Applications of first ODE

Question

I'm having a little trouble with this question:

The population of a certain animal species is governed by the differential equation $$ 1000 {dp\over dt} = p(100 − p)$$ where $p$ is the number of individuals in the colony at $t$ years. The initial population is known to be $200$ individuals.

Find $p(t)$ and sketch the population–time graph.

This is what I have tried so far:

$$\int {1\over p(100-p)} dp = \int {1\over 1000} dt $$

I did the necessary steps and got it reduced to $$ \log |p|-\log|p-100| ={1\over10} t + C $$

I then got $$ \log \left({p\over p-100}\right)={1\over10}t + 100C $$

I reduced it further and got: $e^{100c} = 2$

Thus,

$${p\over p-100} = 2 e^{{1\over 10}t} $$

I'm not sure how to solve it from here

Answer 1

$$\frac{p}{ p-100} = 2 e^{{1\over 10}t}$$

$$\frac{ p-100}{p} = \frac12 \exp(-\frac1{10}t)$$

$$1-\frac{ 100}{p} = \frac12 \exp(-\frac1{10}t)$$

$$1- \frac12 \exp(-\frac1{10}t)= \frac{ 100}{p}$$

$$p= \frac{ 100}{1- \frac12 \exp(-\frac1{10}t)}$$

Answer 2

You need to solve for p

$$ \frac {p}{p-100} = 2e^{{1\over 10}t}$$

cross multiply to get

$$p= (p-100) 2e^{{1\over 10}t}$$

$$ p(1-2e^{{1\over 10}t})=-200e^{{1\over 10}t}$$

$$p= \frac {-200e^{{1\over 10}t}}{1-2e^{1\over 10}t}$$

Answer 3

Your second step is incorrect.

$$\int {1\over \color{red}{p(100-p)}} dp = \int {1\over 1000} dt$$

The red part is quadratic in $p$.

Since

$$\mathcal{I}(p)=\int {dp\over{p(100-p)}}=-\int\frac{dp}{(p-50)^2-50^2}$$

A proper substitution here is $p-50=50\sec\theta, dp=50\tan^2\theta d\theta$, such that

$$\mathcal{I}=-\int\frac{50\tan^2\theta d\theta}{50\tan^2\theta}=-\theta+C=-\sec^{-1}(\frac p{50}-1)+C$$

You may continue here.