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Topics - Junjie Zhang

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Quiz-1 / TUT 0601
« on: January 26, 2018, 01:18:34 PM »
Find the general solution of the given function.
$$\frac{dy}{dx} =  -\frac{(4x+3y)}{(2x+y)}$$

Let y = xv,
$\frac{dy}{dx} = v + v'x$
$v+v'x = -\frac{4x+3xv}{2x+xv}$
$v+v'x = -\frac{(4+3v)}{(2+v)}$
$v'x = \frac{-v^2-5v-4}{2+v}$
$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{x} dx $
Let u = $(v^2+5v+4)$, du = (2v+5)dv
$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{2} \frac{(2v+5)}{(v^2+5v+4)}dv - \int \frac{0.5}{(v^2+5v+4)}dv$
$\frac{1}{2} ln(v^2+5v+4)dv - \frac{1}{2}*\frac{1}{3} \int(\frac{1}{v+1}-\frac{1}{v+4})dv$
$3ln(v^2+5v+4)-ln(\frac{(v+1)}{(v+4)}) = c-6ln(x)$
$(v+1)^2(v+1)^4 = x^6$
Therefore, we can get to know that
$(v+4)^2(v+1) = \frac{c}{x^{3}}$
plug y = xv into this equation
Therefore, we can get
$$(4x+y)^{2} (x+y) = c$$

MAT244--Misc / The form of Online Bonus Question and Quiz Answer
« on: January 25, 2018, 06:24:19 PM »
What should be the form of the posting quiz and the bonus question? Should it be typed? Or Handwriting can be accepted? Cause I saw some people just take a picture/scan their answer in the quiz section

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