1. ## First order differentials

$\displaystyle \frac {dy}{dx} = \frac{2x}{x+y}$ 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.

2. 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.

The I-factor only works on linear ODE's

From here the equation is seperable

$\displaystyle \frac {du}{dx} + u = \frac {2}{1+u}$

$\displaystyle \frac {du}{dx} = \frac {2-u(1+u)}{1+u}=-\frac{u^2+u-2}{u+1} =-\frac{(u+2)(u-1)}{u+1}$

Now seperate to get

$\displaystyle \frac{u+1}{(u+2)(u-1)}du=-dt$

From here use partial fractions to integrate.

I hope this helps

3. Ah thanks for that, turns out I differentiated wrongly at the beginning when I used the product rule so my final answer was wrong, but you helped me out anyway so thanks