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Messages - zeyang zhang

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Test 2 / Topics covered in TT2; schedule of the 4 sittings; and Final arrangement
« on: March 18, 2020, 10:20:15 PM »
1.The old coverage was:
It will be 20 pts worth, containing 4 problems, each 5 pts worth, based on Chapters 4--6 of the online Textbook. and a 2.5pt bonus
Will we follow this?

2.What is the schedule of the 4 sittings in TT2?

3.BTW, what will be the format of Final exam? Will it be the same with Quiz and TT2? We should have the info by today......

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Quiz-1 / Re: Q1: TUT 0601
« on: September 29, 2018, 12:08:03 AM »
I think Start from here : $-\frac{2}{1+v} - \ln (1+v) + c = \ln x$
All thing inside ln() should be given absolute value.
Answer was like
$\frac{2x}{x+y} + \ln (|x+y|) = c$

Quote from: Yiting Zhang on September 28, 2018, 06:30:48 PM
$\frac{dy}{dx} = \frac{x + 3y}{x - y}$

$\frac{dy}{dx} = \frac{x+3y}{x-y} = \frac{1+3(\frac{y}{x})}{1 - (\frac{y}{x})}$

Let $v = \frac{y}{x}, y=xv$

$\frac{dy}{dx} = v+x \frac{dv}{dx}$

$x \frac{dv}{dx} = \frac{dy}{dx} -v = \frac{1+3v}{1-v} - v = \frac{(1+v)^2}{1-v}$

$\frac{1-v}{(1+v)^2}dv = \frac{1}{x}dx$

$\int \frac{1-v}{(1+v)^2}dv = \int \frac{1}{x}dx$

$-\frac{2}{1+v} - \ln (1+v) + c = \ln x$

$-\frac{2}{1+(\frac{y}{x})} - \ln (1+(\frac{y}{x})) + c = \ln x$

$c-\frac{2x}{x+y} = \ln (x (1+\frac{y}{x}))$

$c-\frac{2x}{x+y} = \ln (x+y)$

$\frac{2x}{x+y} + \ln (x+y) = c$

$x + y = Ce^{-\frac{2x}{x+y}}$

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