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Chapter 11: Problem 8

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d I}{d x}=0.2 I, \quad I=6 \text { where } x=-1$$

Short Answer

Expert verified
The solution is \( I = 6e^{0.2(x+1)} \).

Step by step solution

01

Identify the Equation Type

The given differential equation is \( \frac{dI}{dx} = 0.2I \). This is a first-order linear differential equation that can be solved using the method of separation of variables.
02

Separate Variables

Rewrite the equation by separating variables to get \( \frac{1}{I} \frac{dI}{dx} = 0.2 \). Next, express it as \( \frac{dI}{I} = 0.2 \, dx \).
03

Integrate Both Sides

Integrate both sides of the equation. The left side, \( \int \frac{dI}{I} \), results in \( \ln |I| \), and the right side, \( \int 0.2 \, dx \), gives \( 0.2x + C \). So, the equation becomes \( \ln |I| = 0.2x + C \).
04

Solve for I

Exponentiate both sides to solve for \( I \). This gives \( |I| = e^{0.2x + C} \). Let \( K = e^C \), a constant, then \( I = Ke^{0.2x} \).
05

Apply Initial Condition

Use the initial condition \( I = 6 \) when \( x = -1 \) to find \( K \). Substitute to get \( 6 = Ke^{-0.2} \). Thus, \( K = 6e^{0.2} \).
06

Write the General Solution

Substitute the value of \( K \) back into the equation to find the solution: \( I = 6e^{0.2}e^{0.2x} \), which simplifies to \( I = 6e^{0.2(x+1)} \).

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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 technique used to solve differential equations. It involves rewriting the equation so that each variable appears on a different side of the equation. This process simplifies the task of integrating the equation as you can focus on one variable at a time.

In our problem, we started with the differential equation \( \frac{dI}{dx} = 0.2I \). To use separation of variables effectively, we reorganized the equation to isolate \( I \) and its differential \( dI \) on one side, and \( x \) and its differential \( dx \) on the other side.
  • This step led us to \( \frac{1}{I} \frac{dI}{dx} = 0.2 \)
  • Further simplification gave us \( \frac{dI}{I} = 0.2 \, dx \)
Once variables are separated, each side of the equation can be integrated independently, rendering the process much more intuitive.
Initial Conditions
Initial conditions are essential in solving differential equations, particularly when you're looking to find a specific solution rather than a general one. They are given values for the function or its derivatives at particular points that are used to determine the constants that appear after integration.

In the exercise we solved, the initial condition provided was \( I = 6 \) when \( x = -1 \). After integrating the separated equation, we were left with \( \ln |I| = 0.2x + C \), which simplified to a general form: \( I = Ke^{0.2x} \).
  • The initial condition \( I( -1 ) = 6 \) allows us to find the constant \( K \).
  • Substitute \( x = -1 \) and \( I = 6 \) into the equation, resulting in: \( 6 = Ke^{-0.2} \).
  • Solving gives \( K = 6e^{0.2} \).
Using the initial condition ensures the solution reflects the specific behavior of the system at known points.
First-Order Linear Differential Equations
First-order linear differential equations are a common type of differential equation. They typically take the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). These equations can model a variety of real-world phenomena such as population growth, radioactive decay, or fluid flow.

In our example, \( \frac{dI}{dx} = 0.2I \) fits this form with \( P(x) = -0.2 \) and \( Q(x) = 0 \).
  • First-order equations can often be solved using separation of variables when they can be manipulated to isolate the variables on different sides of the equation.
  • This method leads to solutions of exponential form when the function is proportional to its rate of change, reflecting growth or decay processes.
Understanding these equations is crucial as they frequently appear in mathematics, physics, engineering, and other fields.

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Most popular questions from this chapter

An item is initially sold at a price of \(p\text{dollars}\) per unit. Over time, market forces push the price toward the equilibrium price, \(p \text{dollars}^{*},\) at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, \(p\text{dollars},\) is proportional to the difference between the market price and the equilibrium price. (a) Write a differential equation for \(p\) as a function of \(t\) (b) Solve for \(p\) (c) Sketch solutions for various different initial prices, both above and below the equilibrium price. (d) What happens to \(p\) as \(t \rightarrow \infty ?\)

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 $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Are the statements in Problems \(59-62\) true or false? Give an explanation for your answer. The differential equation \(d y / d x-x y=x\) can be solved by separation of variables.

Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. If \(f(1)=5,\) then (1,5) could be a critical point of \(f\)

Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. $$f^{\prime}(x)=2 x-f(x)$$
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