Question

g Suppose that the populationy(t)of a certain kind of fish is given by the logistic modely′= 3y−2y2(a) Solve the Bernoulli equation. (Must be done using Bernoulli’s Equation method)(b) Suppose fish are added to the pond at a constant rate of 2 (Basically add 2 to your equation).Set up the Riccati equation, find a particular solution by inspection (it should be obvious if youlook for a constantyfunction), and then solve the equation.

Answer 1

y =
`(3)/(2 + 3Ce^(-3t) )`

Explanation:

As given , y' = 3y - 2y²

⇒y' - 3y = -2y²

Divide by y² in the above equation

⇒
`(y')/(y^(2) ) - (3y)/(y^(2) ) = -(2y^(2) )/(y^(2) )`

⇒
`(y')/(y^(2) ) - (3)/(y ) = -2` ........(1)

Now , let
`(1)/(y)` = u

⇒-
`(1)/(y^(2) ) (dy)/(dt) = (du)/(dt)`

⇒-
`(1)/(y^(2) )y' = (du)/(dt)`

∴ equation (1) becomes

-
`(du)/(dt)` - 3u = -2

⇒
`(du)/(dt)` + 3u = 2

It is a linear differential equation

Now,

Integrating factor = I.F =
`e^{\int\limits {3} \, dt }` =
`e^(3t)`

∴ The solution becomes

u.(I.F) =
`\int\limits {2.(I.F)} \, dt` + C

⇒u.(
`e^(3t)`) =
`\int\limits {2.(e^(3t) )} \, dt` + C

⇒u.(
`e^(3t)`) =
`(2e^(3t) )/(3)` + C

⇒u =
`(2)/(3) + Ce^(-3t)`

As
`(1)/(y)` = u

⇒
`(1)/(y)` =
`(2)/(3) + Ce^(-3t)`

⇒ y =
`(1)/((2)/(3) + Ce^(-3t) ) = (3)/(2 + 3Ce^(-3t) )`

⇒ y =
`(3)/(2 + 3Ce^(-3t) )`