Problem 21 - Advanced Engineering Mathematics

Question

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \frac{d r}{d \theta}+r \sec \theta=\cos \theta $$

Step-by-Step Solution

Answer

The general solution is \( r = \frac{\tan \theta + C}{|\sec \theta + \tan \theta|} \) on \( \theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) \) with no transient terms.

Step 1: Identify the Type of Differential Equation

The given equation is a first-order linear differential equation of the form \( \frac{dr}{d\theta} + P(\theta)r = Q(\theta) \), where \( P(\theta) = \sec \theta \) and \( Q(\theta) = \cos \theta \).

Step 2: Determine the Integrating Factor

For a linear differential equation, the integrating factor \( \mu(\theta) \) is given by \( \mu(\theta) = e^{\int P(\theta)\,d\theta} \). Here, \( \int P(\theta)\,d\theta = \int \sec \theta \,d\theta = \ln |\sec \theta + \tan \theta| \). Thus, \( \mu(\theta) = |\sec \theta + \tan \theta| \).

Step 3: Solve the Differential Equation Using the Integrating Factor

Multiply the entire differential equation by the integrating factor:\[|\sec \theta + \tan \theta| \left( \frac{dr}{d\theta} + r \sec \theta \right) = |\sec \theta + \tan \theta| \cos \theta. \]This simplifies to:\[\frac{d}{d\theta} \left( r \cdot |\sec \theta + \tan \theta| \right) = |\sec \theta + \tan \theta| \cos \theta. \]Integrate both sides wrt \( \theta \). The solution is \( r \cdot |\sec \theta + \tan \theta| = \int |\sec \theta + \tan \theta| \cos \theta \, d\theta + C \), where \( C \) is the constant of integration.

Step 4: Simplify the Integral

Solving the integral, observe that \(|\sec \theta + \tan \theta|\cos \theta = \sec \theta + \tan \theta|\) and its integral is a simplified form:\[\int \cos \theta (|\sec \theta + \tan \theta|) d\theta = \int \frac{|\sec \theta + \tan \theta| \cos \theta}{\cos \theta} d\theta = \tan \theta. \]So \( r \cdot |\sec \theta + \tan \theta| = \tan \theta + C \).

Step 5: Solve for the General Solution

Divide both sides by \( |\sec \theta + \tan \theta| \) to solve for \( r \):\[r = \frac{\tan \theta + C}{|\sec \theta + \tan \theta|}. \]This represents the general solution.

Step 6: Determine Largest Interval for Solution

The expression for \( |\sec \theta + \tan \theta| \) is undefined at points where \( \theta = \frac{\pi}{2} + k\pi \), \( k \in \mathbb{Z} \), because \( \sec \theta \) and \( \tan \theta \) are undefined. The largest interval is \( \theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) \).

Step 7: Check for Transient Terms

Transient terms are those that vanish as \( \theta \to \pm\infty \), usually involving terms like exponentials. Since \( C \) is a constant, the term \( \tan \theta \) does not vanish with increasing or decreasing \( \theta \) over the interval; thus, there are no transient terms.

Key Concepts

Integrating Factor

An integrating factor is a clever mathematical tool used to simplify solving first-order linear differential equations. It's used to turn these equations into something more manageable. To find an integrating factor, consider a standard form of a differential equation: - \( \frac{dr}{d\theta} + P(\theta)r = Q(\theta) \).- The task is to identify the function \( P(\theta) \). Once you know \( P(\theta) \), you compute the integrating factor as follows:

• Calculate \( \mu(\theta) = e^{\int P(\theta) \, d\theta} \).

In our specific exercise, we identified \( P(\theta) = \sec \theta \). The integral of \( \sec \theta \) gives us \( \ln |\sec \theta + \tan \theta| \). Hence, the integrating factor is \( \mu(\theta) = |\sec \theta + \tan \theta| \).
This integrating factor, when multiplied to all terms in the differential equation, allows us to rewrite it into a form where we can easily integrate both sides, leading directly to finding a solution.

General Solution

The **general solution** to a differential equation is a solution that contains all possible solutions for the given differential equation. The key idea is that the general solution incorporates a constant of integration, typically denoted as \( C \).
The provided solution process demonstrates how using the integrating factor transforms our differential equation. We multiplied the equation by \( |\sec \theta + \tan \theta| \), which simplifies to:

• \( \frac{d}{d\theta} \left( r \cdot |\sec \theta + \tan \theta| \right) = |\sec \theta + \tan \theta| \cos \theta \).

After integrating both sides, you'll find:

• \( r \cdot |\sec \theta + \tan \theta| = \tan \theta + C \).

Finally, to express \( r \) and solve the differential equation fully, divide by \( |\sec \theta + \tan \theta| \):- \( r = \frac{\tan \theta + C}{|\sec \theta + \tan \theta|} \).
This equation constitutes the general solution for this particular first-order linear differential equation.

Transient Terms

Transient terms in a mathematical context refer to parts of a solution that vanish as the independent variable approaches infinity. They are temporary parts of the solution, becoming negligible over time.
In the exercise, transient terms usually imply exponentials, which decay to zero with time in a physical system. By studying the general solution derived

• \( r = \frac{\tan \theta + C}{|\sec \theta + \tan \theta|} \),

we analyze the behavior as \( \theta \to \pm\infty \). Here, \( \tan \theta \) does not naturally decrease to zero at infinity for finite \( \theta \); it has a cyclic nature due to \( \theta \)'s periodicity.
Thus, contrary to other differential equations with exponential components, our solution does not possess terms that fade away as \( \theta \) becomes large or small in this interval. Therefore, no transient terms exist for this specific example.