Hello Em Yeu Anh Originally Posted by

**Em Yeu Anh** Thanks for the help.

Okay I've gotten it down to these steps:

$\displaystyle \frac{ln|y|}{2}-\frac{ln|2+y|}{2}=\frac{x^2}{2}+C $

Assuming I've made no errors in simplifying,

$\displaystyle \frac{y}{2+y} = Ae^{x^2} $ where A=__+__$\displaystyle e^C$

And here I don't know how to isolate for y...Is it allright to split up $\displaystyle \frac{y}{2+y}$ into $\displaystyle \frac{y}{2}+1 $ and go from there?

$\displaystyle \frac{y}{2+y} = Ae^{x^2} $

$\displaystyle \Rightarrow y= (2+y)Ae^{x^2}$$\displaystyle =2Ae^{x^2}+yAe^{x^2}$

$\displaystyle \Rightarrow y(1-Ae^{x^2})=2Ae^{x^2}$

$\displaystyle \Rightarrow y=\frac{2Ae^{x^2}}{(1-Ae^{x^2})}$

which you could write as:

$\displaystyle y=\frac{2e^{x^2}}{(A'-e^{x^2})}$, replacing $\displaystyle \frac1A$ by $\displaystyle A'$.

Grandad