Question

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. $$2 x y^{\prime \prime}+y^{\prime}+2 y=0$$

Short answer

\( r = \frac{1}{4} \pm i \sqrt{\frac{15}{16}} \)

Step-by-step solution

Rewrite the Differential Equation

First, write the given differential equation in the standard form suitable for the Frobenius method: \[ 2x y'' + y' + 2y = 0 \] Divide through by 2x to make the coefficient of \( y'' \) equal to 1: \[ y'' + \frac{1}{2x} y' + \frac{y}{x} = 0 \]

Assume a Power Series Solution

Assume a solution of the form: \[ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} \] where \( r \) is the indicial exponent to be determined.

Compute Derivatives

Find the first and second derivatives of the power series: \[ y'(x) = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} \] \[ y''(x) = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} \]

Substitute into the Differential Equation

Substitute the series representations of \( y \), \( y' \), and \( y'' \) into the differential equation: \[ \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} + \frac{1}{2x} \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} + \frac{1}{x} \sum_{n=0}^{\infty} a_n x^{n+r} = 0 \]

Collect and Simplify Terms

To combine the series into one, adjust the indices of the sums where necessary so that all are expressed in terms of the same power of \( x \): \[ \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} + \sum_{n=0}^{\infty} a_n \frac{(n+r)}{2} x^{n+r-2} + \sum_{n=0}^{\infty} a_n x^{n+r-2} = 0 \]

Factor Out Common Terms

Factor out \( x^{n+r-2} \) and combine like terms: \[ \sum_{n=0}^{\infty} \left[ a_n (n+r)(n+r-1) + a_n \frac{(n+r)}{2} + a_n \right] x^{n+r-2} = 0 \]

Solve the Indicial Equation

Set the coefficient of the lowest power of \( x \) equal to zero to find the indicial equation: \[ (r)(r-1) + \frac{r}{2} + 1 = 0 \] Solving this quadratic equation gives: \[ r^2 - \frac{1}{2} r + 1 = 0 \] The roots of this equation are: \[ r = \frac{1}{4} \pm i \sqrt{\frac{15}{16}} \]

Find Recurrence Relation

With the roots found, use them to determine the recurrence relation for \( a_n \). The recurrence relation will allow you to determine all the coefficients \( a_n \) in terms of \( a_0 \), depending on the initial conditions.

Key concepts

Power Series Solutions

Power series solutions are a fundamental concept in solving differential equations, especially where the solutions are not straightforward. The method involves expressing the solution as an infinite sum of terms (a series) in powers of the variable, typically denoted as \(x\). For example, in our differential equation:

• We assume the solution is of the form \( y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} \), where \(a_n\) are the coefficients to be determined, and \(r\) is the indicial exponent.
• This method is particularly useful in handling linear differential equations with variable coefficients.
• Each term in the series needs to satisfy the differential equation when plugged back in, leading to a system of equations for the coefficients \(a_n\).

Using a power series, we convert a complex differential equation into simpler algebraic equations.

Differential Equations

Differential equations are equations that involve the derivatives of a function and the function itself. They are essential in modeling various real-world phenomena where change rates are important, such as in physics, engineering, and economics.

• The goal is to find a function \(y(x)\) that satisfies the given equation.
• In our example, we dealt with a second-order linear differential equation with variable coefficients: \( 2xy'' + y' + 2y = 0 \).
• Such equations can often be simplified and solved using the Frobenius method by rewriting them in a standard form and assuming a power series solution.
• The process involves manipulating derivatives and coefficients into a manageable form to solve for the unknowns step by step.

This systematic approach helps break down complex problems into simpler, solvable parts.

Indicial Exponent

The indicial exponent, \( r \), is a key component of the Frobenius method. It represents the exponent of the first term in the power series expansion of the solution.

• Finding \( r \) involves solving the indicial equation, which arises from the lowest power of \( x \) in the series.
• In our example, we derive the indicial equation from the term: \( (r)(r-1) + \frac{r}{2} + 1 = 0 \).
• Solving this quadratic equation gives the possible values for \( r \), which may be real or complex.
• The roots of the indicial equation help determine the structure of the series solution and ensure that all terms in the series properly balance.

Understanding the indicial exponent is crucial as it influences the form and convergence of the solution.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence: each term of the sequence is a function of the preceding terms. In the context of power series solutions for differential equations, the recurrence relation helps determine the coefficients \(a_n\) of the series.

• Once the indicial exponent \( r \) is found, it can be used to derive the recurrence relation.
• We substitute the power series back into the differential equation and collect like terms to form the recurrence relation for \( a_n \).
• For our example, the recursion would determine each coefficient in terms of the previous ones, ensuring all terms satisfy the differential equation.
• This relation is essential for generating the entire series from the initial conditions or given values.

Mastering recurrence relations allows for systematic construction of power series solutions, providing a clear methodical path to solving otherwise complex differential equations.