Author Topic: TUT0101 QUIZ1  (Read 705 times)

Yiheng Bian

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TUT0101 QUIZ1
« on: September 27, 2019, 03:02:56 PM »
answer

Yiheng Bian

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Re: TUT0101 QUIZ1
« Reply #1 on: September 27, 2019, 03:55:09 PM »
$$
\frac{dy}{dx}= \frac{x+3y}{x-y}
$$
$$
\frac{dy}{dx}=\frac{1+3\frac{y}{x}}{1-\frac{y}{x}}
$$
$$
Let \frac{y}{x}=u,y=xu
$$
$$
\frac{dy}{dx}=\frac{d(xu)}{dx}=\frac{1+3u}{1-u}
$$
$$
u+x\frac{du}{dx}=\frac{1+3u}{1-u}
$$
$$
u+x\frac{du}{dx}=\frac{1+3u}{1-u}
$$
$$
x\frac{du}{dx}=\frac{1+3u}{1-u}-u=\frac{(1+u)^2}{1-u}
$$
$$
\frac{1-u}{(1-u)^2}=\frac{1}{x}dx
$$
$$
\int\frac{1-u}{(1+u)^2}du=\int\frac{1}{x}dx
$$
$$
\int\frac{2}{(1+u)^2}-\frac{1+u}{(1+u)^2du}=lnx+c
$$
$$
-\frac{2}{(1+u)^2}-ln(1+u)=lnx+c
$$
$$
-\frac{2x}{x+y}-c+ln(x(1+\frac{y}{x}))
$$
$$
-\frac{2x}{x+y}-c=ln(x+y)
$$
$$
ln(x+y)+\frac{2x}{x+y}=-c
$$
$$
ln(x+y)=-c-\frac{2x}{x+y}
$$
$$
x+y=e^{-c-\frac{2x}{x+y}}=Ce^{-\frac{2x}{x+y}}
$$




$$

lyujiahe

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Re: TUT0101 QUIZ1
« Reply #2 on: October 03, 2019, 10:20:03 AM »
In the 10th line, by formula $\int \frac{1}{x} = ln|x|+C$, how can we identify $u+1 \geq 0, x \geq 0$?
« Last Edit: October 03, 2019, 10:24:26 AM by lyujiahe »

Yiheng Bian

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Re: TUT0101 QUIZ1
« Reply #3 on: October 03, 2019, 03:26:11 PM »
Sure, you can add absolute value sign. But it doesn’t matter actually. Because whatever + or- . At final step. It is a part of constant C.

lyujiahe

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Re: TUT0101 QUIZ1
« Reply #4 on: October 07, 2019, 10:59:55 AM »
Thank you so much!  ;D