Problem 13 - Thomas Calculus

Question

Solve the differential equations in Exercises \(1-14\) $$\sin \theta \frac{d r}{d \theta}+(\cos \theta) r=\tan \theta, \quad 0<\theta<\pi / 2$$

Step-by-Step Solution

Answer

The solution is \( r = -\cot \theta + C \csc \theta \).

Step 1: Recognize the form of the equation

Notice the given differential equation: \( \sin \theta \frac{dr}{d\theta} + (\cos \theta)r = \tan \theta \). This is a linear first-order differential equation of the form \( P(\theta) \frac{dr}{d\theta} + Q(\theta) r = R(\theta) \).

Step 2: Simplify the equation

Divide the entire equation by \( \sin \theta \) to isolate the derivative and transform the equation into standard linear form: \[ \frac{dr}{d\theta} + \frac{\cos \theta}{\sin \theta}r = \frac{\tan \theta}{\sin \theta} \]. The equation becomes \( \frac{dr}{d\theta} + \cot \theta \, r = 1 \).

Step 3: Identify the integrating factor

For linear equations of the form \( \frac{dr}{d\theta} + P(\theta)r = Q(\theta) \), the integrating factor \( \mu(\theta) \) is given by \( e^{\int P(\theta) \, d\theta} \). Here, \( P(\theta) = \cot \theta \), so we compute \( \mu(\theta) = e^{\int \cot \theta \, d\theta} = e^{\ln |\sin \theta|} = \sin \theta \).

Step 4: Multiply by the integrating factor

Multiply the entire differential equation by the integrating factor \( \sin \theta \): \[ \sin \theta \frac{dr}{d\theta} + \sin \theta \cot \theta \, r = \sin \theta \]. The left side becomes: \( \frac{d}{d\theta}(r \sin \theta) \).

Step 5: Integrate both sides

The equation \( \frac{d}{d\theta}(r \sin \theta) = \sin \theta \) implies \( r \sin \theta = \int \sin \theta \, d\theta \). Integrate the right side: \( \int \sin \theta \, d\theta = -\cos \theta \). So, \( r \sin \theta = -\cos \theta + C \) where \( C \) is a constant of integration.

Step 6: Solve for r

Solve for \( r \) by dividing both sides by \( \sin \theta \): \[ r = \frac{-\cos \theta}{\sin \theta} + \frac{C}{\sin \theta} \], or \[ r = -\cot \theta + C \csc \theta \].

Key Concepts

Linear First-Order Differential Equations

A linear first-order differential equation is a type of equation that can be expressed in the form:

• \[ P(x) \frac{dy}{dx} + Q(x)y = R(x) \]

Here, \(y\) is the dependent variable, and \(x\) is the independent variable. The functions \(P(x)\), \(Q(x)\), and \(R(x)\) are given, which are typically differentiable.
These equations are "first-order" because the highest derivative is first order \( \left( \frac{dy}{dx} \right) \), and "linear" due to the linearity in the dependent variable \(y\).
In solving such equations, the aim is to find a function \(y(x)\) that satisfies the equation over a range of \(x\).
In this way, it provides a useful approach for solving real-world problems modeled by differential equations, as they often appear in physics, engineering, and other fields.

Integrating Factor

An integrating factor is a function used to simplify the process of solving linear first-order differential equations. When we have a linear equation in standard form:

• \[ \frac{dy}{dx} + P(x)y = Q(x) \]

The integrating factor \( \mu(x) \) is determined by:

• \[ \mu(x) = e^{\int P(x) \, dx} \]

By multiplying every term in the differential equation by this integrating factor, we can convert the left-hand side into a single derivative:

• \[ \frac{d}{dx}(\mu(x)y) = \mu(x)Q(x) \]

This transformation is key because it allows us to integrate both sides easily with respect to \(x\), enabling us to find the solution \(y(x)\).
In solving the original equation, the identifying and applying of an integrating factor is often a critical step.

Trigonometric Functions

Trigonometric functions are often encountered when dealing with differential equations, especially those involving angular changes, like in physics. The basic trigonometric functions are sine (\(\sin\)), cosine (\(\cos\)), tangent (\(\tan\)), and their reciprocals: cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)).
Each function has unique properties and graphs, useful in modeling periodic behavior and circular motion.
Within differential equations, trigonometric functions can represent oscillations or waves. They can also be part of the integration process as you transform and solve linear differential equations, just as seen in the step-by-step solution. Here, integrating functions like \(\sin\) involved using identities and properties of these functions to simplify and solve the given equations.

Solution Techniques for Differential Equations

Various techniques can be used to solve differential equations depending on their type. For linear first-order differential equations, the integrating factor is often the technique of choice. In general:

• Identify the linear form.
• Simplify the equation if needed to isolate the derivative.
• Find the integrating factor based on the given functions.
• Multiply the whole equation by this integrating factor.

After transforming the equation, integrate both sides to find the solution.
Sometimes, using trigonometric identities or calculus can also assist in further simplification or integration. The objective is always to rearrange and simplify until the equation can be integrated to solve for the unknown function. Each technique is selected based on the equation's form and the functions involved, ensuring a flexible approach to different problems.