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.
1 answer