zgribestika

Answered

2022-01-17

How do you solve the Initial value probelm

$\frac{dp}{dt}=10p(1-p)$ ,

$p\left(0\right)=0.1$

Solve and show that$p\left(t\right)\to 1\text{}\text{as}\text{}t\to \mathrm{\infty}$ .

Solve and show that

Answer & Explanation

Jenny Sheppard

Expert

2022-01-18Added 35 answers

This is the so-called logistic equation, which occurs often in population dynamics and many other contexts. There's a trick which works for this particular equation and is much simpler than separation of variables (in my opinion): change variables to $y\left(t\right)=\frac{1}{p}\left(t\right)$ . Then the nonlinear equation for p turns into an inhomogeneous linear equation for y, which can be solved immediately by the usual ''$\text{homogeneous}+\text{particular solution}$ '' method (the homogeneous solution is an exponential, and the particular solution is a constant). Since this is tagged as homework, I'll let you have a go at the details yourself.

Philip Williams

Expert

2022-01-19Added 39 answers

The method we can use here is called Separation of Variables. Take all the ps

alenahelenash

Expert

2022-01-24Added 366 answers

From your comment, it looks you have been able to integrate correctly, following Ragib's Hint and Gourtaur comment. But now your problem is (to finish the solution) to express p(t). This rest part is a simple algebra. Let me express p(t) in terms of t:$\frac{p}{1-p}={e}^{10t+10c}={e}^{10t}\cdot {e}^{10c}=k.{e}^{10t}$ (where $k={e}^{10c}$ is a new constant)$\Rightarrow \frac{p}{(1-p)+p}=\frac{k{e}^{10t}}{1+k{e}^{10t}}$ (I applied $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b+a}=\frac{c}{d+c}$ . You can just multiply both sides by $(1-p)$ , or cross-multiply and solve for p)$\Rightarrow p=p(t)=\frac{1}{1+{k}^{\prime}{e}^{-10t}}$ (dividing numerator and denominator of the fraction on RHS by $k{e}^{10t}$ and writing $k{e}^{10t}$ and writing ${k}^{\prime}=\frac{1}{k}$ )Now, from using the condition $p(0)=0.1=\frac{1}{10}$ , we get $\frac{1}{10}=\frac{1}{1+{k}^{\prime}}\Rightarrow {k}^{\prime}=9$

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