/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 \(P(t)=C e^{k t}+\frac{E}{k}\), ... [FREE SOLUTION] | 91影视

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Chapter 31: Problem 18

\(P(t)=C e^{k t}+\frac{E}{k}\), where \(C\) is a constant, is the general solution to the differential equation \(\frac{d P}{d t}=k P-E .\) Below is the slope eld for \(\frac{d P}{d t}=2 P-6\). (a) i. Find the particular solution that corresponds to the initial condition \(P(0)=2\). ii. Sketch the solution curve through \((0,2)\). (b) i. Find the particular solution that corresponds to the initial condition \(P(0)=3\). ii. Sketch the solution curve through \((0,3)\). (c) i. Find the particular solution that corresponds to the initial condition \(P(0)=4\). ii. Sketch the solution curve through \((0,4)\).

Short Answer

Expert verified
The particular solutions for the given initial conditions are \(P(t)=-e^{2t}+3\), \(P(t)=3\), \(P(t)=e^{2t}+3\). Sketching these solution curves is tiest.

Step by step solution

01

Solving for the Particular Solution when \(P(0)=2\)

The given initial condition is \(P(0)=2\), and the general solution to the differential equation is given by \(P(t)=C e^{kt}+E/k\). Plugging these values into the general solution gives the equation \(2=C e^{k(0)}+E/k\), which simplifies to \(2=C +E/k\). However, from the differential equation we know that \(k=2\) and \(E=6\), hence \(2=C+6/2= C+3\). Solving for \(C\) gives \(C=-1\) the particular solution for these conditions is thereby \(P(t)=-e^{2t}+3\)
02

Solving for the Particular Solution when \(P(0)=3\)

Given the initial condition \(P(0)=3\), we can plug these values into the general solution, resulting in the equation \(3=C e^{k(0)}+E/k\), which simplifies to \(3=C +E/k\). However, from the differential equation we know that \(k=2\) and \(E=6\), hence \(3=C+6/2= C+3\). Solving for \(C\) gives \(C=0\) the particular solution for these conditions is thereby \(P(t)=3\)
03

Solving for the Particular Solution when \(P(0)=4\)

Finally, for the initial condition \(P(0)=4\), plug these values into the general solution, which gives us the equation \(4=C e^{k(0)}+E/k\), which simplifies to \(4=C +E/k\). However, from the differential equation we know that \(k=2\) and \(E=6\), hence \(4=C+6/2= C+3\). Solving for \(C\) gives \(C=1\) the particular solution for these conditions is thereby \(P(t)=e^{2t}+3\)

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Key Concepts

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

Particular Solution
The term particular solution refers to a specific solution of a differential equation that not only satisfies the differential equation itself but also meets a specified set of initial conditions. The process of finding a particular solution begins with the general solution, which includes arbitrary constants, and then uses initial conditions to determine the exact values of those constants.

In our example, we work with the general solution of the differential equation \( \frac{d P}{d t} = k P - E \) in the form \( P(t) = Ce^{kt} + \frac{E}{k} \). We find the particular solutions for different initial conditions, such as \( P(0) = 2 \), \( P(0) = 3 \), and \( P(0) = 4 \). By plugging the initial conditions into the general formula, we can solve for the constant \( C \) and obtain the precise expression for each scenario.
Initial Conditions
In the context of differential equations, initial conditions provide a starting point for the system described by the equation. They are values of the function and/or its derivatives at a specific point, and they are essential for finding a particular solution that uniquely satisfies both the differential equation and the given conditions.

In the given exercise, we see initial conditions such as \( P(0) = 2 \) representing the value of \( P(t) \) at \( t = 0 \). The inclusion of initial conditions allows us to narrow down the infinite set of possible solutions from the general form to the one particular solution that fits the given scenario. For each different initial value provided, the corresponding value of \( C \) is calculated to achieve the particular solutions \( P(t) = -e^{2t} + 3 \) for \( P(0) = 2 \) and so on. This is a critical step in understanding the system鈥檚 behavior at a given starting point.
Slope Field
A slope field, also known as a direction field, is a graphical representation of the solutions to a first-order differential equation. It consists of small line segments or arrows drawn at various points in the plane, where the slope of these segments represents the value of the derivative of the function at those points.

To visualize a slope field, one could imagine being on a hiking trail where the slope field provides hints at each point on how the trail goes uphill or downhill. Similarly, in differential equations, the slope field illustrates the trend of solutions without actually solving the equation.

In the exercise, the differential equation is represented by the slope field for \( \frac{d P}{d t} = 2P - 6 \). By sketching solution curves through the given points, we can see how different initial conditions influence the path of solutions on the slope field. Each curve represents a particular solution and will adhere to the implied slopes given by the field. This offers us not only a solution but also a visually intuitive understanding of how different solutions behave within the system modeled by the differential equation.

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Most popular questions from this chapter

For each differential equation below, sketch the slope eld and nd the general solution. (a) \(\frac{d y}{d t}=-y\) (b) \(\frac{d y}{d t}=-t\) (c) \(\frac{d y}{d t}=e^{-t}\)

Suppose we modify the Volterra predator-prey equations to reflect competition among prey for limited resources and competition among predators for limited resources. The equations would be of the form $$ \left\\{\begin{array}{l} \frac{d x}{d t}=k_{1} x-k_{2} x^{2}-k_{3} x y \\ \frac{d y}{d t}=-k_{4} y-k_{5} y^{2}+k_{6} x y \end{array}\right. $$ where \(k_{1}, k_{2}, \ldots, k_{6}\) are positive constants. Consider the system $$ \left\\{\begin{array}{l} \frac{d x}{d t}=x(1-0.5 x-y) \\ \frac{d y}{d t}=y(-1-0.5 y+x) \end{array}\right. $$ (a) Find the equilibrium points. (b) Do a qualitative phase-plane analysis. (In fact, solution trajectories will spiral in toward the non-trivial equilibrium point.)

The population in a certain country grows at a rate proportional to the population at time \(t\), with a proportionality constant of \(0.03 .\) Due to political turmoil, people are leaving the country at a constant rate of 6000 people per year. Assume that there is no immigration into the country. Let \(P=P(t)\) denote the population at time \(t\). (a) Write a differential equation re ecting the situation. (b) Solve the differential equation for \(P(t)\) given the information that at time \(t=0\) there are 3 million people in the country. In other words, nd \(P(t)\), the number of people in the country at time \(t\).

There are many places in the world where populations are changing and immigration and/or emigration play a big role. People may move to nd food, or to nd jobs, or to ee political or religious persecution. Pick a situation that interests you. You could look at the number of Tibetans in Tibet, or the number of Tibetans in India, or the number of lions in the Serengeti, or the number of tourists in Nepal. Get some data and try to model the population dynamics using a differential equation. What simplifying assumptions have you made?

A lake contains \(10^{10}\) liters of water. Acid rain containing \(0.02\) milligrams of pollutant per liter of rain falls into the lake at a rate of \(10^{3}\) liters per week. An outlet stream drains away \(10^{3}\) liters of water per week. Assume that the pollutant is always evenly distributed throughout the lake, so the runoff into the stream has the same concentration of pollutant as the lake as a whole. The volume of the lake stays constant at \(10^{10}\) liters because the water lost from the runoff balances exactly the water gained from the rain. (a) Write a differential equation whose solution is \(P(t)\), the number of milligrams of pollutant in the lake as a function of \(t\) measured in weeks. (b) Find any equilibrium solutions. (c) Sketch some representative solution curves. (d) How would you alter the differential equation if there was a dry spell and rain was falling into the lake at a rate of only \(10^{2}\) liters per week.
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