# 1st order diff. eq.

• Feb 2nd 2010, 04:28 AM
vonflex1
1st order diff. eq.
hi,

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

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

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

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

• Feb 2nd 2010, 04:46 AM
Calculus26
dx/dy + (2/y)x = cos(y)/y^2

now use the integrating factor of y^2
• Feb 2nd 2010, 05:16 AM
vonflex1
thanks.