Question

Use the method of Problem 33 to find a second independent solution of the given equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x\)

Short answer

Question: Find a second linearly independent solution for the given differential equation \(x^2y'' + xy' + (x^2 - 0.25)y = 0\), using the method of Problem 33 and considering the known solution \(y_1(x) = x^{-\frac{1}{2}} \sin{x}\). Answer: The second linearly independent solution is given by \(y_2(x) = x^{-\frac{1}{2}} \sin{x} \cdot F(x)\) where \(F(x)\) is an antiderivative of \(x\sin^2{x}\).

Step-by-step solution

Calculate p(x)

First, calculate \(p(x)\) using \(a(x)\) and \(b(x)\): \(p(x) = \frac{b(x)}{a(x)} = \frac{x}{x^2} = \frac{1}{x}\).

Calculate the integral of p(x)

Integrate \(p(x)\) with respect to x: \(\int p(x) \, dx = \int \frac{1}{x} \, dx = \ln{x}\)

Calculate the exponential of the integral of p(x)

Calculate \(e^{-\int p(x) \, dx}\): \(e^{-\int p(x) \, dx} = e^{-\ln{x}} = x^{-1}\)

Calculate the fraction term of the formula

Calculate \(\frac{1}{y_1^2(x)}\): \(\frac{1}{y_1^2(x)} = \frac{1}{\left(x^{-\frac{1}{2}} \sin{x}\right)^2} = \frac{1}{x^{-1} \sin^2{x}} = \frac{x \sin^2{x}}{1}\)

Calculate the integral term of the formula

Calculate the integral term of the formula (we only need to calculate an antiderivative): \(\int \frac{1}{y_1^2(x)} \cdot e^{-\int p(x) \, dx} \, dx = \int x \sin^2{x} \, dx\) The result of this integral is not easily expressed in terms of elementary functions. However, the important point to note is that this integral represents an antiderivative of \(x\sin^2{x}\). Nonetheless, we can represent the result of this integral as an unknown function \(F(x)\), and express the second solution in terms of this function.

Find the second independent solution

Finally, the second linearly independent solution, \(y_2(x)\), is given by: \(y_2(x) = y_1(x) \int \frac{1}{y_1^2(x)} \cdot e^{-\int p(x) \, dx} \, dx = x^{-\frac{1}{2}} \sin{x} \cdot F(x)\) where \(F(x)\) is an antiderivative of \(x\sin^2{x}\).

Key concepts

Method of Independent Solutions

The Method of Independent Solutions provides a systematic way to determine a second linearly independent solution to a second-order linear differential equation when one solution is already known. In the context of differential equations, independence of solutions is essential because it ensures that the solutions form a basis for the space of all possible solutions.First, we utilize a given solution, say, \(y_1(x)\). We then transform the differential equation into an easy-to-manage form that allows us to find the second solution. This typically involves calculating specific terms like \(p(x)\) and forming integrals involving \(y_1(x)\).The real beauty of this method is that once we have the structure given by \(y_1(x)\), the second solution, \(y_2(x)\), can be expressed as:

• \(y_2(x) = y_1(x) \int \frac{1}{y_1^2(x)} \cdot e^{-\int p(x) \, dx} \, dx\)

This formula combines integration techniques with the known solution \(y_1(x)\) to construct a second, independent solution. The flexibility in expressing \(y_2(x)\) means that even if integrals don't simplify neatly, they can be expressed in terms of unknown but computable antiderivatives.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that revert differentiation. If you have a function \(f(x)\), the antiderivative is another function whose derivative is \(f(x)\). This process is crucial for finding solutions to differential equations, especially when calculating integrals involving solutions of these equations.In our exercise, the antiderivative is involved when integrating \(x \sin^2{x}\). Although finding an analytical expression in terms of elementary functions might not be possible, understanding that it can be represented by an unknown function, such as \(F(x)\), provides valuable insight.

• The antiderivative tells us how to reverse the differentiation process.
• Such functions serve as building blocks for finding solutions to more complex relationships in differential equations.

Understanding the role of antiderivatives enables students to grasp how integrals relate to differential equations and apply this concept even when the results cannot be exactly expressed.

Integration Techniques

Integration techniques are methods and strategies to find antiderivatives or definite integrals of functions. In the context of solving second-order differential equations, these techniques are crucial to finding expressions like \(\int \frac{1}{y_1^2(x)} \cdot e^{-\int p(x) \, dx} \, dx\).Some common strategies include:

• Integration by parts: Useful when functions are products of algebraic and trigonometric parts.
• Substitution: Changes variables to make integral easier.
• Partial fraction decomposition: Simplifies complex rational expressions.

In our specific case, when dealing with \(x \sin^2{x}\), selecting the appropriate technique is key to simplifying or approximating the integral. Even if the integral cannot be simplified, recognizing the type of functions involved helps us determine if a combination of techniques might be needed.

Special Functions

Special functions arise frequently in the study of differential equations, especially when solutions cannot be expressed as elementary functions. These include functions like Bessel functions, Legendre polynomials, and others often encountered in physics or engineering contexts.In the exercise, an antiderivative of \(x \sin^2{x}\) doesn’t yield a simple closed-form solution, hinting that special functions may be involved. In practice, when such outcomes occur, we often use numerical methods or look up tables to approximate their values.

• Special functions help manage solutions to differential equations that resist simplification.
• They often arise from boundary conditions or specific function behaviors.

Recognizing when special functions are applicable allows students to understand the broader context of their solutions and appreciate the complexity inherent in certain differential equations.