<s>[INST] 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?

 [/INST] 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 $e^C$.
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