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### Messages - Xuewen Yang

Pages: [1] 2
1
##### Quiz 6 / Re: problem 2, night section
« on: November 27, 2013, 10:04:38 PM »
see attachment

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##### Quiz 6 / problem 2, night section
« on: November 27, 2013, 10:04:15 PM »
section 7.9 question 3

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##### Quiz 6 / Re: problem 2, night section
« on: November 27, 2013, 09:58:43 PM »
see attachment

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##### Quiz 6 / problem 1, night section
« on: November 27, 2013, 09:58:10 PM »
7.8 question #2

5
##### MidTerm / Re: Mt, P3
« on: October 10, 2013, 12:54:34 AM »
By the way Mr Ivrii, is it possible for us to get 4.5 out of 5 marks on the midterm questions? When can we know the result?

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##### MidTerm / Re: MT, P1
« on: October 09, 2013, 10:57:34 PM »
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. d(ln(1-y))/dt = -1/1-y

You are right! My bad

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##### MidTerm / Re: MT, P4
« on: October 09, 2013, 10:49:45 PM »
Just wondering, for this question, do we need to draw the direction field?

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##### MidTerm / Re: MT, P1
« on: October 09, 2013, 10:47:45 PM »
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

9
##### MidTerm / Re: MT, P2
« on: October 09, 2013, 10:27:34 PM »
As promised:

$$dy/dt + 2y/t = 1 \\ \mu (dy/dt) + 2y\mu/t = \mu \\ d/dt (\mu y) = d\mu/dt\cdot y + \mu\cdot dy/dt \\ \implies d\mu/dt\cdot y = 2y\mu/t \\ \implies \mu = t^2 \\ d/dt(t^2 y) = t^2 \\ \implies t^2 y = (1/3)t^3 + c \\ \implies y = (1/3)t + c/t^2$$

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##### MidTerm / Re: MT, P5
« on: October 09, 2013, 10:04:35 PM »
see attach

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##### MidTerm / Re: Mt, P3
« on: October 09, 2013, 09:56:07 PM »
see attachment

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##### MidTerm / Re: MT, P2
« on: October 09, 2013, 09:51:41 PM »
I will try to type this afterwards. See attachment

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##### MAT244 Misc / Style and coverage of midterm?
« on: October 04, 2013, 09:44:21 PM »
I'm just wondering about the coverage on the midterm. So only chapter 1 and 2 are covered right? Because I looked at the past term test 1 and found questions on second order linear equations, so I just want to make sure if that's gonna be on the midterm or not.

And also, according to some of the past tests, there are 4 questions in total, out of 20 points?

Thanks!

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##### MAT244 Misc / Will some of past midterms be posted?
« on: October 03, 2013, 12:55:48 PM »
Since our midterm is coming very soon, I am just wondering if some of the past/sample midterms will be posted here? Thanks

15
##### Quiz 1 / Re: Q1, P2 Night sections
« on: October 03, 2013, 12:52:26 PM »
First I apologize that the format is bad, because it's my first time typing math equations using this forum, so I am not sure how it works.

Typing is always better than scanning. Please try to modify post to see how I did it. Observe special meaning of \$...\$ for inline math and \$\$...\$\$ for display math (occupying its own lines) in the source. -- V.I.

Second, wherever it says Q(x,y), it is not Q actually, it's that Greek letter phi which I don't know how to type. OK, I will change $Q$ to $\Phi$ but I am not sure why you don't like $Q$.
So here we go.

Solution: Rewrite function equation as
$$( 9x^2 + y - 1 ) + ( x - 4y )y' = 0$$
So we have  $M_y(x,y) = 1 = N_x(x,y)$, so this function equation  is exact.
\left\{\begin{aligned} &\Phi_x(x,y) = 9x^2 + y - 1,\\ &\Phi_y(x,y) = x - 4y. \end{aligned}\right.
Above system was more complicated to type.
By integrating $\Phi_x(x,y)$, we have  $\Phi (x,y) = 3x^3 + xy - x + h(y)$ with arbitrary $h(y)$.
By differentiating derived $\Phi (x,y)$ with respect to $y$ we have $\Phi_y(x,y) = x + h'(y)$
Note that   $\Phi_y(x,y) = x + h'(y) = x - 4y$.
Therefore we have  $h'(y) = -4y$, and $h(y) = -2y^2$.
So
$$\Phi (x,y) = 3x^3 + xy - x - 2y^2 = c.$$
Now we plug $y(1) = 0$ into this equation, we have  $3(1^3) + (1)(0) - 1 - 2(0^2) = c\implies c = 2$.
So the solution is
$$3x^3 + xy - x - 2y^2 = 2.$$
Good Job -- V.I.

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