Problem 46 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \( x \).

Step 2: 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 \( dt/t \):\[\frac{1}{\tan x}dx = \frac{1+2\ln t}{t}dt.\]

Step 3: 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.

Step 4: 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 \( u = \ln t \):\[\int \frac{1 + 2\ln t}{t} \, dt = \ln t + (\ln t)^2 + C_2.\]

Step 5: 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 \( K = e^C \).

Key Concepts

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.