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 [tex]\frac{dy}{dt} = 0[/tex]
1. Let's look at the first equation:
[tex]\frac{dy}{dt} = \frac{(y+3)}{(1-y)}[/tex]
equating [tex]\frac{dy}{dt} = 0[/tex] to the expression [tex]\frac{(y+3)}{(1-y)}[/tex] gives
y + 3 = 0
y = -3
Therefore, the equilibrium solution occurs at y = -3
2. Let's look at the second solution:
[tex]\frac{dy}{dy} = \frac{(t^{2}- 1) (y^{2}-2) }{y^{2}-4 }[/tex]
dividing each side by (t²-1) gives
1/(t²-1)[tex]\frac{dy}{dt}[/tex] = (y²-2)/ (y²-4)
factorizing the right hand side gives:
at equilibrium: [tex]\frac{dy}{dt} = 0[/tex], then
y² - 2 = 0
solving for y gives y = ±√2
= ±1.414
