Question

In Exercises 1 and 2, find the equilibrium solutions of the differential equation specified.1) dy/dt = (y + 3)/(1 - y)2) dy/dt = (t² - 1)(y² - 2)/(y ² - 4)

Answer 1

Answer:

1. Equilibrium solution: y= -3

2. Equilibrium solution y= ±1.414

Step-by-step explanation:

Thinking process:

The equilibrium solution can only be derived when $\frac{dy}{dt} = 0$

1. Let's look at the first equation:

$\frac{dy}{dt} = \frac{(y+3)}{(1-y)}$

equating $\frac{dy}{dt} = 0$ to the expression $\frac{(y+3)}{(1-y)}$ gives

y + 3 = 0

     y = -3

Therefore, the equilibrium solution occurs at y = -3

2. Let's look at the second solution:

$\frac{dy}{dy} = \frac{(t^{2}- 1) (y^{2}-2) }{y^{2}-4 }$

dividing each side by (t²-1) gives

1/(t²-1)$\frac{dy}{dt}$ = (y²-2)/ (y²-4)

factorizing the right hand side gives:

at equilibrium:  $\frac{dy}{dt} = 0$, then

y² - 2 = 0

solving for y gives y = ±√2

                                 = ±1.414