Problem 18 - 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. $$ \cos ^{2} x \sin x \frac{d y}{d x}+\left(\cos ^{3} x\right) y=1 $$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{1}{\cos x} + \frac{C}{\sin x} \), valid on \( (n\pi, (n+1)\pi) \) with no transient terms.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \cos^2 x \sin x \frac{dy}{dx} + \left(\cos^3 x\right) y = 1 \). It is a first-order linear differential equation of the form \( a(x) \frac{dy}{dx} + b(x) y = g(x) \) where \( a(x) = \cos^2 x \sin x \), \( b(x) = \cos^3 x \), and \( g(x) = 1 \).

Step 2: Simplify the Equation

Divide the entire equation by \( \cos^2 x \sin x \) to make it standard:\[\frac{dy}{dx} + \frac{\cos x}{\sin x} y = \frac{1}{\cos^2 x \sin x}\]This yields a simplified equation \( \frac{dy}{dx} + \cot x \, y = \frac{1}{\cos^2 x \sin x} \).

Step 3: Find the Integrating Factor

The equation is in the standard linear form. The integrating factor \( \mu(x) \) is given by:\[ \mu(x) = e^{\int P(x) \, dx} \]where \( P(x) = \cot x \). Therefore:\[ \mu(x) = e^{\int \cot x \, dx} = e^{\ln |\sin x|} = |\sin x| \].

Step 4: Multiply Through by the Integrating Factor

Multiply every term of the differential equation by the integrating factor \( |\sin x| \):\[|\sin x| \frac{dy}{dx} + |\sin x| \cot x \, y = \frac{|\sin x|}{\cos^2 x \sin x}\]This simplifies to:\[\sin x \frac{dy}{dx} + y = \frac{1}{\cos^2 x}\] as \( |\sin x| \cot x = \cos x \).

Step 5: Solve the Resulting Exact Equation

The equation can be rewritten as a derivative of a product:\[ \frac{d}{dx}(y \sin x) = \frac{1}{\cos^2 x} \]Integrate both sides with respect to \( x \):\[ y \sin x = \int \sec^2 x \, dx = \tan x + C \]where \( C \) is the constant of integration.

Step 6: Solve for \( y \)

Rearrange to find \( y \):\[ y = \frac{\tan x + C}{\sin x} = \frac{\tan x}{\sin x} + \frac{C}{\sin x}\]Simplify to:\[ y = \frac{1}{\cos x} + \frac{C}{\sin x} \].

Step 7: Determine the Largest Interval for the Solution

The solution involves \( \sin x \) and \( \cos x \), which are undefined for \( x = n\pi \) and \( x = \frac{(2n+1)\pi}{2} \) respectively. The largest interval where the solution is defined is between these undefined points, i.e., on intervals of the form \( (n\pi, (n+1)\pi) \).

Step 8: Check for Transient Terms

Transient terms are those that tend to zero as \( x \to \pm \infty \). In this context, since \( \frac{1}{\cos x} \) and \( \frac{C}{\sin x} \) do not tend towards zero, the solution does not contain transient terms.

Key Concepts

Integrating Factor

First-order linear differential equations come in a standard form which makes solving them efficient through a simple tool called the integrating factor. The integrating factor is a function used to multiply through a differential equation to make it easily integrable.
To determine the integrating factor, you need to rearrange the differential equation into the form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

In our exercise, this step involved simplifying the equation to:

• \( \frac{dy}{dx} + \cot x \, y = \frac{1}{\cos^2 x \sin x} \)

The function \( P(x) = \cot x \) helps us find the integrating factor \( \mu(x) \), through the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

The integral of \( \cot x \) is \( \ln |\sin x| \), so \( \mu(x) = |\sin x| \). By applying the integrating factor to the entire differential equation, it becomes possible to express one side as a derivative, paving the way for straightforward integration.

General Solution

When working with differential equations, the goal is often to find the general solution. This is a solution that encompasses all possible particular solutions.
From our exercise, after applying the integrating factor, the equation gets simplified to:

• \( \frac{d}{dx}(y \sin x) = \frac{1}{\cos^2 x} \)

This form allows us to integrate both sides easily.
The integration of \( \frac{d}{dx}(y \sin x) \) with respect to \( x \) gives:

• \( y \sin x = \tan x + C \)

where \( C \) is the constant of integration. Solving for \( y \) gives:

• \( y = \frac{1}{\cos x} + \frac{C}{\sin x} \)

The function \( y \) here is a general solution because it includes the arbitrary constant \( C \), representing an entire family of solutions, depending on the value of \( C \). This general solution indicates the behavior of solutions over the largest possible interval of definition as determined by the properties of sine and cosine functions.

Transient Terms

When exploring solutions to differential equations, transient terms play a significant role in determining the stability or eventual behavior of the solution over time.
A transient term typically describes a component of the solution that diminishes or goes to zero as \( x \) approaches a certain bound. For the exercise at hand, we examine the general solution:

• \( y = \frac{1}{\cos x} + \frac{C}{\sin x} \)

Neither \( \frac{1}{\cos x} \) nor \( \frac{C}{\sin x} \) go to zero as \( x \) tends to infinity or other asymptotic points.
Since neither term vanishes, there are no transient terms in this particular general solution. It's important to recognize that the solution's behavior is dominated by the non-vanishing terms here, as they are defined within the bounds of the interval \( (n\pi, (n+1)\pi) \). Understanding whether a solution includes transient terms helps in knowing if the corresponding physical system's response will stabilize over time.