TUT0101 QUIZ1

[Yiheng Bian]

answer

[Yiheng Bian]

$$
\frac{dy}{dx}= \frac{x+3y}{x-y}
$$
$$
\frac{dy}{dx}=\frac{1+3\frac{y}{x}}{1-\frac{y}{x}}
$$
$$
Let \frac{y}{x}=u,y=xu
$$
$$
\frac{dy}{dx}=\frac{d(xu)}{dx}=\frac{1+3u}{1-u}
$$
$$
u+x\frac{du}{dx}=\frac{1+3u}{1-u}
$$
$$
u+x\frac{du}{dx}=\frac{1+3u}{1-u}
$$
$$
x\frac{du}{dx}=\frac{1+3u}{1-u}-u=\frac{(1+u)^2}{1-u}
$$
$$
\frac{1-u}{(1-u)^2}=\frac{1}{x}dx
$$
$$
\int\frac{1-u}{(1+u)^2}du=\int\frac{1}{x}dx
$$
$$
\int\frac{2}{(1+u)^2}-\frac{1+u}{(1+u)^2du}=lnx+c
$$
$$
-\frac{2}{(1+u)^2}-ln(1+u)=lnx+c
$$
$$
-\frac{2x}{x+y}-c+ln(x(1+\frac{y}{x}))
$$
$$
-\frac{2x}{x+y}-c=ln(x+y)
$$
$$
ln(x+y)+\frac{2x}{x+y}=-c
$$
$$
ln(x+y)=-c-\frac{2x}{x+y}
$$
$$
x+y=e^{-c-\frac{2x}{x+y}}=Ce^{-\frac{2x}{x+y}}
$$




$$

[lyujiahe]

In the 10th line, by formula $\int \frac{1}{x} = ln|x|+C$, how can we identify $u+1 \geq 0, x \geq 0$?

[Yiheng Bian]

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]

Thank you so much!