# Thread: First order differential equation

1. ## First order differential equation

Hi,

I'm new to this forum, so please let me know if I'm posting in the wrong place and I'll move this thread.

I have the equation:
(x-y)dx + (x+y)dy = 0

So far I've rearranged it to get:
y' = (-x+y) / (x+y)

But I'm not sure where to go from here.

Help would be appreciated,

Thanks

2. ## Re: First order differential equation

Originally Posted by dwally89
Hi,

I'm new to this forum, so please let me know if I'm posting in the wrong place and I'll move this thread.

I have the equation:
(x-y)dx + (x+y)dy = 0

So far I've rearranged it to get:
y' = (-x+y) / (x+y)

But I'm not sure where to go from here.

Help would be appreciated,

Thanks
The DE equation...

$y^{'} = \frac{y-x}{y+x}$ (1)

... is of the 'Homogeneous type'. Setting $\frac{y}{x}= t \implies \frac{dy}{dx}= t + x \frac{dt}{dx}$ You arrive at
the DE...

$x\ \frac{dt}{dx} = -\frac{1+t^{2}}{1+t}$ (2)

... where the variables are separated...

Kind regards

$\chi$ $\sigma$

3. ## Re: First order differential equation

Originally Posted by chisigma
Setting $\frac{y}{x}= t \implies \frac{dy}{dx}= t + x \frac{dt}{dx}$ You arrive at
the DE...

$x\ \frac{dt}{dx} = -\frac{1+t^{2}}{1+t}$ (2)
After you set t=y/x, I don't understand how you get to the next part,

Thanks

EDIT: Understand now, thanks

4. ## Re: First order differential equation

Originally Posted by dwally89
After you set t=y/x, I don't understand how you get to the next part,

Thanks

EDIT: Understand now, thanks
$\displaystyle t = \frac{y}{x} \implies y = tx$. This is a product of functions, so to get $\displaystyle \frac{dy}{dx}$ you need to use the product rule on $\displaystyle tx$.