanted was to use "align*". Turns out mathjax supports this, so I'll be more latexy in the future.

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##### MAT 244 Misc / Re: Advice on how to get faster?
« on: February 14, 2013, 12:08:00 AM »
Haha, are you another theory dude?

This is my first "computation" course. It's taking getting used to!

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##### Term Test 1 / Re: TT1--Problem 4
« on: February 14, 2013, 12:06:58 AM »
for double roots y2 and y4: y2=xy1and y4=xy3

y=c1 cos(2x) + c2 sin(2x) +c3 xcos(2x) + c4 xsin(2x)

solve I.C: y(o)=1 => c1=1
y'(0)=y''(0)=y(0)'''=0   =>  c2=c3=0 , c4=1

y=cos(2x) + xsin(2x)
Aw man, using the real solutions / trigonometric decomposition would've really sped things up. I wish I'd thought of that. Clever thinking, dude!

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##### MAT 244 Misc / Re: Advice on how to get faster?
« on: February 14, 2013, 12:01:40 AM »
There's not finishing and then there's not finishing.

I completed #1, #3, and #2a, but #1 had a mistake I couldn't find. I did not get anywhere (other than to state the roots) with #4, and I was in the middle of a long (and probably erroneous) computation for #2b.

Overall, not feeling so good about my grade. The way I figure it, in the best case I get maybe a 60%, and that's if people are nice marking -- but it's more likely I'll get a 40%, considering how harshly the quizzes were graded with regards to computation errors.

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##### Term Test 1 / Re: TT1--Problem 3
« on: February 13, 2013, 11:30:33 PM »
Funny story: after I posted this I also checked my work against wolfram alpha, which said I got it wrong for the homogeneous part. I spent some time trying to spot any error, but there isn't any. I am smarter than the machine!

(You can press "show step by step solution" and wolfram alpha will come up with what I came up with, so that's why I'm so confident. I filed a bug report, maybe someday this will be fixed...)

$\renewcommand{Re}{\operatorname{Re}}$

Here we use the method of undetermined coefficients. First we find the characteristic equation and its derivative (which will definitely be useful):

$$Q(r) = r^2 + 4r + 5$$
$$Q^\prime(r) = 2r + 4$$

Now we solve for $te^{-2t}$, for which our exponent is $r_1 = -2$.

$$Q(r_1) = 4 - 8 + 5 = 1$$
$$Q^\prime(r_1) = -4 + 4 = 0$$

So we want to use $t e^{-2t}$. If we plug that in we get $L[t e^{-2t}] = e^{-2t}(t Q(-2) + Q^\prime(-2)) = t e^{-2t}$. This is what we want, so we're done the first part.

Now we solve for $e^{-2t} cos(t)$, for which the exponent is $r_2 = -2 + i$.

$$Q(r_2) = (-2+i)^2 + 4(-2 + i) +5 = 4 - 4i - 1 - 8 + 4i + 5 = 0$$
$$Q^\prime(r_2) = 2(-2+i) + 4 = -4 + 2i + 4 = 2i$$

So we want to use $t e^{(-2 + i)t}$. If we plug that in, we get $L[t e^{(-2 + i)t}] = e^{(-2 + i)t}(t Q(-2 + i) + Q^\prime(-2 + i)) = (2i)e^{(-2 + i)t}$.

We don't want that 2i, so let's divide it out in the input. It turns out that $1/2i = i/-2 = -i/2$. So $L[\frac{-i}{2} t e^{(-2 + i)t}] = e^{(-2 + i)t}$, which is nearly right.

Note:
$$\Re e^{(-2 + i)t}) = \Re (e^{-2t}e^{it}) = \Re (e^{-2t}(\cos t + i \sin t)) = e^{-2t} \cos t$$.

Since $L$ is linear, if we take only the real part of the input, we'll get only the real part of the output. So we need to compute the real part of $\frac{-i}{2} t e^{(-2 + i)t}$.

To do that let's multiply it out:

$$\frac{-i}{2} t e^{(-2 + i)t}$$
$$\frac{-i}{2} t e^{-2t}(\cos t + i \sin t)$$
$$t e^{-2t}(\frac{-i}{2}\cos t + \frac{-i}{2}i \sin t)$$
$$t e^{-2t}(\frac{-i}{2}\cos t + \frac{1}{2}\sin t)$$

The real part of that is $\frac{1}{2}t e^{-2t}\sin t$, so that $L[\frac{1}{2}t e^{-2t}\sin t] = e^{-2t} \cos t$.

So then we put our two subproblem solutions together and we get a particular solution, thanks to the linearity of $L$. Our particular solution is:

$$Y(t) = t e^{-2t} + \frac{1}{2}t e^{-2t}\sin t$$

What remains is to compute a general solution for the homogenous part, and we can combine that to find the general solution for the non-homogenous ODE.

During the above work we found that $-2 + i$ was a root of the characteristic equation. So the conjugate is also a root, $-2 - i$, and we can take the real and imaginary parts of $e^{-2 \pm i}$ to find the solution, which is:

$$y_{gen[homogeneous]}(t) = c_1 e^{-2t}\cos(t) + c_2 e^{-2t}\sin (t)$$

So the general solution for the nonhomogeneous ODE is:

$$y(t) = c_1 e^{-2t}\cos(t) + c_2 e^{-2t}\sin (t) + t e^{-2t} + \frac{1}{2}t e^{-2t}\sin (t)$$

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##### MAT 244 Misc / Advice on how to get faster?
« on: February 13, 2013, 09:58:24 PM »
Evening section/Lash-Miller person here. I did not come close to finishing the term test. I knew how to solve every problem, given enough time, but that's not really good enough.

It's not a very specific question, but does anyone have any advice for how to get faster at this for the midterm? Like, uh, maybe it would help if you said how much time you spend per week doing questions?

I'll find out if there were any technique differences when others post their answers to the problems. Like, I know I worked faster by using the in-class version of undetermined coefficients, but maybe I missed out on something when I missed the lecture that covered the wronskian... I don't know.

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##### MAT 244 Misc / Re: What has been covered so far?
« on: January 27, 2013, 02:41:44 AM »
Ack, I'm sorry, I made you go and look up which section I'm in. Thank you for that effort, and for your helpful reply. I will be continuing to attend the night section, since I need to be up at 6PM for two other classes anyway.

I take it the material about reducing the order of second-order ODEs is in the textbook somewhere (e.g. 3.4?)  If not I'll try and poke some of the other students after this week's lecture.

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##### MAT 244 Misc / What has been covered so far?
« on: January 27, 2013, 12:48:37 AM »
Due to a terrible illness -- stupidity -- I inverted my sleep schedule and missed two weeks of classes. I am catching up with MAT244 now, but am not sure where to stop in the textbook and homework assignments (somewhere in chapter 3, I gather from reading the boards).

I can't tell which homework assignments come from which week. They are listed according to chapter and section, rather than by week. Can anyone help me here?

For reference, assignment page: http://www.math.toronto.edu/courses/mat244h1/20131/homeassignments.html

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