First order differential equation

• Jul 4th 2011, 04:32 AM
dwally89
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
• Jul 4th 2011, 06:05 AM
chisigma
Re: First order differential equation
Quote:

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...

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

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

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

... where the variables are separated...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jul 4th 2011, 06:23 AM
dwally89
Re: First order differential equation
Quote:

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

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

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

Thanks

EDIT: Understand now, thanks
• Jul 4th 2011, 06:54 AM
Prove It
Re: First order differential equation
Quote:

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

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