# First order differentials

• Jan 28th 2010, 10:04 AM
alrightgeez
First order differentials
$\frac {dy}{dx} = \frac{2x}{x+y}$ using the substitution $y = xu$

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

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

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

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

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

Firstly, am I right so far?
Secondly, if I am, I need help integrating $\frac {2e^x}{1+u}$ 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.
• Jan 28th 2010, 10:41 AM
TheEmptySet
Quote:

Originally Posted by alrightgeez
$\frac {dy}{dx} = \frac{2x}{x+y}$ using the substitution $y = xu$

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

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

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

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

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

Firstly, am I right so far?
Secondly, if I am, I need help integrating $\frac {2e^x}{1+u}$ 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

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

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

Now seperate to get

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

From here use partial fractions to integrate.

I hope this helps
• Jan 29th 2010, 02:51 AM
alrightgeez
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 :)