/*! 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 43 Solve the differential equations... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d L}{d x}=k(x+a)(L-b)$$

Short Answer

Expert verified
The general solution is \( L = b \pm C_3 e^{\frac{k(x+a)^2}{2}} \), where \( C_3 \) is an arbitrary constant.

Step by step solution

01

Set Up the Differential Equation

The given differential equation is \( \frac{dL}{dx} = k(x+a)(L-b) \). This is a separable differential equation, which means we can rearrange the terms to isolate \( L \) and \( x \) on opposite sides.
02

Separate the Variables

Rearrange the equation to separate the variables. Divide both sides by \( (L-b) \) and multiply both sides by \( dx \). This gives us: \[ \frac{1}{L-b} \, dL = k(x+a) \, dx \]
03

Integrate Both Sides

Integrate both sides of the equation. The left side becomes \( \int \frac{1}{L-b} \, dL \) and the right side becomes \( \int k(x+a) \, dx \).
04

Perform the Integrations

Calculate each integral. For the left side: \( \int \frac{1}{L-b} \, dL = \ln|L-b| + C_1 \). For the right side: \( \int k(x+a) \, dx = \frac{k(x+a)^2}{2} + C_2 \).
05

Combine Integrations

Equate the integrated expressions: \( \ln|L-b| = \frac{k(x+a)^2}{2} + C \), where \( C = C_2 - C_1 \).
06

Solve for L

To find \( L \), exponentiate both sides: \( |L-b| = e^{\frac{k(x+a)^2}{2} + C} \). Let's denote \( e^C \) as a new constant \( C_3 \), so: \( |L-b| = C_3 e^{\frac{k(x+a)^2}{2}} \).
07

Express L in Terms of x

Finally, solve for \( L \). We can drop the absolute value by considering positive and negative cases, resulting in: \( L = b + C_3 e^{\frac{k(x+a)^2}{2}} \) or \( L = b - C_3 e^{\frac{k(x+a)^2}{2}} \).

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Key Concepts

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

Separable Equations
Separable equations are a class of differential equations that allow us to solve by separating variables into different sides of the equation. In a separable equation, we can rearrange terms such that all terms involving the dependent variable (like \(L\) in our problem) are on one side, and all terms involving the independent variable (like \(x\)) are on the other. This makes it possible to integrate both sides separately.

To solve the given equation \( \frac{dL}{dx} = k(x+a)(L-b) \), we notice it's separable because we can express it as \(\frac{1}{L-b} \, dL = k(x+a) \, dx\). This simple rearrangement is at the heart of solving separable equations. It allows us to work with each variable independently, making integration possible.
Integration
Integration is the mathematical process of finding the antiderivative, or the reverse of differentiation. When dealing with separable differential equations like \(\frac{1}{L-b} \, dL = k(x+a) \, dx\), integration helps us solve for the variables.

- **Left Side Integration:** For the term \( \int \frac{1}{L-b} \, dL \), use the natural logarithm because the integral of \( \frac{1}{u} \, du \) is \( \ln|u| \). Therefore, \(\int \frac{1}{L-b} \, dL = \ln|L-b| + C_1\).

- **Right Side Integration:** Here, the term is \( \int k(x+a) \, dx \). You distribute the linear term, finding that \( \int k(x+a) \, dx = \frac{k(x+a)^2}{2} + C_2 \). This is a standard polynomial integration result.

Integrating both sides finds expressions related to \(L\) and \(x\) that can be equated to solve the separable differential equation.
Exponential Functions
Exponential functions describe situations where a quantity grows or decays at a rate proportional to its current value. They appear in solutions like our differential equation, which yields an exponential function when solved.

Following integration, we find \(\ln|L-b| = \frac{k(x+a)^2}{2} + C\). Exponentiating both sides solves for \(|L-b|\) and involves exponential functions:
  • Taking the natural exponential \( e \) on both sides gives \(|L-b| = e^{\frac{k(x+a)^2}{2} + C}\).
  • Notably, let \(e^C \) be a constant \( C_3\), leading to \(|L-b| = C_3 e^{\frac{k(x+a)^2}{2}}\).
The base \(e\) is particularly common in these scenarios due to its unique properties in growth models.
Constants in Equations
Constants often emerge in differential equations during integration. They represent fixed values that can adjust based on initial conditions or boundaries. Let's explore their role and significance.

- **Arbitrary Constants:** During integration, constants like \( C_1 \) and \( C_2 \) represent unknowns from indefinite integrals. They adjust based on specific contexts.

- **Combined Constants:** When both integrations introduce constants, they are typically combined into a single, arbitrary constant, such as \( C = C_2 - C_1 \). This combined constant helps simplify solutions.

- **Exponential Constant:** In exponential solutions, constants often appear as multipliers of these functions. For instance, our term \( C_3 = e^C \) is derived post-integration, acting as an adjuster for the equation's flexibility.

Understanding constants is crucial. They allow the differential equation to adapt to a variety of initial conditions and contexts, providing a wide range of potential solutions.

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Most popular questions from this chapter

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. $$f^{\prime}(x)=2 x-f(x)$$

(a) Define the variables. (b) Write a differential equation to describe the relationship. (c) Solve the differential equation. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality 0.347 if the amount of nicotine is in \(\mathrm{mg}\) and time is in hours. The amount of nicotine in the body immediately after smoking a cigarette is \(0.4 \mathrm{mg}\).

Are the statements true or false? Give an explanation for your answer. The solutions of the differential equation \(d y / d x=x^{2}+\) \(y^{2}+1\) are increasing at every point.

At 1: 00 pm one winter afternoon, there is a power failure at your house in Wisconsin, and your heat does not work without electricity. When the power goes out, it is \(68^{\circ} \mathrm{F}\) in your house. At 10: 00 pm, it is \(57^{\circ} \mathrm{F}\) in the house, and you notice that it is \(10^{\circ} \mathrm{F}\) outside. (a) Assuming that the temperature, \(T,\) in your home obeys Newton's Law of Cooling, write the differential equation satisfied by \(T\) (b) Solve the differential equation to estimate the temperature in the house when you get up at 7: 00 am the next morning. Should you worry about your water pipes freezing? (c) What assumption did you make in part (a) about the temperature outside? Given this (probably incorrect) assumption, would you revise your estimate up or down? Why?

Give the solution to the logistic differential equation with initial condition. $$\frac{d P}{d t}=0.04 P(1-0.0001 P) \text { with } P_{0}=200$$

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