# 13.1: Linear and nonlinear differential equations (2023)

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##### learning goals
1. Identify the difference between infinite and density-dependent population growth. Explain the terms of the logistic equation in its original version, Equation (13.1), and in its modified version, Equation (13.3).
2. Give the definition of a linear differential equation.
3. Explain the law of mass action and derive simple differential equations for interacting species based on this law.

In the population growth model in Chapter 11, we find the differential equation

$\frac{d N}{d t}=k N, \nonumber$

##### Mastered material tests
1. What does analytic solution of a differential equation mean?
2. What other solutions are possible?
3. Give an example of a non-linear function $$f(y)$$. where $$N(t)$$ is the population size at time $$t$$ and $$k$$ is a constant per capita growth rate. We show that this differential equation has exponential solutions. This means that generically two behaviors are conserved: explosive growth if $$k>0$$ or extinction if $$k<0$$.

The case of $$k>0$$ is unrealistic, as real populations cannot grow explosively and exponentially indefinitely. Eventually, when space or resources are depleted, population growth slows and the population reaches a static level instead of expanding forever. This motivates us to revise our previous model to represent density-dependent growth.

## The logistic equation for population growth

As before, let $$N(t)$$ be the size of a population at time $$t$$. Consider the differential equation

$\frac{d N}{d t}=r N \frac{(K-N)}{K} \text {. } \label{Logistics}$

We call this differential equationlogistic equationThe logistic equation has a long history in modeling population growth in humans, microorganisms, and animals. Here the parameter $$r$$ is theintrinsic growth rateand $$K$$ is thebattery capacity. Both $$r, K$$ are assumed to be positive constants for a given population in a given neighborhood.

In the form written above, we could interpret the logistic equation as

$\frac{d N}{d t}=R(N) \cdot N \nonumber$

Wo

$R(N)=\left[r \frac{(K-N)}{K}\right] . \no number$

The term $$R(N)$$ is a function of $$N$$ substituting the constant growth rate $$k$$ (found in the unrealistic model of infinite population growth). $$R$$ is called the density-dependent growth rate.

(Video) Linear versus Nonlinear Differential Equations

## Linear versus non-linear

The logistic equation presents the first example of a nonlinear differential equation. We explain the distinction between linear and nonlinear differential equations and why it is important.

##### Definition: Linear Differential Equation

A first-order differential equation is called linear if it is a linear combination of terms of the form

$\frac{d y}{d t}, \quad y, \quad 1 \nonumber$

That is, it can be written in the form

$\alpha \frac{d y}{d t}+\beta y+\gamma=0 \nonumber$

where $$\alpha, \beta, \gamma$$ do not depend on $$y$$. Note that "first order" means that only the first derivative (or no derivative at all) can appear in the equation.

##### Mastered material tests
1. What happens if $$k=0$$ ? Explain under what conditions this can occur and what happens to the population $$N(t)$$ in this case.
##### Mastered material tests
1. Can the differential equation $$\frac{d y}{d t}=a-b y$$ be written in the form (13.2)? If so, what are the values ​​of $$\alpha, \beta, \gamma ?$$ So far, we've seen several of these examples with constant coefficients $$\alpha, \beta, \gamma$$. For example $$\alpha=1, \beta=-k$$ and $$\gamma=0$$ in the equation $$11.2$$, while $$\alpha=1, \gamma=-a$$ , and $$\beta=b$$ in equation (12.4). A differential equation that does not have this form is called nonlinear.
##### Example 13.1

(Linear vs. non-linear differential equations) Which of the following differential equations are linear and which are non-linear? (a) $$\frac{d y}{d t}=y^{2}$$ (b) $$\frac{d y}{d t}-y=5$$ (c) $$y \frac{d y {dt}=-1$$.

###### Solution

Each term of the form $$y^{2}, \sqrt{y}, 1 / y$$ etc. is non-linear in $$y$$. A product like $$y \frac{d y}{d t}$$ is also nonlinear in the independent variable. Therefore, equations (a), (c) are non-linear while (b) is linear.

The importance of distinguishing between linear and nonlinear differential equations is that nonlinearities make it much more difficult to find a systematic solution to the differential equation given by "analytical" methods. Most linear differential equations have solutions consisting of exponential functions or expressions containing such functions. This does not apply to nonlinear equations.

(Video) ODE | Linear versus nonlinear

However, as we will see shortly, geometric methods are very useful for understanding the behavior of such nonlinear differential equations.

## mass action law

Nonlinear terms in differential equations arise naturally in a variety of ways. A common source comes from describing interactions between individuals, as the following example shows.

In a chemical reaction, molecules of types $$A$$ and $$B$$ combine and react to form the product $$P$$. Let $$a(t), b(t)$$ be the concentrations of $$A$$ and $$B$$. These concentrations depend on time, as the chemical reaction consumes both species in the manufacture of the product.

The reaction only occurs when $$A$$ and $$B$$ molecules "collide" and stick together. The collisions happen randomly, but when the concentrations are higher, more collisions occur and the reaction is faster. If the concentration $$a$$ or $$b$$ is doubled, the reaction rate also doubles. But if $$a$$ and $$b$$ are doubled, then the reaction speed should be four times faster based on the higher probability of collisions between $$A$$ and $$B$$. The simplest assumption that captures this dependency is

$\text{Reaction rate} \proptoa \cdot b\nonumber$

o

$\text{Reaction speed} =k \cdot to \cdot b \nonumber$

where $$k$$ is a constant that represents the reactivity of molecules.

We can formally state this result, which is known asmass action lawAs follows:

$\frac{d y}{d t}=I . \none number$

When the chemical reaction occurs, the depletion of $$A$$ depends on the interactions of the pairs of molecules. According to the law of mass action, the reaction rate has the form $$k \cdot y \cdot y=k y^{2}$$, and as the concentration decreases, it appears in the DE with a sign of less. Therefore

$\frac{d y}{d t}=I-k y^{2} . \sem number$

This is a non-linear differential equation: it contains a term of the form $$y^{2}$$. $$\Diamond$$

##### Mastered material tests
1. For what values ​​of $$\alpha, \beta$$ and $$\gamma$$ can Example 13.1(b) be converted to form (13.2)?
(Video) First Order Linear Differential Equations
##### Mastered material tests
1. If the concentration of $$A$$ is tripled and that of $$B$$ is doubled, how much faster would we expect the rate of the reaction to be?
2. Why does the product $$a \cdot b$$ appear in the law of mass action and not the sum $$a+b$$? The Law of Mass Action: The rate of a chemical reaction involving two or more chemical species is proportional to the product of the concentrations of the given species.
##### Example 13.2

(Differential equation for interacting chemicals) Substance A is added to a 1 liter flask at a constant rate of 1 mol per hour. Pairs of molecules $$A$$ chemically interact to form a product $$P$$. Write a differential equation that tracks the concentration of A, denoted $$y(t)$$.

###### Solution

First, consider the case of no response. So adding $$A$$ to the reactor at a constant rate leads to a change in $$y(t)$$, described by the differential equation

##### Example $$\PageIndex{1}$$: reinterpreted logistic equation

Rewrite the logistic equation in the form

$\frac{d N}{d t}=r N-b N^{2} \nonumber$

(where $$b=r / K$$ is a positive quantity).

a) Interpret the meaning of this reformulated form of the equation by explaining what each of the terms on the right can represent.

b) Which of the two terms dominates for small or large population levels?

###### Solution

a) This form of the equation has a growth term $$r N$$ proportional to the size of the population, as found earlier with infinite population growth. But there is also a quadratic (non-linear) loss rate (note the minus sign) $$-b N^{2}$$. This term can describe interactions between individuals that result in death, e.g. through struggle or competition.

b) From familiarity with power functions (in this case, the functions of $$N$$ that form the two terms, $$r N$$ and $$b N^{2}$$ ), we can deduce the second , the quadratic term dominates for larger values ​​of $$N$$, and this means that in a populated population, the loss of individuals is greater than the reproduction rate.

##### Mastered material tests
1. In each of the examples $$13.2$$ and 13.3, clearly identify the constant quantities.
(Video) ❖ First Order Linear Differential Equations ❖

## Scaling the Logistic Equation

Consider the units involved in the logistic equation (13.1):

$\frac{d N}{d t}=r N \frac{(K-N)}{K} . \nhum number$

This equation has two parameters, $$r$$ and $$K$$. Since the units on each side of an equation must balance and be the same for added or subtracted terms, we can conclude that $$K$$ has the same units as $$N$$ and therefore is a population density. If $$N=K$$, the population growth rate is zero $$(d N / d t=0)$$.

It turns out that we can understand the behavior of the logistic equation by converting it to a "generic" form that does not depend on the constant $$K$$. We do this by transforming variables, which boils down to choosing a convenient way to measure population size.

##### Example $$\PageIndex{4}$$: Change the scale

Define a new variable

$y(t)=\frac{N(t)}{K}, \nonúmero$

with $$N(t)$$ and $$K$$ as in the logistic equation. So $$N(t)=Ky(t)$$.

1. Interpret what the transformed variable y represents.
2. Rewrite the logistic equation in terms of this variable.
###### Solution

a) The variable $$y(t)$$ represents a scaled version of the population density. Rather than measuring population in some arbitrary units, such as the number of individuals per acre or the number of bacteria per $$\mathrm{ml}-y(t)$$, population is measured in "multiples of carrying capacity ". .

For example, if the environment can support 1000 aphids per plant (i.e. $$K=1000$$ individuals per plant) and the current population size on a given plant is $$N=950$$, then the aphid value scaled variable is \ (y=950 / 1000=\) 0.95. We would say "the aphid population is at $$95\%$$ of its carrying capacity in the plant".

b) Since $$K$$ is assumed to be constant, it follows

$N(t)=K y(t) \quad \Rightarrow \quad \frac{d N}{d t}=K \frac{d y}{d t} . \wk number$

We can simplify the logistic equation as follows:

(Video) Identifying Linear Ordinary Differential Equations

\begin{aligned} \frac{d N}{d t}=r N \frac{(K-N)}{K}, & \Rightarrow K \frac{d y}{d t}=r(K y) \frac {(K-K y)}{K}, \\ & \Rightarrow \frac{d y}{d t}=r y(1-y) . \end{alinhado} \número

Equation (13.3) "looks simpler" than Equation (13.1) because it only depends on one parameter, $$r$$. Furthermore, by understanding this equation and transforming it back to the original logistic in terms of $$N(t)=K y(t)$$, we can interpret the results of the original model. While it doesn't go further with the variable transformation for the time being, it turns out that scaled logistics can be further reduced to an equation where $$r=1$$ by "scaling units of time".

##### Exercise $$\PageIndex{1}$$
1. Suppose an environment can support 2000 aphids per plant and the current population size of a given plant is 1700. What are $$K,N$$ and $$y$$ based on this information?
2. This population is at what percentage of its carrying capacity?

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