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.
To show that y = 1 and y = -1 are solutions to the given differential equation, you can substitute these values into the equation and check if it holds true.
Let's start with y = 1:
Substituting y = 1 into the differential equation, we get: dy/dx = (1^2 - 1)/x = 0/x = 0.
This implies that the derivative of y with respect to x is zero when y = 1.
Therefore, y = 1 satisfies the differential equation.
Now let's check y = -1:
Substituting y = -1 into the differential equation, we get: dy/dx = (-1^2 - 1)/x = -2/x.
This implies that the derivative of y with respect to x is -2/x when y = -1.
Therefore, y = -1 satisfies the differential equation as well.
By substituting the values of y into the original equation and verifying that they satisfy it, you have shown that y = 1 and y = -1 are indeed solutions to the given differential equation.
Regarding your general equation, it seems there is a small error. The correct general equation that satisfies the given differential equation is:
(x^2 + C)(y-1) = y+1
where C is the constant of integration. This expression can be simplified further if desired.
To determine if any of the curves found in question 1 intersect the line y = 1, you can substitute y = 1 into the general equation and check if it holds true. If it does, it means that the curve intersects the line at some point.