Originally Posted by

**alrightgeez** $\displaystyle \frac {dy}{dx} = \frac{2x}{x+y}$ using the substitution $\displaystyle y = xu$

I started by stating that $\displaystyle dy/dx = du/dx + u$ (product rule) and subbing so get the equation:

$\displaystyle \frac {du}{dx} + u = \frac {2x}{x+xu} $

Simplifying to get: $\displaystyle \frac {du}{dx} + u = \frac {2}{1+u} $

Then found my integrating factor $\displaystyle (e^(p(x)))$ which was $\displaystyle e^x $ and multiplied, leaving me with:

$\displaystyle \frac {d(ue^x)}{dx} = \frac {2e^x}{1+u} $

Firstly, am I right so far?

Secondly, if I am, I need help integrating $\displaystyle \frac {2e^x}{1+u}$ with respect to x, having the 2 unknowns is giving me second thoughts about integration by parts so I'm not entirely sure what to do.