How do you do these questions?

I'll do the first problem to get you started.

Part (a)

We have a separable equation. Get the y term to the left side and then integrate to get

[tex]\frac{dy}{dt} = ky^{1+c}\\\\\frac{dy}{y^{1+c}} = kdt\\\\\displaystyle \int\frac{dy}{y^{1+c}} = \int kdt\\\\\displaystyle \int y^{-(1+c)}dy = \int kdt\\\\\displaystyle -\frac{1}{c}y^{-c} = kt+D\\\\\displaystyle -\frac{1}{c*y^{c}} = kt+D\\\\[/tex]

I'm using D as the integration constant rather than C since lowercase letter c was already taken.

Let's use initial condition that [tex]y(0) = y_0[/tex]. This means we'll plug in t = 0 and [tex]y = y_0[/tex]. After doing so, solve for D

[tex]\displaystyle -\frac{1}{c*y^{c}} = kt+D\\\\\displaystyle -\frac{1}{c*(y_0)^{c}} = k*0+D\\\\\displaystyle D = -\frac{1}{c*(y_0)^{c}}\\\\[/tex]

Let's plug that in and isolate y

[tex]\diplaystyle -\frac{1}{c}y^{-c} = kt+D\\\\\\\diplaystyle -\frac{1}{c}y^{-c} = kt-\frac{1}{c*(y_0)^{c}}\\\\\\\diplaystyle y^{-c} = -ckt+\frac{1}{(y_0)^{c}}\\\\\\\diplaystyle y^{-c} = \frac{1-c*(y_0)^{c}kt}{(y_0)^{c}}\\\\\\\diplaystyle \frac{1}{y^{c}} = \frac{1-c*(y_0)^{c}kt}{(y_0)^{c}}\\\\\\\diplaystyle y^{c} = \frac{(y_0)^{c}}{1-c*(y_0)^{c}kt}\\\\\\\diplaystyle y = \left(\frac{(y_0)^{c}}{1-c*(y_0)^{c}kt}\right)^{1/c}\\\\\\\diplaystyle y = \frac{y_0}{\left(1-c*(y_0)^{c}kt\right)^{1/c}}\\\\\\[/tex]

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We end up with [tex]\displaystyle y(t) = \frac{y_0}{\left(1-c*(y_0)^{c}kt\right)^{1/c}}\\\\[/tex] as our final solution. There are likely other forms to express this equation.

========================================================

Part (b)

We want y(t) to approach positive infinity.

Based on the solution in part (a), this will happen when the denominator approaches 0 from the left.

So [tex]y(t) \to \infty[/tex] as [tex]1-c*(y_0)^{c}kt \to 0[/tex] in which we can effectively "solve" for t showing that [tex]t \to \frac{1}{c*(y_0)^{c}k}[/tex]

If we define [tex]T = \frac{1}{c*(y_0)^{c}k}[/tex] , then approaching T from the left side will have y(t) approach positive infinity.

This uppercase T value is doomsday. This the time value lowercase t approaches from the left when the population y(t) explodes to positive infinity.

Effectively t = T is the vertical asymptote.

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Part (c)

We're told that the initial condition is y(0) = 5 since at time 0, we have 5 rabbits. This means [tex]y_0 = 5[/tex]

Another fact we know is that y(3) = 35 because after three months, there are 35 rabbits.

Lastly, we know that c = 0.01 since the exponent of dy/dt = ky^(1.01) is 1.01; so we solve 1+c = 1.01 to get c = 0.01

We'll use y(3) = 35, c = 0.01 and [tex]y_0 = 5[/tex] to solve for k

Doing so leads to the following:

[tex]\displaystyle y(t) = \frac{y_0}{\left(1-c*(y_0)^{c}kt\right)^{1/c}}\\\\\\\displaystyle y(3) = \frac{5}{\left(1-0.01*(5)^{0.01}k*3\right)^{1/0.01}}\\\\\\\displaystyle 35 \approx \frac{5}{\left(1-0.0304867k\right)^{100}}\\\\\\\displaystyle 35\left(1-0.0304867k\right)^{100} \approx 5\\\\\\\displaystyle \left(1-0.0304867k\right)^{100} \approx \frac{1}{7}\\\\\\[/tex]

[tex]\displaystyle \left(1-0.0304867k\right)^{100} \approx 7^{-1}\\\\\\\displaystyle 1-0.0304867k \approx \left(7^{-1}\right)^{1/100}\\\\\\\displaystyle 1-0.0304867k \approx 7^{-0.01}\\\\\\\displaystyle k \approx \frac{7^{-0.01}-1}{-0.0304867}\\\\\\\displaystyle k \approx 0.63211155281122\\\\\\\displaystyle k \approx 0.632112\\\\\\[/tex]

We can now compute the doomsday time value

[tex]T = \frac{1}{c*(y_0)^c*k}\\\\\\T \approx \frac{1}{0.01*(5)^{0.01}*0.632112}\\\\\\T \approx \frac{1}{0.00642367758836}\\\\\\T \approx 155.674064621806\\\\\\T \approx 155.67\\\\\\[/tex]

The answer is approximately 155.67 months

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