Consider anew the logistic equation, (d P)/(d t)=r P(1-(P)/(K)), P(t0)=P0 . Typically, mathema

Consider anew the logistic equation, (d P)/(d t)=r...

Consider anew the logistic equation, $$ \frac{d P}{d t}=r P\left(1-\frac{P}{K}\right), \quad P\left(t_{0}\right)=P_{0} . $$ Typically, mathematical ecologists will introduce dimensionless variables to reduce the number of parameters in the logistic equation before proceeding with their analysis. (a) Show that the substitutions $\omega=\alpha P$ and $s=\beta t$ transform equation (1.29) into $$ \frac{d \omega}{d s}=\frac{r}{\beta} w-\frac{r}{\alpha \beta K} \omega^{2} . $$ (b) Find values of $\alpha$ and $\beta$ that transform equation (1.30) into $$ \frac{d \omega}{d s}=\omega-\omega^{2} $$ (c) Note that equation (1.31) is a variant of Bernoulli's equation. Use the technique of Exercise 22 in Section $2.4$ to show that equation (1.31) has the solution $$ \omega=\frac{1}{1-C e^{-s}} \text {. } $$

Consider anew the logistic equation,
$$
\frac{d P}{d t}=r P\left(1-\frac{P}{K}\right), \quad P\left(t_{0}\right)=P_{0} .
$$
Typically, mathematical ecologists will introduce dimensionless variables to reduce the number of parameters in the logistic equation before proceeding with their analysis.
(a) Show that the substitutions $\omega=\alpha P$ and $s=\beta t$ transform equation (1.29) into
$$
\frac{d \omega}{d s}=\frac{r}{\beta} w-\frac{r}{\alpha \beta K} \omega^{2} .
$$
(b) Find values of $\alpha$ and $\beta$ that transform equation (1.30) into
$$
\frac{d \omega}{d s}=\omega-\omega^{2}
$$
(c) Note that equation (1.31) is a variant of Bernoulli's equation. Use the technique of Exercise 22 in Section $2.4$ to show that equation (1.31) has the solution
$$
\omega=\frac{1}{1-C e^{-s}} \text {. }
$$

Key Concepts

Logistic Equation

The logistic equation is a fundamental model in population dynamics, representing the growth of a population with a carrying capacity. It features a growth term that is proportional to the current population and a saturation term that limits growth as the population approaches its carrying capacity. Understanding this equation is crucial as it encapsulates both exponential growth and the eventual leveling off of the population due to environmental constraints.

Non-Dimensionalization

Non-dimensionalization is a mathematical technique used to reduce the number of parameters in a differential equation by introducing dimensionless variables. This method involves scaling variables and time so that the equation is expressed in a simpler form, often highlighting the essential dynamics without the clutter of units or extraneous parameters. This process aids in analyzing and comparing systems by focusing on the inherent characteristics of the model.

Change of Variables

Change of variables is a method used in differential equations to simplify the form of the equation or to transform it into a standard form that is easier to solve. By introducing new variables that are functions of the original variables, one can often reduce complexity, eliminate parameters, or reveal underlying structures in the equation. This technique is especially useful when dealing with non-dimensionalization and when attempting to match a differential equation to a known solvable form.

Bernoulli's Equation

Bernoulli's equation refers to a class of nonlinear ordinary differential equations that, despite their nonlinearity, can be transformed into linear differential equations using an appropriate substitution. Recognizing an equation as a variant of Bernoulli's form enables the use of established solution techniques such as finding an integrating factor or making a variable change. This is particularly valuable when the equation already exhibits a structure that is similar to the standard Bernoulli form.

Transcript

00:01 We have given a logistic equation y dash of p is equal to r into y of p one minus y of p but m also we have given the value of r and m here also we have given the initial value y of u is equal to 2 into 10k power 7 kilo derive now as you know this is the logistic equation and the solution about this equation is why is equal to this one so first we'll convert the equation in this form then you'll write the directive of this differential equation so first to write here y -dash is equal to y -dash is equal to r into y into y now simplify this one then you get m but a y divided by m now we write here y -dash is equal to r into y into y minus y divided by m now divide both sides by y into m minus y then you'll get one by y into m minus y into m minus y into y dash is equal to r by m now compare the given equation here so here in the place of m here is m you are same thing but here is r but a m there is r but m and here is k so we can write k is equal to r but a m so k is equal to here r but a m and the value of r is zero five seven one divided by and the value of m is eight into ten k power 7 now we'll write the solution of this differential equation is equal to y is equal to a m into e k into k into mt divided by 1 plus a into e k power k m into t now plug in the value of m and k in this equation then we'll get value of m is 8 into 10k power 7a into the value of k is 0 .7 .1 divided by 8 into 10 k power 7 and the value of here m is 8 into 10 k power 7 divided by 1 plus a into e k power 0 .7 divided by 1 plus a into e k power value of k is 0 .71 divided by 8 10k power 7 into 8 into 10 k power 7 into t now cancel that this one from the numerator and denominator then we'll get y is equal to 8 into 10 k power 7 a e k power 0 .7 1 into t divided by 1 plus a into e k power 0 .7 1 into p now we solve the value of a using the initial value we have a given initial value y of 0 is equal to 2 into 10k power 7 which means at t is equal to 0 the value of y is 2 into 10k power 7 now this is the equation 1 8 into 10k power 7 a when you are we will plug at t is equal to 0 then you get e k power 0 as in another value of e k power 0 is 1 so here we'll write 1 single leave 1 plus a into 1 now multiply the do the cross multiplication here so multiply one this one then you'll get 8 into 10k power 7 into a here 2 into 10k power 7 plus 2 into 10k power 7 into a now subtract both sides 2 into 10 k power 7 then here will get 6 into 10 k power 7 a is equal to 2 into 10 k power 7 now divide both sides by 6 into 10k power 7 then we'll get the value of a is equal to 1 by 3 now plug in the value of a in equation one in this equation then we'll get y is equal to 8 10k power 7 into 1 by 3 into ek power 0 7 1 2 t divided by 1 plus 8 the value of age 1 by 3 into e k power 0 .7 1 2 1 .71 into t now multiplied by 3 in the numerator and denominator then we'll get here 8 into k power 7 e k power 0 .7 into 3 0 .7 1 .0 .7 .p...

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