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

Answer:

y = $\frac{3}{2 + 3Ce^{-3t} }$

Step-by-step explanation:

As given , y' = 3y - 2y²

⇒y' - 3y = -2y²

Divide by y² in the above equation

⇒$\frac{y'}{y^{2} } - \frac{3y}{y^{2} } = -\frac{2y^{2} }{y^{2} }$

⇒$\frac{y'}{y^{2} } - \frac{3}{y } = -2$       ........(1)

Now , let $\frac{1}{y}$ = u

⇒-$\frac{1}{y^{2} } \frac{dy}{dt} = \frac{du}{dt}$

⇒-$\frac{1}{y^{2} }y' = \frac{du}{dt}$

∴ equation (1) becomes

-$\frac{du}{dt}$ - 3u = -2

⇒$\frac{du}{dt}$ + 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.($e^{3t}$) = $\frac{2e^{3t} }{3}$ + C

⇒u = $\frac{2}{3} + Ce^{-3t}$

As $\frac{1}{y}$ = u

⇒$\frac{1}{y}$  = $\frac{2}{3} + Ce^{-3t}$

⇒ y = $\frac{1}{\frac{2}{3} + Ce^{-3t} } = \frac{3}{2 + 3Ce^{-3t} }$

⇒ y = $\frac{3}{2 + 3Ce^{-3t} }$