Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}+6 x y^{\prime}-y=0\)

Short answer

Question: Given the second-order linear homogeneous differential equation \(x^2 y'' + 6x y' - y = 0\), find its general solution using power series. Answer: The general solution to the given differential equation is \(y(x) = a_1\left(x - \frac{x^3}{3\cdot 4} + \frac{x^5}{5\cdot 6}\cdots\right)\), where \(a_1\) is an arbitrary constant.

Step-by-step solution

Rewrite the differential equation

Write the given differential equation in a more familiar form: \(x^2 y'' + 6x y' - y = 0\)

Power series solution attempt

Assume a power series solution of the form: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\)

Calculate the first and second derivatives

Calculate the first and second derivatives of \(y(x)\) with respect to \(x\): \(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\) \(y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\)

Substitute the power series and its derivatives into the differential equation

Plug \(y(x)\), \(y'(x)\), and \(y''(x)\) into the differential equation: \(x^2\left(\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\right) + 6x\left(\sum_{n=1}^{\infty} n a_n x^{n-1}\right) -\left(\sum_{n=0}^{\infty} a_n x^n\right) = 0\)

Simplify and combine the power series

Simplify the power series expressions in the differential equation and combine them into a single power series: \(\sum_{n=2}^{\infty} n(n-1) a_n x^n + \sum_{n=1}^{\infty} 6n a_n x^n - \sum_{n=0}^{\infty} a_n x^n = 0\)

Equate coefficients

Since this equation is true for all \(x\) in the valid interval, we can equate the coefficients of the power series terms: For \(n = 0\): \(-a_0 = 0\) (this implies that \(a_0 = 0\)). For \(n \ge 1\): \(n(n-1)a_n + 6na_n - a_n = 0\)

Find the recurrence relation

Solve the coefficients equation for \(a_n\): \(a_n = \frac{1}{n(n+1)}a_{n-2}\) for \(n \ge 1\)

Construct the general solution

Use the recurrence relation to find the general power series solution to the differential equation: \(y(x) = a_0 \left(1 - \frac{x^2}{2\cdot 3} + \frac{x^4}{4\cdot 5}\cdots\right) + a_1\left(x - \frac{x^3}{3\cdot 4} + \frac{x^5}{5\cdot 6}\cdots\right)\) Since \(a_0 = 0\), the general solution is: \(y(x) = a_1\left(x - \frac{x^3}{3\cdot 4} + \frac{x^5}{5\cdot 6}\cdots\right)\) This solution is valid in any interval not including the singular point.

Key concepts

Power Series Solution

A power series solution is a method used to solve differential equations by expressing the unknown function as an infinite sum of powers. This approach is often used when the differential equation has variable coefficients or a non-ordinary point, which makes finding a closed-form solution difficult.

The power series solution begins by assuming that the solution can be written in the form
\(y(x) = \textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\textstyle\)This involves calculating the derivatives of \(y(x)\) and plugging them into the given differential equation. After simplification, the resulting equation's coefficients can be set to zero to find relations between the terms of the power series. This method converts the problem of solving a differential equation into a problem of identifying a series of coefficients.

Recurrence Relation

A recurrence relation in the context of differential equations is a formula that relates each term in the power series with one or several of its predecessors. Its purpose is to establish a systematic way to determine all the coefficients of the power series.

In the given exercise, after the power series and its derivatives are substituted into the differential equation, a reorganized series leads to a recurrence relation: \(a_n = \frac{1}{n(n+1)}a_{n-2}\) for \(n \ge 1\). This allows us to find all terms of the power series by successively applying this formula, starting from a known term. In this case, since \(a_0 = 0\), all even-index terms disappear, making the computation more straightforward.

General Solution

The general solution of a differential equation is an expression that contains all possible solutions of the equation. It usually includes arbitrary constants, which can be determined by initial conditions or other constraints.

In power series solutions, these arbitrary constants are the first terms of the series, \(a_0\) and \(a_1\). By applying the recurrence relation, the general solution for the specific exercise given is represented by a single power series that encapsulates all possible solutions, depending on the value of \(a_1\). The final expression simplifies to a series in terms of \(a_1\) because \(a_0 = 0\), and the series inherits the convergence properties from the assumptions made while constructing the series.

Differential Equation Coefficients

In differential equations, the coefficients are the functions that multiply the various derivatives of the unknown function. These coefficients play a pivotal role in determining the nature of the equation and the appropriate methods for finding a solution.

In the case of a power series solution, equating the coefficients of like powers of \(x\) from both sides of the equation leads to a system of equations. Solving this system gives the specific values or relations for these coefficients, which are fundamental in constructing the power series solution. For the exercise provided, the coefficients of the power series are derived by equating the series resulted from the substitution of the derivatives back into the original equation, leading to a recurrence relation that is used to find the value of all the subsequent coefficients in terms of the initial ones.