Question

Consider the general linear homogeneous second order equation $$ P(x) y^{\prime \prime}+Q(x) y^{\prime}+R(x) y=0 $$ $$ \begin{array}{l}{\text { We seck an integrating factor } \mu(x) \text { such that, upon multiplying Eq. (i) by } \mu(x) \text { , the resulting }} \\\ {\text { equation can be written in the form }}\end{array} $$ $$ \left[\mu(x) P(x) y^{\prime}\right]+\mu(x) R(x) y=0 $$ $$ \text { (a) By equating coefficients of } y \text { , show that } \mu \text { must be a solution of } $$ $$ P \mu^{\prime}=\left(Q-P^{\prime}\right) \mu $$ $$ \text { (b) Solve Eq. (iii) and thereby show that } $$ $$ \mu(x)=\frac{1}{P(x)} \exp \int_{x_{0}}^{\pi} \frac{Q(s)}{P(s)} d s $$ $$ \text { Compare this result with that of Problem } 27 \text { in Section } 3.2 . $$

Short answer

Question: Find the integrating factor µ(x) for a linear homogeneous second-order equation given by P(x)y'' + Q(x)y' + R(x)y = 0, such that, upon multiplying the given equation by µ(x), the resulting equation can be written in the form [µ(x)P(x)y']' + µ(x)R(x)y = 0. Answer: The integrating factor µ(x) is given by: $$ \mu(x) = \frac{1}{P(x)} \exp \int_{x_{0}}^{x} \frac{Q(s)}{P(s)} ds $$

Step-by-step solution

Write the given equation in the specified form

To find the integrating factor µ(x), multiply the given equation by µ(x): $$ \mu(x)\left[P(x) y^{\prime \prime}+Q(x) y^{\prime}+R(x) y\right] = 0 $$ Now, rearrange the terms to obtain: $$ \mu(x)P(x) y^{\prime \prime} + \mu(x)Q(x) y^{\prime} + \mu(x)R(x) y = 0 $$ We seek to write this equation in the form: $$ \left[\mu(x) P(x) y^{\prime}\right]^{\prime} + \mu(x) R(x) y = 0 $$

Equate coefficients of y

To write the equation in the specified form, equate the coefficients of y in both equations: $$\mu(x)R(x) = \mu(x)R(x)$$ This equation holds true, so we proceed to the next step.

Equate coefficients of y'

Now, equate the coefficients of y' in both equations: $$ \frac{d}{dx}(\mu(x) P(x) y^{\prime}) = \mu(x)Q(x) y^{\prime} $$ Using the product rule, we get: $$ \left[P(x) \mu^{\prime}(x) + P^{\prime}(x) \mu(x)\right] y^{\prime} = \mu(x)Q(x) y^{\prime} $$ Since this equation must hold for all y', we can equate the terms inside the brackets: $$ P(x) \mu^{\prime}(x) + P^{\prime}(x) \mu(x) = \mu(x)Q(x) $$ Rearrange to match the given equation in part (a): $$ P(x) \mu^{\prime}(x) = \mu(x) \left[Q(x) - P^{\prime}(x) \right] $$

Solve the equation for µ(x)

First, we rewrite the equation as a first-order linear differential equation: $$ \frac{\mu^{\prime}(x)}{\mu(x)} = \frac{Q(x) - P^{\prime}(x)}{P(x)} $$ Integrate both sides with respect to x: $$ \int \frac{\mu^{\prime}(x)}{\mu(x)} dx = \int \frac{Q(x) - P^{\prime}(x)}{P(x)} dx $$ Now, on the left side, use the substitution method with u = µ(x), du = µ'(x) dx: $$ \int \frac{1}{u} du = \int \frac{Q(x) - P^{\prime}(x)}{P(x)} dx $$ Integrate both sides: $$ \ln|\mu(x)| = \int_{x_{0}}^{x} \frac{Q(s) - P^{\prime}(s)}{P(s)} ds + C $$ Apply the exponent to both sides: $$ \mu(x) = C \exp \int_{x_{0}}^{x} \frac{Q(s) - P^{\prime}(s)}{P(s)} ds $$ From the comparison with problem 27 in section 3.2, we know that the integrating factor is: $$ \mu(x) = \frac{1}{P(x)} \exp \int_{x_{0}}^{x} \frac{Q(s)}{P(s)} ds $$ Thus, we have found the integrating factor µ(x) for the given equation, the solution for part (b).

Key concepts

Integrating Factor

An integrating factor is a function used in differential equations to simplify the process of finding solutions. It tailors the equation to make integration easier by turning it into an exact differential.

• The idea is to multiply the entire differential equation by a function, \( \mu(x) \), known as the integrating factor, so that the equation can be rewritten in a format that's easier to solve.
• For linear first-order equations, the integrating factor transforms the equation into one that can be directly integrated, allowing for a straightforward solution process.

For example, in the exercise, the integrating factor is found by addressing a form of equation \( \left[ \mu(x) P(x) y^{\prime} \right]^{\prime} + \mu(x) R(x) y = 0\). This transforms the differential equation into a form where standard methods of calculus like integration can be applied easily.
The integrating factor \( \mu(x) \) must satisfy \( P \mu^{\prime} = (Q - P^{\prime}) \mu\). Solving this provides a pathway to find the solution to the differential equation.

Linear Homogeneous Equation

A Linear Homogeneous Equation is a type of differential equation where the function and its derivatives are linearly dependent. In simple terms, it means that all the terms involve the unknown function or its derivatives, and there's no constant term on the right side of the equation.

• Such equations take the form \( P(x)y^{\prime\prime} + Q(x)y^{\prime} + R(x)y = 0 \), indicating that the solution consists entirely of components of the function \( y \) and its derivatives.
• The nature of the coefficients, which can be functions of \( x \), determines the form of solutions and whether special methods like integrating factors are required.

In the context of our exercise, the goal was to transform the linear homogeneous equation by finding the integrating factor that meets certain conditions, thereby aiding in solving the equation neatly.
This transformation often simplifies work involving solution processes of these equations.

Second Order Equation

A Second Order Equation involves the second derivative of the unknown function. Such equations are significant in physics and engineering, describing systems like oscillations in springs or electrical circuits. The general form is \( P(x) y^{\prime\prime} + Q(x) y^{\prime} + R(x) y = 0 \).

• These equations involve terms up to the second derivative like \( y^{\prime\prime} \).
• They often require complex solution techniques, as they describe systems with curvature, acceleration, or similar properties.
• The coefficients \( P(x), Q(x), \text{ and } R(x) \) can be constants or functions, changing the solution technique dramatically.

A crucial part of solving second order equations is transforming them into forms that are easier to manage, often employing methods like integrating factors. Such transformation, as demonstrated in the original task, aligns the equation with forms amendable to integration or other solving techniques.
Recognizing the character and form of a second order equation is pivotal in determining the correct approach to finding a solution.

First-order Linear Differential Equation

First-order linear differential equations are the simplest type of differential equations which involve only the first derivative of the unknown function. Represented in the form: \( y^{\prime} + P(x)y = Q(x) \), recognizing such form is crucial for using integrating factors method effectively.

• These equations serve as foundational examples for applying integrating factors due to their approachable structure.
• The simplicity in their structure allows for intuitive application of calculus techniques once rewritten appropriately.

In solving such equations, finding an integrating factor significantly simplifies the process. It transforms the equation to one that is immediately integrable, leading to straightforward solutions.
The exercise demonstrated rearranging complex second-order problems into forms comparable to these simpler structures, allowing the solution to unfold naturally by leveraging the principle of integrating factors. Understanding their methods provides solid perspective for approaching more complex equations.