/*! 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 46 Solve the differential equations... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the differential equations in Problems \(44-47 .\) Assume \(x, y, t > 0\). $$t \frac{d x}{d t}=(1+2 \ln t) \tan x$$

Short Answer

Expert verified
The solution is \( \sin x = K e^{\ln t + (\ln t)^2} \), where \( K \) is a constant.

Step by step solution

01

Write Down the Given Differential Equation

We are given the differential equation \( t \frac{dx}{dt} = (1 + 2 \ln t) \tan x \). This is a first-order differential equation involving the variables \( t \) and \( x \).
02

Separate Variables

To solve the differential equation, we first need to separate the variables. We move all terms involving \( x \) to one side and terms involving \( t \) to the other side. Divide both sides by \( \tan x \) and multiply both sides by \( dt/t \):\[\frac{1}{\tan x}dx = \frac{1+2\ln t}{t}dt.\]
03

Simplify and Prepare for Integration

We simplify the equation by noting that \( \frac{1}{\tan x} = \cot x \). So, the equation becomes:\[\cot x \, dx = \frac{1 + 2\ln t}{t} \, dt.\]Now, we are ready to integrate both sides.
04

Integrate Both Sides

We integrate both sides of the equation:- The left side: \( \int \cot x \, dx = \ln |\sin x| + C_1 \).- The right side: Use substitution \( u = \ln t \), then \( du = \frac{1}{t} dt \), thus \( \int (1 + 2\ln t) \frac{1}{t} dt = \int (1 + 2u) du = u + u^2 + C_2 \).Substituting back for \( u = \ln t \):\[\int \frac{1 + 2\ln t}{t} \, dt = \ln t + (\ln t)^2 + C_2.\]
05

Combine the Results

After integrating, equate the two sides:\[\ln |\sin x| = \ln t + (\ln t)^2 + C.\]Here, \( C = C_2 - C_1 \). Solve for \( x \) in terms of \( t \):\[|\sin x| = e^{\ln t + (\ln t)^2 + C}.\]This gives \( \sin x = Ke^{\ln t + (\ln t)^2} \) where \( K = e^C \).

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Key Concepts

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

First-Order Differential Equation
First-order differential equations involve expressions where the highest derivative is the first derivative. In our exercise, the equation is in the form \( t \frac{dx}{dt} = (1 + 2 \ln t) \tan x \). This tells us we are dealing with a relationship that connects the rate of change of \( x \) with \( t \) and \( x \) itself.

Such equations are foundational in mathematical modeling. They express how a quantity changes with respect to another variable, essential in physics, biology, and engineering for modeling systems. Understanding the nature of first-order equations helps to predict the behavior of dynamic systems, simulating real-world phenomena across disciplines.
Separation of Variables
Separation of variables is a powerful method for solving differential equations. The technique involves rearranging the equation to isolate all terms involving one variable on one side and terms involving the other variable on the opposite side.

In our exercise, we separated \( x \) and \( t \) by rearranging terms as \( \frac{1}{\tan x}dx = \frac{1+2\ln t}{t}dt \). This separation makes each side of the equation depend only on one variable, enabling us to integrate both sides independently.

Mastering this technique is valuable. It often simplifies complex problems, leading to straightforward integrations and solutions for many types of differential equations. Keep in mind, though, not every differential equation can be solved by this method.
Integration Techniques
Integration is crucial in solving differential equations. It is how we reverse the process of differentiation, finding the function that represents the original problem.

For our exercise, solving \( \int \cot x \, dx \) involves recognizing the need for a logarithmic antiderivative, resulting in \( \ln |\sin x| + C_1 \). The right-hand side \( \int (1 + 2\ln t) \, dt \) involves substitution, identifying \( u = \ln t \). This simplifies the integral to \( \int (1 + 2u) \, du \), giving us \( u + u^2 + C_2 \).

Learning diverse integration techniques is essential. Recognizing patterns and applying substitution or partial fraction decomposition, among others, enables solving a wide range of complex integrals.
Logarithmic Functions
Logarithmic functions are essential in many calculus problems, especially involving growth and decay models. They provide solutions to differential equations where logarithms manifest naturally during integration, as seen in the exercise.

We encountered \( \ln |\sin x| \) from integrating \( \cot x \, dx \). Similarly, understanding \( \ln t \) facilitated integrating terms involving natural logs\( (1 + 2\ln t) / t \).

Logarithms translate multiplicative dynamics into additive scales, offering a simplified view in many mathematical contexts. A firm grasp of their properties is fundamental, serving as a bridge to understanding exponential functions and aiding in the simplification of complex expressions.

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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. There could be more than one value of \(x\) such that \(f^{\prime}(x)=1\) and \(f(x)=5\)

The total number of people infected with a virus often grows like a logistic curve. Suppose that time, \(t,\) is in weeks and that 10 people originally have the virus. In the early stages, the number of people infected is increasing exponentially with \(k=1.78 .\) In the long run, 5000 people are infected. (a) Find a logistic function to model the number of people infected. (b) Sketch a graph of your answer to part (a). (c) Use your graph to estimate the length of time until the rate at which people are becoming infected starts to decrease. What is the vertical coordinate at this point?

(a) Sketch the slope field for \(y^{\prime}=-y / x\) (b) Sketch several solution curves. (c) Solve the differential equation analytically.

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(\lim _{x \rightarrow \infty} g(x)=0,\) then \(\lim _{x \rightarrow \infty} f(x)=0.\)

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.

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