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

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)

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Eloise 2 weeks 2022-01-12T10:11:11+00:00 1 Answer 0 views 0

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    2022-01-12T10:12:21+00:00

    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

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