Calculus
posted by Raj on .
Solve the differential and initial value problem:
x(dy/dx) + 1 = y^2
y(1)=0
I tried using bernoulli and it didn't quite work.

xy' + 1 = y^2
xy' = y^21
y' = (y^21)/x
dy/(y^21) = dx/x
arctanh(y) = log(x)
or,
log (1y)/(1+y) = 2log(x)+c
and you can massage that into exponentials and wind up with
y = 1e^(2cx^2) / 1+e^(2cx^2)