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

The series solution is obtained by solving the recurrence relation formed from the coefficients of the generalized power series.

Step-by-step solution

Identify the Differential Equation

Given the differential equation: \[2 x y^{\times \times} - y^{\times} + 2 y = 0.\] Identify it as a second-order linear differential equation.

Transform to Standard Form

Divide the entire equation by 2x to transform it into standard form: \[y^{\times \times} - \frac{1}{2x} y^{\times} + \frac{1}{x} y = 0.\]

Assume Series Solution

Assume a solution of the form:\[y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}.\] This is a generalized power series for the solution.

Compute Derivatives

Compute the first and second derivatives of the assumed series solution:\[y^{\times} (x) = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1}, \]\[y^{\times \times} (x) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2}.\]

Substitute Series into Differential Equation

Substitute the series and its derivatives into the standard form of the differential equation:\[ \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} - \frac{1}{2x} \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} + \frac{1}{x} \sum_{n=0}^{\infty} a_n x^{n+r} = 0.\]

Combine into a Single Series

Combine the series into a single summation and match the powers of x:\[ \sum_{n=0}^{\infty} \left[ (n+r)(n+r-1) a_n - \frac{(n+r)}{2} a_n x + 2 a_n \right] x^{n+r-2} = 0.\]

Set Coefficients to Zero

Set the coefficients of each power of x to zero to form a recurrence relation:\[ (n+r)(n+r-1) a_n - \frac{1}{2} (n+r) a_n + 2 a_n = 0.\]

Solve Recurrence Relation

Simplify the recurrence relation to find the coefficients. Factor out common terms and solve for \(a_n\): \[ (n+r)[(n+r-1) - \frac{1}{2}] a_n + 2 a_n = 0.\]

Determine Series Solutions

By solving the recurrence relation, find the values of r and consequently the general series solutions.

Key concepts

second-order linear differential equations

Let's start by understanding what a second-order linear differential equation is. These equations are of the form:
\[ a(x) y'' + b(x) y' + c(x) y = 0. \]
Here, the highest derivative is the second derivative \( y'' \). These equations often describe physical systems where the rate of change of a quantity depends on the quantity itself and its rate of change.

• Homogeneous vs Inhomogeneous: The equation we are dealing with is homogeneous because it is set equal to zero.
• Linear: It’s linear because the dependent variable \( y \) and its derivatives appear to the power of one and are not multiplied together.
• Transform to Standard Form: For solving, we transform it to \( y'' + p(x)y' + q(x)y = 0 \).

Understanding the basics of second-order linear differential equations helps lay the groundwork for using methods like the Frobenius Method.

generalized power series

To solve a differential equation using the Frobenius Method, we assume a solution in the form of a generalized power series:

• Form: \( y(x) = \sum_{n=0}^{\text{\textbackslash infinity}} a_n x^{n+r} \).
• Coefficients: Here, \( a_n \) are the coefficients that will be determined, and \( r \) is a constant that we also need to find.
• Why Use It: This form generalizes the simple power series by including an additional exponent \( r \), which helps to solve around singular points.

The generalized power series is crucial because it captures solutions that might not be evident through simpler methods.
You begin by assuming this form and then proceed to find its derivatives for substitution into the differential equation.

recurrence relation

A key step in the Frobenius Method is obtaining and solving a recurrence relation. This relation helps determine the coefficients \( a_n \) of the series.

• Setting Up: Substitute the generalized power series back into the differential equation.
• Combine Terms: Combine all terms to express the equation as a single series.
• Match Powers: Match coefficients for each power of \( x \), equating them to zero to get the recurrence relation.

The recurrence relation looks like:
\[ (n+r)[(n+r-1) - \frac{1}{2}] a_n + 2 a_n = 0 \]
Solving this recurrence relation helps you find the coefficients \( a_n \), which are then used to write the final solution.
This step requires careful algebra and is crucial for pulling out the correct series solution.

series solutions

Finally, we arrive at the series solutions, which are the core result of applying the Frobenius Method. Once all coefficients \( a_n \) are found, the solution to the differential equation is expressed through these series.

• Forming the Solution: Substitute the coefficients \( a_n \) back into the generalized power series \( y(x) = \sum_{n=0}^{\text\textbackslash{infinity}} a_n x^{n+r} \).
• Convergence: The series solution generally has a radius of convergence to be determined.
• Multiple Solutions: Often, two independent series solutions exist, corresponding to different values of \( r \).

Understanding series solutions allows you to see how complex differential equations can be tamed into an infinite but manageable sum. Each term of the series offers a part of the solution.
By combining all steps, you transform the differential equation into a series that represents the complete solution for the problem at hand.