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Chapter 11: Problem 34

(a) Solve the differential equation $$ \frac{d P}{d t}=0.2 P-10 $$ Write the solution \(P\) as an explicit function of \(t\) (b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane. $$ P(0)=40, \quad P(0)=50, \quad P(0)=60 $$

Short Answer

Expert verified
\( P(t) = -10 e^{0.2t} + 50 \) for \( P(0)=40 \); \( P(t) = 50 \) for \( P(0)=50 \); \( P(t) = 10 e^{0.2t} + 50 \) for \( P(0)=60 \).

Step by step solution

01

Write the Differential Equation

Given the differential equation \( \frac{dP}{dt} = 0.2P - 10 \). This is a first-order linear differential equation.
02

Rewrite the Equation for Separation of Variables

Rearrange the equation to separate variables: \( \frac{dP}{0.2P - 10} = dt \).
03

Integrate Both Sides

Integrate both sides of the equation. The left side becomes \( \int \frac{1}{0.2P - 10} dP \), and the right side becomes \( \int dt \).
04

Solve the Integrals

The integral of the left side is \( \frac{1}{0.2} \ln|0.2P - 10| + C_1 \), and the integral of the right side is \( t + C_2 \). Combine these to get \( \frac{1}{0.2} \ln|0.2P - 10| = t + C \).
05

Solve for P

Multiply both sides by 0.2 and exponentiate to solve for \(P\):\[ 0.2P - 10 = Ce^{0.2t} \]where \( C \) is a constant. Solve for \(P\):\[ P(t) = \frac{Ce^{0.2t} + 10}{0.2} \].
06

Substitute Initial Condition \(P(0)=40\)

Substitute \( t = 0 \) and \( P = 40 \) into the equation: \[ 40 = \frac{C + 10}{0.2} \]Calculate \( C \): \[ 40 \times 0.2 = C + 10 \Rightarrow C = 40 \times 0.2 - 10 = -2 \].
07

Substitute Initial Condition \(P(0)=50\)

Substitute \( t = 0 \) and \( P = 50 \) into the equation: \[ 50 = \frac{C + 10}{0.2} \]Calculate \( C \): \[ 50 \times 0.2 = C + 10 \Rightarrow C = 50 \times 0.2 - 10 = 0 \].
08

Substitute Initial Condition \(P(0)=60\)

Substitute \( t = 0 \) and \( P = 60 \) into the equation: \[ 60 = \frac{C + 10}{0.2} \]Calculate \( C \): \[ 60 \times 0.2 = C + 10 \Rightarrow C = 60 \times 0.2 - 10 = 2 \].

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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 method employed to solve first-order differential equations. It consists of organizing the equation so that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the opposite side. This is essential because it sets the stage for integrating each side.
In our example, the differential equation given is \( \frac{dP}{dt} = 0.2P - 10 \). The goal is to place every term with \(P\) together and those with \(t\) on the other side. This results in \( \frac{dP}{0.2P - 10} = dt \). Now each side of the equation can be independently integrated.
  • This step allows for easier handling of complex relationships between two variables.
  • It transforms a differential equation into two separate integrals, which are typically simpler to solve.
Initial Conditions
Initial conditions are given values that provide specific solutions to differential equations. Without them, the solution is general and includes an arbitrary constant.
When solving differential equations, initial conditions help determine the particular solution from the general form by setting the unknown constants.
In the exercise, different initial conditions are explored: \( P(0) = 40 \), \( P(0) = 50 \), and \( P(0) = 60 \). By substituting each into the general solution, different constants are calculated, giving unique solutions that map onto graphs.
  • Initial conditions make solutions applicable to real-world scenarios, where exact starting values are known.
  • They are used to create accurate models in a wide variety of scientific applications.
Integration
Integration is a fundamental mathematical process which is often used for solving differential equations. It involves finding a function whose derivative matches the original function's differential form.
In separation of variables, after arranging the terms, each side of the equation is integrated. For our example:
The left side becomes \( \int \frac{1}{0.2P - 10} \, dP \), simplifying to \( \frac{1}{0.2} \ln|0.2P - 10| \).
The right side simplifies to \( \int dt \), returning \( t + C_2 \).
  • Integration transforms complicated expressions into usable solutions.
  • It is crucial for determining the general form of the solution, which can later be refined with initial conditions.
First-Order Linear Differential Equations
First-order linear differential equations are differential equations of the form \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( y \) is dependent on a single independent variable. They are central in modelling real-time dynamic systems.
In our problem, the equation \( \frac{dP}{dt} = 0.2P - 10 \) follows this form and can be tackled using techniques like separation of variables.
Such equations often describe exponential growth or decay, processes found in economics, biology, physics, and more.
  • Understanding their structure helps students establish strategies for finding solutions.
  • Recognize them by their distinctive format, facilitating a wide range of applicable solution techniques.

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Most popular questions from this chapter

A differential equation for a quantity that is increasing and grows fastest when the quantity is small and grows more slowly as the quantity gets larger.

Find the general solution to the given differential equation. $$y^{\prime \prime}+6 y^{\prime}+8 y=0$$

The system of differential equations models the interaction of two populations \(x\) and \(y\) (a) What kinds of interaction (symbiosis, \(^{34}\) competition, predator-prey) do the equations describe? (b) What happens in the long run? Your answer may depend on the initial population. Draw a slope field. $$\begin{aligned} &\frac{1}{x} \frac{d x}{d t}=y-1\\\ &\frac{1}{y} \frac{d y}{d t}=x-1 \end{aligned}$$

\(p\) and \(q\) are the number of individuals in two interacting populations with \(p, q>0\) satisfying the system of equations $$ \begin{array}{l} \frac{1}{p} \frac{d p}{d t}=0.01 q-0.3 \\ \frac{1}{q} \frac{d q}{d t}=0.02 p-0.2 \end{array} $$ Give a description of how these populations interact.

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester's analysis to this battle, with \(x\) representing the number of US troops and \(y\) the number of Japanese troops, it has been estimated \(^{35}\) that \(a=0.05\) and \(b=0.01\) (a) Using these values for \(a\) and \(b\) and ignoring reinforcements, write a differential equation involving \(d y / d x\) and sketch its slope field. (b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500 draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted?
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