# Thread: 1st order diff. eq.

1. ## 1st order diff. eq.

hi,

i have a little problem figuring out what to do with this equation:

$\displaystyle y` = \frac{y^2}{cos(y)-2xy}$

i think the way to go is to inverse the differentials and thus get:

$\displaystyle \frac{dx}{dy} = \frac{cos(y)-2xy}{y^2}$

2. dx/dy + (2/y)x = cos(y)/y^2

now use the integrating factor of y^2

3. thanks.