# First Order Differential Equation (Homogeneous?)

• Mar 6th 2011, 03:26 AM
Natalie11391
First Order Differential Equation (Homogeneous?)
So I'm stuck on:

(1-x²)y' = 1-y²

I think it's homogeneous but even that I'm not sure of. (I'm pretty new to first and second order ODEs so I'm finding it really hard at the moment!)

I simply took the (1-x²) to the other side to make it an equation in y'(x) but I have no idea on what to do now.

I know for homogeneous equations you can say that y=xv and let
y'(x) = xdv/dx + v for example but I can't get that to work here (suggesting that I might be wrong in thinking that it's a homogeneous equation....).

Any help is really appreciated!
• Mar 6th 2011, 03:33 AM
FernandoRevilla
It is not homogeneus, it is a separated variables equation.
• Mar 6th 2011, 03:34 AM
girdav
It's not a homogeneous equation because $y=0$ is not a solution.
You can first solve the equation $(1-x^2)y'=y^2$ (which is homogeneous) then try to find a particular solution, which is not difficult.
• Mar 6th 2011, 03:41 AM
FernandoRevilla
Quote:

Originally Posted by girdav
It's not a homogeneous equation because $y=0$ is not a solution.
You can first solve the equation $(1-x^2)y'=y^2$ (which is homogeneous) then try to find a particular solution, which is not difficult.

More directly:

$(1-x^2)dy=(1-y^2)dx\Leftrightarrow \dfrac{dy}{1-y^2}=\dfrac{dx}{1-x^2}\Leftrightarrow\displaystyle\int \dfrac{dy}{1-y^2}=\displaystyle\int\dfrac{dx}{1-x^2}\Leftrightarrow \ldots$