Solve this Differential Equation in y and x. 
Solve the following DE:

$$y'=\frac{y+y^2}{x+x^2}$$

in particular

$$y(2)=1$$$$\frac{y'}{y+y^2}=\frac{1}{x+x^2}$$$$\int \frac{dy}{y+y^2}=\int \frac{dx}{x+x^2}$$$$y'dx=dy$$$$ln(y)-ln(y+1)=ln(x)-ln(x+1)+C$$

but where to from here?

 A: This gives
\begin{align*}
\frac{y}{y+1} &=e^{C}\frac{x}{x+1}\\
2y&=\frac{e^Cx+x+1}{e^Cx-x-1}.
\end{align*}
Now use $y(2)=1$, to get $e^C$.
A: Well, when we solve:
$$\text{y}\space'\left(x\right)=\frac{\text{y}\left(x\right)+\text{y}\left(x\right)^2}{x+x^2}\space\Longleftrightarrow\space\int\frac{\text{y}\space'\left(x\right)}{\text{y}\left(x\right)+\text{y}\left(x\right)^2}\space\text{d}x=\int\frac{1}{x+x^2}\space\text{d}x\tag1$$
So, for the intergals we get:
$$\ln\left|\frac{\text{y}\left(x\right)}{1+\text{y}\left(x\right)}\right|=\ln\left|1+\frac{1}{x}\right|+\text{C}\tag2$$
Using $\text{y}\left(2\right)=1$ we get:
$$\ln\left|\frac{1}{1+1}\right|=\ln\left|1+\frac{1}{2}\right|+\text{C}\space\Longleftrightarrow\space\text{C}=-\ln\left(3\right)\tag3$$
So, we get:
$$\ln\left|\frac{\text{y}\left(x\right)}{1+\text{y}\left(x\right)}\right|=\ln\left|1+\frac{1}{x}\right|-\ln\left(3\right)\tag4$$
Take the $\exp$ of both sides:
$$\left|\frac{1}{1+\frac{1}{\text{y}\left(x\right)}}\right|=\frac{1}{3}\cdot\left|1+\frac{1}{x}\right|\space\Longleftrightarrow\space\left|1+\frac{1}{\text{y}\left(x\right)}\right|=\frac{3}{\left|1+\frac{1}{x}\right|}\tag5$$
