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Consider the following differential equation, which is important in population genetics: $$a(x) g(x)-\frac{1}{2} \frac{d}{d x}[b(x) g(x)]=0$$ Here, \(b(x)>0\). (a) Define \(y=b(x) g(x)\), and show that \(y\) satisfies $$\frac{a(x)}{b(x)} y-\frac{1}{2} \frac{d y}{d x}=0.$$ (b) Separate variables in \((8.50)\), and show that if \(y>0\), then $$y=C \exp \left[2 \int \frac{a(x)}{b(x)} d x\right].$$

Short Answer

Expert verified
(a) Define \( y = b(x) g(x) \) and simplify to show \( \frac{a(x)}{b(x)} y - \frac{1}{2} \frac{dy}{dx} = 0 \). (b) Separate variables and integrate to show \( y = C \exp\left[2 \int \frac{a(x)}{b(x)} dx \right] \) if \( y > 0 \).

Step by step solution

01

Substitution

First, define the substitution as given in the problem: let \( y = b(x) g(x) \). The goal is to rewrite the original differential equation in terms of the new variable \( y \).
02

Differentiate the Substituted Function

To substitute into the original equation, calculate \( \frac{d}{dx}[y] = \frac{d}{dx}[b(x) g(x)] \). Using the product rule, \[ \frac{d}{dx}[b(x) g(x)] = b'(x) g(x) + b(x) g'(x) \].
03

Substitute into the Original Equation

Replace \( b(x)g(x) \) with \( y \) and \( \frac{d}{dx}[b(x)g(x)] \) with the derivative derived in Step 2 into the original equation:\[ a(x)g(x) - \frac{1}{2}(b'(x)g(x) + b(x)g'(x)) = 0 \].
04

Express \( g(x) \) in Terms of \( y \) and Simplify

Rewrite \( g(x) \) and \( g'(x) \) using \( y = b(x) g(x) \): \( g(x) = \frac{y}{b(x)} \) and \( \frac{d}{dx}[g(x)] = \frac{d}{dx}\left(\frac{y}{b(x)}\right) \). Using the quotient rule, \( g'(x) = \frac{1}{b(x)}\frac{dy}{dx} - \frac{y b'(x)}{b^2(x)} \). Substitute these into the equation.
05

Simplification to Target Equation

After substituting and simplifying, obtain the equation:\[ \frac{a(x)}{b(x)} y - \frac{1}{2} \frac{dy}{dx} = 0 \]. This satisfies the required form of differential equation.
06

Separating Variables

Rearrange the equation \( \frac{a(x)}{b(x)} y - \frac{1}{2} \frac{dy}{dx} = 0 \) to separate variables:\[ \frac{dy}{dx} = 2 \frac{a(x)}{b(x)} y \].
07

Integrate Both Sides

To solve the separable differential equation, integrate both sides:\[ \int \frac{1}{y} \, dy = \int 2 \frac{a(x)}{b(x)} \, dx \]. This gives:\[ \ln |y| = 2 \int \frac{a(x)}{b(x)} \, dx + C \].
08

Exponentiate to Solve for \( y \)

Exponentiating both sides obtains:\[ |y|= e^{2 \int \frac{a(x)}{b(x)} \, dx + C} \]. Simplifying, we find:\[ y = C e^{2 \int \frac{a(x)}{b(x)} \, dx} \]. Since \( y > 0 \), we can remove the absolute value, showing the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Genetics
Population genetics is a fascinating area that explores how genetic traits are inherited in populations over time. Differential equations, such as the one provided in the exercise, are essential tools in this field. They help model the change in genetic trait frequencies within a population.

Such equations often incorporate factors like natural selection, genetic drift, and mutation rates to predict future changes. The differential equation provided plays a crucial role in understanding how these variables interact and influence genetic variation in populations. By transforming the equation using substitutions, we can gain a clearer understanding of genetic dynamics. This transformation allows for more straightforward solutions using techniques like separation of variables.
Separation of Variables
The method known as 'separation of variables' is a common technique for solving differential equations, including those in population genetics. This process involves rearranging the equation so that each variable and its derivatives are on opposite sides of the equation.

In our exercise, by transforming the variables and separating them, we can simplify the differential equation from its initial complicated form into a solvable one.
  • Begin by isolating the derivative term on one side: \( \frac{dy}{dx} = 2 \frac{a(x)}{b(x)} y \)
  • Rearrange it to get terms involving \( y \) on one side and terms involving \( x \) on the other: \( \int \frac{1}{y} \, dy = \int 2 \frac{a(x)}{b(x)} \, dx \)

This clever manipulation opens the door to straightforward integration, leading to the final solution.
Integration in Calculus
Integration is a fundamental piece of calculus and is central to solving differential equations like the one in the exercise. When we perform the integration step after separating variables, we're essentially finding the antiderivative or the sum of the function over an interval.

For our separated equation:
  • On the left, we integrate \( \int \frac{1}{y} \, dy \) which results in \( \ln |y| \)
  • On the right, we integrate \( \int 2 \frac{a(x)}{b(x)} \, dx \) which remains as an indefinite integral \( 2 \int \frac{a(x)}{b(x)} \, dx \)

By solving these integrals, we move towards expressing \( y \) explicitly, helpful in interpreting genetic models. The final step involves exponentiating to solve for \( y \), revealing the population dynamics modeled by the equation.
Product Rule
The product rule is a crucial differentiation technique used in calculus, particularly when dealing with products of two functions. In the context of the original equation, it is used to differentiate \( b(x) g(x) \), rewriting it in terms of \( y \).

The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then their derivative is \( u'(x)v(x) + u(x)v'(x) \). Applying this:
  • For \( y = b(x) g(x) \), the derivative is \( b'(x) g(x) + b(x) g'(x) \)

This step is essential for correctly substituting back into the differential equation and further simplifying it.
Quotient Rule
This rule is essential when differentiating a ratio of two functions and was used to find \( g'(x) \) in the given solution. The quotient rule states that for functions \( u(x) \) and \( v(x) \), \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

In the context of this problem, after defining \( y = b(x) g(x) \), we need to express \( g(x) \) and differentiate it. By substitution and applying the quotient rule:
  • \( g(x) = \frac{y}{b(x)} \)
  • Its derivative is \( g'(x) = \frac{1}{b(x)}\frac{dy}{dx} - \frac{yb'(x)}{b^2(x)} \)

Understanding this differentiation step is key to transforming the original equation into a more solvable form, ultimately leading to understanding genetic change over time.

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Most popular questions from this chapter

Suppose that the size of a population, denoted by \(N(t)\), satisfies $$\frac{d N}{d t}=0.7 N\left(1-\frac{N}{35}\right)$$ (a) Determine all equilibria by solving \(d N / d t=0\). (b) Solve \((8.46)\) for (i) \(N(0)=10\), (ii) \(N(0)=35\), (iii) \(N(0)=50\), and (iv) \(N(0)=0\). Find \(\lim _{t \rightarrow \infty} N(t)\) for each of the four initial conditions. (c) Compare your answer in (a) with the limiting values you found in (b).

At the beginning of this section, we modified the exponentialgrowth equation to include oscillations in the per capita growth rate. Solve the differential equation we obtained, namely, $$ \frac{d N}{d t}=2(1+\sin (2 \pi t)) N(t) $$ with \(N(0)=5\)

Solve the given autonomous differential equations. \(\frac{d y}{d x}=3 y\), where \(y_{0}=2\) for \(x_{0}=0\)

Let \(N(t)\) denote the size of a population at time \(t .\) Assume that the population exhibits exponential growth. (a) If you plot \(\log N(t)\) versus \(t\), what kind of graph do you get? (b) Find a differential equation that describes the growth of this population and sketch possible solution curves.

Denote by \(p=p(t)\) the fraction of occupied patches in a metapopulation model, and assume that $$\frac{d p}{d t}=c p(1-p)-p^{2} \quad \text { for } t \geq 0$$ where \(c>0 .\) The term \(p^{2}\) describes the density-dependent extinction of patches; that is, the per-patch extinction rate is \(p\), and a fraction \(p\) of patches are occupied, resulting in an extinction rate of \(p^{2}\). The colonization of vacant patches is the same as in the Levins model. (a) Set \(g(p)=c p(1-p)-p^{2}\) and sketch the graph of \(g(p)\). (b) Find all equilibria of \((8.70)\) in \([0,1]\), and determine their stability. (c) Is there a nontrivial equilibrium when \(c>0 ?\) Contrast your findings with the corresponding results in the Levins model.

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