y'=(1+x^2)*tgy y'=(-1+y^2)/(x*y*(1-x^2)) Thanks.
Last edited by Ackbeet; Jul 25th 2011 at 04:59 AM.
Follow Math Help Forum on Facebook and Google+
Here's a better idea, show some manners, show some effort, show where you are stuck, and you might get some help.
A little help for the first one: write $\displaystyle \dfrac{dy}{\tan y}=(1+x^2)dx$ (separated variables). Now, integrate both sides, let's see what do you obtain.
View Tag Cloud