Thursday

September 29, 2016
Posted by **Raj** on Tuesday, December 10, 2013 at 8:39pm.

x(dy/dx) + 1 = y^2

y(1)=0

I tried using bernoulli and it didn't quite work.

- Calculus -
**Steve**, Tuesday, December 10, 2013 at 11:33pmxy' + 1 = y^2

xy' = y^2-1

y' = (y^2-1)/x

dy/(y^2-1) = dx/x

arctanh(y) = log(x)

or,

log (1-y)/(1+y) = 2log(x)+c

and you can massage that into exponentials and wind up with

y = 1-e^(2cx^2) / 1+e^(2cx^2)