# Calculus

posted by
**Hebe** on
.

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

1. Give the general equation of the curves that satisfy this equation.

2. Show that the straight lines y=1 and y=-1 are also solutions.

3. Do any of the curves you found in 1) intersect y=1?

i started by

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

and found that

[ln(y-1) - ln(y+1)]/2 = lnx

what do i do next?

The left becomes

ln((y-1)/(y+1))

move the 2 to the right to make 2lnx or lnx^2

When y is +- 1, what is the slope? Is it that slope for all x? If so, it is a straight line.

so x^2 + C = (y-1)/(y+1) is the general equation?

how do i show that the straight lines y=1 and y=-1 are also solutions.