# Separation of Variables

• Jan 27th 2014, 04:12 PM
petenice
Separation of Variables
Im having trouble separating functions into the form $\displaystyle q(y)dy=p(x)dx$. It seems like they should be simple but im stuck on these. Any help would be appreciated. Thanks.

Here are two examples

$\displaystyle \frac{dy}{dx}=\frac{y^2+1}{xy+y}$
I can only get this far..
$\displaystyle ydy=\frac{y^2+1}{(x+1)}dx$

and im also having trouble with this one.

$\displaystyle \frac{dy}{dx}=xy^2-x-y^2+1$
• Jan 27th 2014, 04:19 PM
romsek
Re: Separation of Variables
Quote:

Originally Posted by petenice
Im having trouble separating functions into the form $\displaystyle q(y)dy=p(x)dx$. It seems like they should be simple but im stuck on these. Any help would be appreciated. Thanks.

Here are two examples

$\displaystyle \frac{dy}{dx}=\frac{y^2+1}{xy+y}$
I can only get this far..
$\displaystyle ydy=\frac{y^2+1}{(x+1)}dx$

and im also having trouble with this one.

$\displaystyle \frac{dy}{dx}=xy^2-x-y^2+1$

$\displaystyle y\;dy=\frac{y^2+1}{x+1}dx \Rightarrow$

$\displaystyle \frac{y\; dy}{y^2+1}=\frac{dx}{x+1}$

for the second one

$\displaystyle \frac{dy}{dx}=xy^2-x-y^2+1 \Rightarrow$

$\displaystyle \frac{dy}{dx}=y^2(x-1)-(x-1)=(y^2-1)(x-1) \Rightarrow$

$\displaystyle \frac{dy}{y^2-1}=(x-1)\;dx$
• Jan 27th 2014, 04:20 PM
SlipEternal
Re: Separation of Variables
Quote:

Originally Posted by petenice
I can only get this far..
$\displaystyle ydy=\frac{y^2+1}{(x+1)}dx$

Divide both sides by $\displaystyle y^2+1$.

Quote:

Originally Posted by petenice
$\displaystyle \frac{dy}{dx}=xy^2-x-y^2+1$

$\displaystyle xy^2-x-y^2+1 = x(y^2-1)-(y^2-1) = (x-1)(y^2-1)$

Romsek beat me to it.
• Jan 28th 2014, 02:04 PM
petenice
Re: Separation of Variables
I see. Yep i figured it was something simple, i just felt i was making up stuff at some point. Too tired to think straight. Thank you so much guys.