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Chapter 9: Problem 1

Solve the differential equations in Exercises \(1-14\) $$x \frac{d y}{d x}+y=e^{x}, \quad x>0$$

Short Answer

Expert verified
The solution is \(y = \frac{e^x + C}{x}\), where \(C\) is an integration constant.

Step by step solution

01

Identify the Equation

The given differential equation is a first-order linear differential equation: \(x \frac{dy}{dx} + y = e^x\). We will solve it using the method of integrating factors.
02

Rewrite in Standard Form

Rewrite the equation in standard form: \(\frac{dy}{dx} + \frac{1}{x} y = \frac{e^x}{x}\). Now it is in the form \(\frac{dy}{dx} + P(x) y = Q(x)\), where \(P(x) = \frac{1}{x}\) and \(Q(x) = \frac{e^x}{x}\).
03

Determine the Integrating Factor

The integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx}\). Here, \(P(x) = \frac{1}{x}\), so \(\mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = x\).
04

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \(x\):\[x\left( \frac{dy}{dx} + \frac{1}{x} y \right) = x \cdot \frac{e^x}{x}\]This simplifies to \(x \frac{dy}{dx} + y = e^x\). The left side is now the derivative of \(x \cdot y\).
05

Integrate Both Sides

Recognize that the left-hand side can be written as a derivative: \(\frac{d}{dx}(xy)\). Thus,\[\frac{d}{dx}(xy) = e^x\]Integrating both sides with respect to \(x\) gives:\[xy = \int e^x \, dx = e^x + C\] where \(C\) is the constant of integration.
06

Solve for y

Solve for \(y\) by dividing both sides by \(x\):\[y = \frac{e^x + C}{x}\] This is the general solution of the differential equation.

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Key Concepts

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

Integrating Factors
An integrating factor is a crucial tool when solving first-order linear differential equations. It's a function used to simplify the differential equation, making it easier to solve. In general, the integrating factor is denoted as \( \mu(x) \) and is calculated using the expression:
  • \( \mu(x) = e^{\int P(x) \, dx} \)
For our exercise, where \( P(x) = \frac{1}{x} \), the integral becomes \( \int \frac{1}{x} \, dx = \ln|x| \). Substituting this into the expression gives us:
  • \( \mu(x) = e^{\ln|x|} = x \)
By multiplying the entire equation by this factor, we transform the initial equation into a form where the left-hand side becomes a derivative of a product, simplifying the process of finding the solution.
Standard Form
To effectively apply the method of integrating factors, the differential equation should be in what is called standard form. This form is:
  • \( \frac{dy}{dx} + P(x)y = Q(x) \)
Before the step involving integrating factors, our differential equation needs to be converted into this standard form. In this exercise:
  • The original equation was \( x \frac{dy}{dx} + y = e^x \)
Dividing through by \( x \) put it in the form:
  • \( \frac{dy}{dx} + \frac{1}{x}y = \frac{e^x}{x} \)
Here, \( P(x) = \frac{1}{x} \) and \( Q(x) = \frac{e^x}{x} \). This standardization is key to identify our \( P(x) \) needed for calculating the integrating factor.
Differentiation
Differentiation is a fundamental concept used throughout calculus and differential equations. In our context, it helps us recognize and transform parts of the equation. An example from our exercise is the phrase "the left side is now the derivative of \( xy \)."The expression \( \frac{d}{dx}(xy) \) is achieved by noting that multiplying through by our integrating factor (\( x \)) results in:
  • \( x \frac{dy}{dx} + y = e^x \)
Recognizing that the left-hand side represents the derivative of the product \( xy \) ensures that we keep the equation balanced and prepared for the final solution methods. Differentiation here helps transition the expression into a more manageable form for integration.
Integration
Integration is the process of finding an integral, most commonly used to reverse the process of differentiation. It is vital for solving differential equations by allowing us to accumulate functions that have been differentiated. In this exercise, after rewriting and differentiating, integration is used to solve:
  • \( \frac{d}{dx}(xy) = e^x \)
Integrate both sides:
  • \( \int \frac{d}{dx}(xy) \, dx = \int e^x \, dx \)
This means that:
  • \( xy = e^x + C \)
Here, \( e^x \) represents the antiderivative of itself. The constant \( C \) accounts for any constant offset in the indefinite integration, capturing all possible scenarios of the interpreted function.
General Solution of Differential Equations
The general solution of a differential equation encompasses all potential solutions, obtained by solving the equation in the most complete form. For first-order linear differential equations like ours, after successfully applying differentiation and integration, the solution emerges cleanly. Having arrived at:
  • \( xy = e^x + C \)
We then isolate \( y \):
  • \( y = \frac{e^x + C}{x} \)
This function describes the overall behavior of \( y \) relative to \( x \), incorporating the constant \( C \) that represents infinite family members of possible solutions. Each particular value of \( C \) corresponds to a specific initial condition or configuration for real-world problems.

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Most popular questions from this chapter

Use a CAS to find the solutions of \(y^{\prime}+y=f(x)\) subject to the initial condition \(y(0)=0,\) if \(f(x)\) is a. 2\(x \quad\) b. \(\sin 2 x\) \(\quad\) c. 3\(e^{x / 2} \quad\) d. 2\(e^{-x / 2} \cos 2 x\) Graph all four solutions over the interval \(-2 \leq x \leq 6\) to compare the results.

Consider another competitive-hunter model defined by $$ \begin{aligned} \frac{d x}{d t} &=a\left(1-\frac{x}{k_{1}}\right) x-b x y \\\ \frac{d y}{d t} &=m\left(1-\frac{y}{k_{2}}\right) y-n x y \end{aligned} $$ where \(x\) and \(y\) represent trout and bass populations, respectively. $$ \begin{array}{l}{\text { a. What assumptions are implicitly being made about the }} \\ {\text { growth of trout and bass in the absence of competition? }} \\\ {\text { b. Interpret the constants } a, b, m, n, k_{1}, \text { and } k_{2} \text { in terms of the }} \\ {\text { physical problem. }}\\\\{\text { c. Perform a graphical analysis: }} \\ {\text { i) Find the possible equilibrium levels. }} \\ {\text { ii) Determine whether coexistence is possible. }} \\ {\text { iii) Pick several typical starting points and sketch typical }} \\ {\text { trajectories in the phase plane. }} \\ {\text { iv) Interpret the outcomes predicted by your graphical }} \\ {\text { analysis in terms of the constants } a, b, m, n, k_{1}, \text { and } k_{2} \text { . }}\end{array} $$ Note: When you get to part (ii), you should realize that five cases exist. You will need to analyze all five cases.

In 1925 Lotka and Volterra introduced the predator-prey equations, a system of equations that models the populations of two species, one of which preys on the other. Let \(x(t)\) represent the number of rabbits living in a region at time \(t,\) and \(y(t)\) the number of foxes in the same region. As time passes, the number of rabbits increases at a rate proportional to their population, and decreases at a rate proportional to the number of encounters between rabbits and foxes. The foxes, which compete for food, increase in number at a rate proportional to the number of encounters with rabbits but decrease at a rate proportional to the number of foxes. The number of encounters between rabbits and foxes is assumed to be proportional to the product of the two populations. These assumptions lead to the autonomous system $$ \begin{array}{l}{\frac{d x}{d t}=(a-b y) x} \\ {\frac{d y}{d t}=(-c+d x) y}\end{array} $$ where \(a, b, c, d\) are positive constants. The values of these constants vary according to the specific situation being modeled. We can study the nature of the population changes without setting these constants to specific values. What happens to the rabbit population if there are no foxes present?

Solve the differential equations in Exercises \(1-14\) $$(1+x) y^{\prime}+y=\sqrt{x}$$

Use the Euler method with \(d x=0.2\) to estimate \(y(1)\) if \(y^{\prime}=y\) and \(y(0)=1 .\) What is the exact value of \(y(1) ?\)
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