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Chapter 2: Problem 40

The differential equation \(d P / d t=\) \((k \cos t) P\), where \(k\) is a positive constant, is a mathematical model for a population \(P(t)\) that undergoes yearly seasonal fluctuations. Solve the equation subject to \(P(0)=P_{0}\). Use a graphing utility to obtain the graph of the solution for different choices of \(P_{0}\)

Short Answer

Expert verified
The solution is \( P(t) = P_0 e^{k \sin t} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{dP}{dt} = (k \cos t) P \). This is a first-order linear differential equation. It can be solved using the method of separation of variables.
02

Separate the Variables

Rearrange the equation to separate variables such that terms involving \( P \) are on one side and terms involving \( t \) are on the other:\[\frac{1}{P} \frac{dP}{dt} = k \cos t\].
03

Integrate Both Sides

Integrate both sides of the separated equation:\[\int \frac{1}{P} \, dP = \int k \cos t \, dt\].The left side simplifies to \( \ln |P| \) and the right side integrates to \( k \sin t \) (since the integral of \( \cos t \) is \( \sin t \)) plus a constant of integration.
04

Solve for P

After integrating, we have the equation:\[\ln |P| = k \sin t + C\].To solve for \( P \), exponentiate both sides to eliminate the natural logarithm:\[P = e^{k \sin t + C} = e^C e^{k \sin t}\].Let \( C_1 = e^C \), hence \( P = C_1 e^{k \sin t} \).
05

Apply Initial Condition

Use the initial condition \( P(0) = P_0 \) to find \( C_1 \):\[P_0 = C_1 e^{k \sin 0} \Rightarrow P_0 = C_1 e^0 = C_1\].Thus, \( C_1 = P_0 \), and the solution becomes \( P(t) = P_0 e^{k \sin t} \).
06

Graph the Solution

Using a graphing utility, plot the function \( P(t) = P_0 e^{k \sin t} \) for various initial population sizes \( P_0 \). Observe how changing \( P_0 \) affects the population growth with respect to seasonal variations modeled by \( k \cos t \).

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Key Concepts

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

Separation of Variables
Separation of variables is a fundamental method for solving differential equations, particularly useful in first-order types. It involves rearranging the equation so that each variable appears on a different side of the equation. For our population model, the differential equation given is \[\frac{dP}{dt} = (k \cos t) P\]
To separate the variables in this equation, we bring all terms involving \(P\) to one side, and terms involving \(t\) to the other. This results in the form:\[\frac{1}{P} \frac{dP}{dt} = k \cos t\]
Now, each variable is isolated, allowing us to integrate both sides concerning their respective variables. This method effectively breaks down a complex relationship into simpler, integrable parts.
First-Order Linear Differential Equations
First-order linear differential equations, like the one presented in our problem, are equations that involve the first derivative of a function. These equations take the standard form:\[\frac{dy}{dx} + P(x)y = Q(x)\]
In our specific example, the equation is:\[\frac{dP}{dt} = (k \cos t) P\]
Here, \(P\) and \(Q\) from the general form are just constants. First-order linear differential equations are renowned for their straightforward solutions compared to higher-order equations. The ability to separate this type of equation showcases why the two-step process of separation and integration is essential in finding practical solutions.
Initial Conditions
Initial conditions are crucial in solving differential equations as they provide specific solutions tailored to particular scenarios. They tell us where to start and ensure the solution fits the context of the problem.
In this exercise, the initial condition given is:\[P(0) = P_0\]
By applying this condition after integration, we can identify the constant of integration's actual value. After separating variables and integrating, we had:\[\ln |P| = k \sin t + C\]
Once we applied the initial condition, \(P(0) = P_0\), it allowed us to determine \(C_1\) in the expression:\[P(t) = C_1 e^{k \sin t}\]
The initial condition ensures the derived function starts at the measured initial population \(P_0\), providing solutions that can be both calculated and interpreted within the problem's real-world context.

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Most popular questions from this chapter

A mathematical model for the rate at which a drug disseminates into the bloodstream is given by \(d x / d t=r-k x\), where \(r\) and \(k\) are positive constants. The function \(x(t)\) describes the concentration of the drug in the bloodstream at time \(t\). (a) Since the \(\mathrm{DE}\) is autonomous, use the phase portrait concept of Section \(2.1\) to find the limiting value of \(x(t)\) as \(t \rightarrow \infty\) (b) Solve the \(\mathrm{DE}\) subject to \(x(0)=0\). Sketch the graph of \(x(t)\) and verify your prediction in part (a). At what time is the concentration one- half this limiting value?

Solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=1+e^{y-x+5} $$

(a) Census data for the United States between 1790 and 1950 is given in the following table. Construct alogistic population model using the data from 1790,1850 , and 1910 . $$ \begin{array}{lr} \hline \text { Year } & \text { Population (in millions) } \\ \hline 1790 & 3.929 \\ 1800 & 5.308 \\ 1810 & 7.240 \\ 1820 & 9.638 \\ 1830 & 12.866 \\ 1840 & 17.069 \\ 1850 & 23.192 \\ 1860 & 31.433 \\ 1870 & 38.558 \\ 1880 & 50.156 \\ 1890 & 62.948 \\ 1900 & 75.996 \\ 1910 & 91.972 \\ 1920 & 105.711 \\ 1930 & 122.775 \\ 1940 & 131.669 \\ 1950 & 150.697 \\ \hline \end{array} $$ (b) Construct a table comparing actual census population with the population predicted by the model in part (a). Compute the error and the percentage error for each entry pair.

A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2}, $$ where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\) (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the \(\mathrm{DE}\) in Problem 39 in Exercises \(2.1\). (c) If distance \(s\), measured from the point where the mass was released above ground, is related to velocity \(v\) by \(d s / d t=v(t)\), find an explicit expression for \(s(t)\) if \(s(0)=0\).

The number \(N(t)\) of people in a community who are exposed to a particularadvertisement is governed by the logisticequation. Initially \(N(0)=500\), and it is observed that \(N(1)=1000\). Solve for \(N(t)\) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000 .
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