Problem 13 - Thomas Calculus

Question

Solve the differential equations. \(\sin \theta \frac{d r}{d \theta}+(\cos \theta) r=\tan \theta, \quad 0<\theta<\pi / 2\)

Step-by-Step Solution

Answer

The general solution to the differential equation is \( r = 1 + \frac{C}{\sin \theta} \), where \(C\) is a constant.

Step 1: Recognize the Form

Identify the given differential equation \(\sin \theta \frac{d r}{d \theta} + (\cos \theta) r = \tan \theta\) as a linear first-order differential equation in the form of \( \frac{dr}{d\theta} + P(\theta)r = Q(\theta) \). Here, \(P(\theta) = \cot \theta\) and \(Q(\theta) = \frac{1}{\sin \theta} \).

Step 2: Find the Integrating Factor

To solve the linear first-order differential equation, we first find the integrating factor. The integrating factor is given by \(e^{\int P(\theta) d\theta}\). Here, \(P(\theta) = \cot \theta\), thus the integrating factor is \(e^{\int \cot \theta d\theta} = e^{\ln |\sin \theta|} = \sin \theta\).

Step 3: Multiply Through by the Integrating Factor

Multiply every term in the original differential equation by the integrating factor \(\sin \theta\). This gives: \(\sin^2 \theta \frac{dr}{d\theta} + \sin \theta \cos \theta r = \cos \theta\).

Step 4: Simplify to Exact Form

After applying the integrating factor, the equation is now exact and can be rewritten as: \(\frac{d}{d\theta}(r \sin \theta) = \cos \theta\).

Step 5: Integrate the Exact Differential Equation

Integrate both sides with respect to \(\theta\). The left side becomes \(r \sin \theta\) and the integral of the right side is \(\int \cos \theta \, d\theta = \sin \theta\). Thus, \(r \sin \theta = \sin \theta + C\), where \(C\) is the constant of integration.

Step 6: Solve for \(r\)

To solve for \(r\), divide both sides of \(r \sin \theta = \sin \theta + C\) by \(\sin \theta\) (noting \(\sin \theta eq 0\) in \(0 < \theta < \pi/2\)), yielding \(r = 1 + \frac{C}{\sin \theta}\).

Key Concepts

Integrating Factor

Linear differential equations often require a special technique to simplify the solving process, called an integrating factor. This method comes in handy for first-order linear differential equations expressed in the form \( \frac{dr}{d\theta} + P(\theta)r = Q(\theta) \). In our exercise, we identified \( P(\theta) \) as \( \cot \theta \), which is the cotangent of \( \theta \).

The integrating factor is computed by evaluating the exponential of the integral of \( P(\theta) \). The expression becomes:

• \( e^{\int \cot \theta d\theta} \)

To resolve this, recall that the integral of the cotangent function, \( \int \cot \theta d\theta \), is the natural logarithm of the sine function, \( \ln |\sin \theta| \). Thus, the integrating factor simplifies to \( e^{\ln |\sin \theta|} \), which equals \( |\sin \theta| \).

Since \( 0 < \theta < \pi/2 \), the sine of \( \theta \) is always positive, allowing us to safely use \( \sin \theta \) as the integrating factor. This turns the differential equation into a format that can be more easily managed.

Exact Differential Equations

Once the integrating factor \( \sin \theta \) is applied, the equation

• \( \sin \theta \frac{d r}{d \theta} + (\cos \theta)r = \tan \theta \)

transforms into an exact differential form. The term "exact" in differential equations implies that the equation corresponds to the derivative of some function. Essentially, in terms of this exercise, multiplying by the integrating factor helps us express the whole equation as the derivative of a single function.

After using the integrating factor, the problem boils down to:

• \( \frac{d}{d\theta}(r \sin \theta) = \cos \theta \)

This expression is perfect for integration, as it captures the exact change of a function—the heart of solving differential equations effectively. This exact form allows a straightforward integration process, simplifying the solution into a more manageable equation.

Integration

Integration is the fundamental tool that ties differential equations into solutions. For solving exact differential equations, integration simplifies the complexity into solvable expressions. Given our exact differential form \( \frac{d}{d\theta}(r \sin \theta) = \cos \theta \), integrating can directly fetch the solution.

Let's integrate both sides with respect to \( \theta \):

• The left side becomes \( r \sin \theta \) after evaluating the integral \( \int \frac{d}{d\theta}(r \sin \theta) \, d\theta \).
• For the right side, the integral is \( \int \cos \theta \, d\theta = \sin \theta \).

The balanced equation resulting from this is \( r \sin \theta = \sin \theta + C \), where \( C \) denotes the constant of integration. This step reveals how integration returns a function from its derivative, translating into the solution \( r = 1 + \frac{C}{\sin \theta} \), finalizing the understanding and solution of the differential equation. Integration steers us home by tackling the problem piece-by-piece, each step conclusively revealing more of the solution.