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\(5-14\) Solve the differential equation. $$y^{\prime}+2 y=2 e^{x}$$

Short Answer

Expert verified
The solution is \( y = \frac{2}{3}e^{x} + Ce^{-2x} \).

Step by step solution

01

Identify the Equation Type

The given equation is \( y' + 2y = 2e^x \). It is a linear first-order ordinary differential equation.
02

Write in Standard Form

The standard form for a linear differential equation is \( y' + P(x)y = Q(x) \). Here, \( P(x) = 2 \) and \( Q(x) = 2e^x \). The equation is already in standard form.
03

Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x)\, dx} = e^{\int 2 \, dx} = e^{2x} \).
04

Multiply by the Integrating Factor

Multiply every term in the equation by the integrating factor \( e^{2x} \). The equation becomes \( e^{2x}y' + 2e^{2x}y = 2e^{3x} \).
05

Solve by Integration

After multiplying by the integrating factor, the left side of the equation becomes the derivative of \( e^{2x}y \). Thus, \( \frac{d}{dx}(e^{2x}y) = 2e^{3x} \). Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{2x}y) \, dx = \int 2e^{3x} \, dx \]This implies \( e^{2x}y = \frac{2}{3}e^{3x} + C \), where \( C \) is the constant of integration.
06

Solve for y

Solve for \( y \) by dividing through by \( e^{2x} \):\[ y = \frac{2}{3}e^{x} + Ce^{-2x} \]

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Key Concepts

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

linear differential equations
Differential equations are mathematical equations that relate a function with its derivatives, and they appear frequently in both science and engineering. Specifically, a **linear differential equation** involves a function and its first derivative, where the function and its derivative are multiplied by coefficients that can be functions of the independent variable. These equations are linear in the sense that any terms with the dependent variable and its derivative are first degree. A typical first-order linear differential equation looks like this:
  • The general form: \( y' + P(x)y = Q(x) \)
  • Where \( y' \) is the first derivative of \( y \), \( P(x) \) and \( Q(x) \) are functions of the independent variable \( x \).
The crucial aspect that classifies it as linear is the absence of powers or products of the dependent variable and its derivatives. In this exercise, the equation \( y' + 2y = 2e^x \) fits the criteria perfectly, with \( P(x) = 2 \) and \( Q(x) = 2e^x \). Understanding the structure of such equations is pivotal as it helps in determining the method for solving them.
integrating factor
An **integrating factor** is a very powerful tool for solving linear first-order differential equations, especially when placed in standard form. The idea is to transform a non-exact differential equation into an exact one by multiplying it by a strategically chosen function. This function, known as the integrating factor, allows us to rewrite the left-hand side of the equation as a derivative of some expression. The process goes as follows:
  • First, identify \( P(x) \) in the standard form \( y' + P(x)y = Q(x) \).
  • Then, compute the integrating factor \( \mu(x) = e^{\int P(x) dx} \).
In the provided equation, \( P(x) = 2 \), which gives us the integrating factor \( \mu(x) = e^{2x} \). Once multiplied by this integrating factor, the differentiation on the left becomes simple, turning it into \( \frac{d}{dx}(e^{2x}y) \). Thus, we reduce the problem of solving a differential equation into simple integration. It's a neat trick that greatly simplifies finding solutions to these problems.
first-order differential equations
First-order differential equations are equations involving the first derivative of a function and can often be solved via characteristic methods like separation of variables or using an integrating factor, as seen in linear cases. First-order implies that only the first derivative \( y' \) appears in the equation—no second derivatives or higher.These equations are fundamental in modeling real-world phenomena, such as exponential growth, cooling processes, and more. They can typically be categorized into:
  • Separable equations, where the equation can be written as \( g(y) dy = f(x) dx \), allowing for direct integration.
  • Linear equations, similar to our example \( y' + P(x)y = Q(x) \), which are often solved using methods like the integrating factor.
In our exercise, the equation \( y' + 2y = 2e^x \) represents a typical linear first-order differential equation. Understanding its solution involves recognizing its structure and applying methods that simplify and solve it effectively, like the integrating factor used here. Mastery of these concepts is essential for delving into more complex differential equations and their applications.

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Most popular questions from this chapter

Solve the differential equation. $$\frac{d y}{d t}=\frac{t e^{t}}{y \sqrt{1+y^{2}}}$$

Solve the equation \(y^{\prime}=x \sqrt{x^{2}+1} /\left(y e^{y}\right)\) and graph several members of the family of solutions (if your CAS does implicit plots). How does the solution curve change as the constant \(C\) varies?

For each predator-prey system, determine which of the variables, \(x\) or \(y,\) represents the prey population and which represents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. $$\begin{aligned} \text { (a) } \frac{d x}{d t} &=-0.05 x+0.0001 x y \\\ \frac{d y}{d t} &=0.1 y-0.005 x y \\ \text { (b) } \frac{d x}{d t} &=0.2 x-0.0002 x^{2}-0.006 x y \\ \frac{d y}{d t} &=-0.015 y+0.00008 x y \end{aligned}$$

In Exercise 13 in Section 9.1 we formulated a model for learning in the form of the differential equation $$\frac{d P}{d t}=k(M-P)$$ where \(P(t)\) measures the performance of someone learning a skill after a training time \(t, M\) is the maximum level of performance, and \(k\) is a positive constant. Solve this differential equation to find an expression for \(P(t) .\) What is the limit of this expression?

Consider the differential equation $$\frac{d P}{d t}=0.08 P\left(1-\frac{P}{1000}\right)-c$$ as a model for a fish population, where \(t\) is measured in weeks and \(c\) is a constant. (a) Use a CAS to draw direction fields for various values of \(c .\) (b) From your direction fields in part (a), determine the values of \(c\) for which there is at least one equilibrium solution. For what values of \(c\) does the fish population always die out? (c) Use the differential equation to prove what you dis- covered graphically in part (b). (d) What would you recommend for a limit to the weekly catch of this fish population?

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