### Author Topic: TUT0101 QUIZ1  (Read 1907 times)

#### Yiheng Bian

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

#### Yiheng Bian

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• Posts: 29
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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 doesnâ€™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!