Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ y^{\prime \prime}+x y^{\prime}+\left(x^{2}-4\right) y=0 $$

Short answer

The power series solution of the given differential equation \(y'' + xy' + (x^2 - 4)y = 0\) is: \( y(x) = \sum_{n=0}^{\infty} a_n x^n \) Where the coefficients \(a_n\) satisfy the recursion formula: \( a_{n+2} = \frac{4 a_n - a_{n-2}}{(n+1)(n+2)} \) The general solution can be written explicitly in terms of the arbitrary constants a_0 and a_1: \( y(x) = a_0 \sum_{n=0}^{\infty} \frac{(4^n -(-1)^n)}{(2n+1)!} x^{2n} + a_1 \sum_{n=0}^{\infty} \frac{(4^n -(-1)^{n+1})}{(2n+2)!} x^{2n+1} \)

Step-by-step solution

Assume the power series form for y

For any function, the power series can be represented as: \( y(x) = \sum_{n=0}^{\infty} a_n x^n \)

Differentiate the power series to find first and second derivatives

Differentiate y with respect to x to get the first derivative: \( y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \) Now differentiate the first derivative (y') with respect to x to get the second derivative: \( y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \)

Plug the power series of y, its first derivative, and the second derivative back to the given differential equation

To find the power series solutions of this differential equation \(y'' + xy' + (x^2 - 4)y = 0\), we need to plug the power series of y, y', and y'' into the differential equation: \( \left(\sum_{n=2}^{\infty}n(n-1) a_n x^{n-2}\right) + x \left(\sum_{n=1}^{\infty}n a_n x^{n-1}\right) + \left(x^{2} - 4\right) \left(\sum_{n=0}^{\infty} a_n x^n\right) = 0 \)

Collect like terms and find the coefficients

We first multiply the power series by their corresponding coefficients and shift indices to match the power of x: \( \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^{n} + \sum_{n=1}^{\infty}n a_n x^n + \sum_{n=2}^{\infty} a_{n-2} x^{n} - 4\sum_{n=0}^{\infty} a_n x^n = 0 \) Now collect the terms that have the same power of x and rewrite the equation: \( \sum_{n=0}^{\infty} [(n+2)(n+1) a_{n+2} + n a_n + a_{n-2} - 4 a_n] x^n = 0 \) For any x, the power series would be equal to 0 if the coefficient of every power of x is equal to 0. So we have: \( (n+2)(n+1) a_{n+2} + n a_n + a_{n-2} - 4 a_n = 0 \) Rearrange the equation to get the recursion formula for a_n: \( a_{n+2} = \frac{4 a_n - a_{n-2}}{(n+1)(n+2)} \)

Write the general solution of the power series

By solving the given recursion formula, we can find all the coefficients of the power series and then write the general solution in terms of these coefficients by using the assumed power series form for y: \( y(x) = \sum_{n=0}^{\infty} a_n x^n \) Since we know the relation between a_n, a_{n+2}, and a_{n-2}, we can rewrite the solution explicitly in terms of just a_0 and a_1 (two arbitrary constants).

Key concepts

Differential Equations

A differential equation is an equation that involves a function and its derivatives. These equations describe a wide variety of phenomena in engineering, physics, economics, and many other fields.
To solve a differential equation means finding the function or functions that satisfy the equation.

In this context:

• The given differential equation is a second-order equation: \[ y'' + xy' + (x^2 - 4)y = 0 \]
• It indicates that the solution, or the function \( y(x) \), is to be expressed in terms of its second derivative \( y'' \), first derivative \( y' \), and the function itself \( y \).
  

When we assume a power series solution for \( y(x) \), we transform a complex differential equation problem into a simpler algebraic one. This makes it easier to find a solution.

Recurrence Relations

Recurrence relations are equations that define sequences of numbers. Each term is defined as a function of one or more previous terms.

In our exercise:

• We derive a recurrence relation from the differential equation by equating coefficients of like powers of \( x \). This helps find the coefficients of the power series solution.
• The recurrence relation obtained is:
  \[ a_{n+2} = \frac{4 a_n - a_{n-2}}{(n+1)(n+2)} \]

This means that each term \( a_{n+2} \) of the series is calculated based on its relationship with previous terms \( a_n \) and \( a_{n-2} \). Recurrence relations help simplify and solve power series by reducing them down to a systematic formula for the terms.

Homogeneous Equations

A homogeneous differential equation is one where every term is a multiple of the dependent variable or its derivatives. The characteristic of homogeneity is that if a certain \( y(x) \) is a solution, so is any constant multiple of \( y(x) \).

In our example, the equation:
\[ y'' + xy' + (x^2 - 4)y = 0 \] is homogeneous because each term is a product of either \( y \), \( y' \), or \( y'' \) with some function of \( x \).

This leads to solutions that can be expressed as linear combinations of functions, where the series coefficients are derived from a constant or arbitrary value, typically denoted \( a_0 \) and \( a_1 \).
This reflects the fundamental nature of linear homogeneous equations; they inherently possess symmetrical solutions that make operations like power series expansion straightforward.