Answer:
(t, y) = (1, 2), (1, -2), (-1, 2), and (-1, -2).
Explanation:
Let's solve each differential equation and find their equilibrium solutions.
dy/dt = y + 3 / (1 - y)
To find the equilibrium solutions, we set dy/dt equal to zero:
0 = y + 3 / (1 - y)
Now, we can solve for y:
y(1 - y) = -3
y - y^2 = -3
y^2 - y + 3 = 0
This is a quadratic equation. We can apply the quadratic formula:
y = (-(-1) ± √((-1)^2 - 4(1)(3))) / (2(1))
Simplifying further:
y = (1 ± √(1 - 12)) / 2
y = (1 ± √(-11)) / 2
Since the discriminant is negative, there are no real solutions for y. Therefore, this differential equation does not have any equilibrium solutions.
dy/dt = (t^2 - 1)(y^2 - 2) / (y^2 - 4)
Setting dy/dt to zero:
0 = (t^2 - 1)(y^2 - 2) / (y^2 - 4)
This equation is satisfied when either the numerator is zero or when the denominator is zero.
Numerator: t^2 - 1 = 0
This gives two solutions: t = 1 and t = -1.
Denominator: y^2 - 4 = 0
This gives two solutions: y = 2 and y = -2.
Therefore, the equilibrium solutions for this differential equation are:
(t, y) = (1, 2), (1, -2), (-1, 2), and (-1, -2).