Question

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. \(3 x^{2}\left(3+x^{2}\right) y^{\prime \prime}+x\left(3+11 x^{2}\right) y^{\prime}+\left(1+5 x^{2}\right) y=0\)

Short answer

Question: Find a fundamental set of Frobenius solutions for the given differential equation, \(3x^2(3+x^2)y'' + x(3+11x^2)y' + (1+5x^2)y = 0\), and provide explicit formulas for the coefficients. Answer: The fundamental set of Frobenius solutions for the given differential equation is given by \(y(x) = a_0 \sum_{n=0}^{\infty} a_n x^{n+r} + a_1 \sum_{n=0}^{\infty} a_n x^{n+r+1}\). The explicit formula for the coefficients is: \(a_{n+2} = -\frac{(n+r)(n+r+1)}{(n+2+r)(n+1+r)} \frac{1+5x^2}{3x^2(3+x^2)} a_n\).

Step-by-step solution

Transform into normal form

Divide the entire equation by \(3x^2(3+x^2)\): \(\frac{1}{3+x^2}y'' + \frac{1}{3x}\left(\frac{3+11x^2}{3+x^2}\right) y' + \frac{1+5x^2}{3x^2(3+x^2)}y = 0\)

Assume a Frobenius solution

Assume a solution of the form: \(y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\) where \(a_n\) are coefficients to be determined and \(r\) is a constant.

Calculate the derivatives

Calculate the first and second derivatives, and substitute them into the normal form of the ODE: \(y'(x) = \sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}\) \(y''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}\)

Substitute into the ODE

Substitute the derivatives into the normal form equation and group terms to find the recurrence relation: \(\sum_{n=0}^{\infty} [(n+r)(n+r-1)a_n x^{n+r-2} + \frac{1}{3x}\left(\frac{3+11x^2}{3+x^2}\right)(n+r)a_n x^{n+r-1} + \frac{1+5x^2}{3x^2(3+x^2)}a_n x^{n+r}] = 0\) We can notice that all terms have a common factor of \(x^{n+r-2}\), so we can remove this factor from the equation: \(\sum_{n=0}^{\infty} [(n+r)(n+r-1)a_n + \frac{1}{3x}\left(\frac{3+11x^2}{3+x^2}\right)(n+r)a_n x + \frac{1+5x^2}{3x^2(3+x^2)}a_n x^2] = 0\)

Find the recurrence relation

Now, we can find the recurrence relation by looking at the coefficients in front of the powers of \(x\). Note that for the series to be an identity, all the coefficients must be zero. Thus, we get: \((n+r)(n+r-1)a_n + \frac{1}{3x}\left(\frac{3+11x^2}{3+x^2}\right)(n+r)a_n x + \frac{1+5x^2}{3x^2(3+x^2)}a_n x^2 = 0\) This is a recurrence relation for \(a_n\), which allows us to find the coefficients \(a_n\) in terms of \(a_0\) and \(a_1\).

Solve the recurrence relation

Solve the recurrence relation to find explicit formulas for the coefficients \(a_n\): We see that if the power order of a term in the equation \(\ge2\)(and the coefficients are 0), it will cancel the remaining terms. Hence: \(a_{n+2} = -\frac{(n+r)(n+r+1)}{(n+2+r)(n+1+r)} \frac{1+5x^2}{3x^2(3+x^2)} a_n\) Using this relation, we can find the coefficients \(a_n\) for all \(n\) in terms of \(a_0\) and \(a_1\).

Construct the Frobenius solution

Now we can construct the Frobenius solution with the determined coefficients and the index \(r\): \(y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\) Substitute the coefficients found from the recurrence relation: \(y(x) = a_0 \sum_{n=0}^{\infty} a_n x^{n+r} + a_1 \sum_{n=0}^{\infty} a_n x^{n+r+1}\) This is the fundamental set of Frobenius solutions for the given differential equation with explicit formulas for the coefficients.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve functions of only one variable and their derivatives. They are fundamental to many areas of science and engineering because they model a variety of phenomena, from the motion of planets to the spread of disease. The ODE is classified as 'ordinary' to distinguish it from 'partial' differential equations, which involve partial derivatives of functions of more than one variable.
A simple example of an ODE is Newton's second law of motion, which can be written as \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration, which is the second derivative of position \( x \) with respect to time \( t \); this can be written as \( F = m\frac{d^2x}{dt^2} \). Solutions to an ODE are functions that satisfy the equation when substituted into it. Finding these solutions can be straightforward for simple ODEs but may require sophisticated mathematical techniques, like the Frobenius method, for more complicated cases.

Recurrence Relation

A recurrence relation is a mathematical expression that relates terms in a sequence to preceding terms. It's a way of defining a sequence where each term is a function of its predecessors, allowing one to calculate subsequent terms based on initial values. Recurrence relations are pivotal in solving ODEs when trying to find a general solution, particularly when using methods that expand the solution as an infinite series.
For instance, in the solution of our ODE using the Frobenius method, we encountered a recurrence relation that was a key step in finding the coefficients of the series, which could be expressed in terms of the previous coefficients. The result is a chain of coefficients, each linked to the previous one, allowing us to compute as many terms of the series as needed. These relations are not only fundamental in the context of differential equations but also appear in the study of algorithms, computer science, and even in defining sequences like the Fibonacci numbers, where each term is the sum of the two preceding numbers.

Differential Equation Solutions

The solutions of differential equations are the centerpiece of the study in this field. For an ordinary differential equation, a solution is a function that satisfies the equation for all values in a certain domain. In the context of the Frobenius method, solutions are presented in the form of power series.
The power series solution is an expansion that expresses the solution as an infinite sum of terms with increasing powers of the independent variable. It is an extremely powerful approach because it can approximate solutions to a high degree of accuracy, even for equations that are too complex for direct kinds of solution. Once a recurrence relation is established, as seen in the Frobenius method, it serves as a lynchpin to unravel the coefficients of the series that together, form the solution to the differential equation. The success of this method is particularly evident when dealing with differential equations with variable coefficients, where traditional techniques may falter. In such cases, a Frobenius-type solution provides a systematic way to find an explicit, and often convergent, series that describes the behavior of the solution over a range of the independent variable.