Logistic growth model

Question

Find the solution of the model when $y(0)=100$

$\frac{dy}{dt}=2y(y-1)(3-y)$

This is what I have using partial fractions we get

$1=A(y-1)(3-y)+B(2y)(3-y)+C(2y)(y-1)$

We then get that :

$A=\frac{-1}{3}, B=\frac{1}{4}, C=\frac{1}{12}$.

Then I have

$\frac{-1}{3}\int \frac{1}{2y}+ \frac{1}{4}\int \frac{1}{y-1}+\frac{1}{12}\int \frac{1}{3-y}= \int dt$

$\rightarrow ln[\frac{(y-1)^{1/4}}{(2y)^{1/6} \cdot (3-y)^{1/12}}]=t+c$

finally we get that:

$[\frac{(y-1)^{1/4}}{(2y)^{1/6} \cdot (3-y)^{1/12}}]=c \cdot e^t$

I'm stuck in this step, what do I have to do to get $y(0)=100$ Any help would be appreciated.

Answer 1

You can find $c$ (the only thing you need) by just evaluating the condition given.

Let me assume that all your computations are correct, and indeed we only need to determine $c$ from your last equation. But I think that with the $-$ sign outside the last integral, you get a term $y-3$.

Then at $t=0$, you have $y = y(0) = 100$.

So you can plug in $100$ every time you see a $y$ and $0$ every time you see a $t$, as you need to evaluate in the point $(t,y(t)) = (0,100)$.

$$ \frac{(100-1)^{1/4}}{(200)^{1/6} (100-3)^{1/12}} = c e^0 = c \approx 0.89092... $$

And the solution with such $c$ is your final answer. I hope I didn't make a mistake when checking the signs of the integrals, please double check it.

Answer 2

The condition:

$ y(0) = 100 $

basically means that y attains a value of 100 when t attains the value 0. Also, the equation you have just derived contains an arbitrary constant as a result of your integration which can be eliminated by imposing this particular condition in the equation.

Replace t with 0 to get c on the RHS and replace y with 100.

$ \frac{(99)^{\frac{1}{4}}}{(200)^{\frac{1}{6}}(97)^{\frac{1}{12}}} = c $

Hence, calculate the value of c and replace it in the original expression to get the required growth equation.