Calculus

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.

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asked by Hebe

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