## 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.