Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ y^{\prime \prime}+x y^{\prime}+\left(2 x^{2}+1\right) y=0 $$

Short answer

Assume the power series for y(x) and its derivatives: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) \(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\) \(y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\) Substitute these into the given differential equation and match coefficients to obtain the recurrence relation: \(a_{n+2} = \frac{-n}{(n+1)(n+2)}a_n - \frac{2a_{n-2}}{(n+1)(n+2)}\) The general solution is given by \(y(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n\) satisfies the recurrence relation.

Step-by-step solution

1. Assume Power Series for y

: To begin, we assume a power series for the unknown function y(x) and its derivatives as follows: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) \(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\) \(y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\)

2. Substitute Power Series into Differential Equation

: Now, substitute the power series of y(x) and its derivatives into the given differential equation: \(\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x\sum_{n=1}^{\infty} n a_n x^{n-1} + (2x^2 + 1)\sum_{n=0}^{\infty} a_n x^n = 0\)

3. Match Coefficients

: Next, we will combine the three sums and match the coefficients of the resulting power series. The combined power series can be represented as: \(\sum_{n=0}^{\infty} P_n x^n = 0 \) where \(P_n\) denotes the coefficient of each term of the power x^n. For all x, the above equation must be true, so we match coefficients of like powers, which gives the following: \(P_n = (n+2)(n+1) a_{n+2} + n a_n + (2 a_{n-2} + a_n) = 0\)

4. Recurrence Relation

: From the above equation, we can find a recurrence relation for the power series coefficients: \(a_{n+2} = \frac{-n}{(n+1)(n+2)}a_n - \frac{2a_{n-2}}{(n+1)(n+2)}\) The general solution will be formed by \(y(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n\) satisfies the above recurrence relation. The problem is now reduced to determining the coefficients \(a_n\) and plugging them back into the series representation for the function y(x).

Key concepts

Differential Equations

Differential equations play a crucial role in understanding a variety of physical and mathematical phenomena. In this problem, we are dealing with a second-order linear differential equation: \[ y'' + xy' + (2x^2 + 1)y = 0 \] The goal is to find the function \(y(x)\) that satisfies this equation. A differential equation involves derivatives of an unknown function, and solving it typically means finding this function, often using various methods, one of which is power series solutions.

• Second-order: because the highest derivative is the second derivative.
• Linear: as it involves no more than the first powers of \(y\), \(y'\), and \(y''\).

Power series solutions are particularly useful when other methods become cumbersome or impossible. They provide solutions in the form of an infinite series, which can handle more complex behaviors over different intervals.

Recurrence Relation

A recurrence relation is a way of defining the terms in a sequence using the terms that have come before it. In this exercise, it is a crucial step in the power series method. Once you substitute the power series into the differential equation and collect terms, you end up with an equation involving coefficients like \(a_n\). This equation is: \[ a_{n+2} = \frac{-n}{(n+1)(n+2)}a_n - \frac{2a_{n-2}}{(n+1)(n+2)} \] This relation allows the computation of higher order terms based on the known terms. - Each term in the series for \(y(x)\) is expressed in terms of its predecessors.- Solving it involves determining \(a_0\) and \(a_1\) as initial conditions, then computing all higher terms recursively.

Coefficients Matching

When the power series is substituted into the differential equation, the result is a long series where every power of \(x\) must independently satisfy the equation. This means that the coefficients of each power (referred to as \(P_n\)) must all be zero. This is how we arrive at: \[ P_n = (n+2)(n+1) a_{n+2} + n a_n + (2 a_{n-2} + a_n) = 0 \] By matching coefficients, we ensure that the series as a whole satisfies the original differential equation. This often requires:- Aligning terms by their power of \(x\).- Solving for coefficients such as \(a_{n+2}\) in terms of previous coefficients.Without this step, achieving a valid power series solution would be challenging, as the unsatisfied terms would persist to ruin the overall solution.

Power Series Assumption

The power series assumption is the starting point in this method. It proposes that the solution \(y(x)\) can be expressed as an infinite sum of terms involving powers of \(x\): \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \] Here, each \(a_n\) is a coefficient that needs to be found. This assumption turns a potentially complex problem into a more manageable form by converting differentiation into a process of working with coefficients. - The series has no upper limit (infinite) but is often truncated in practice for approximations.- Converts the differential equation into an algebraic problem by substituting derivatives of this power series back into the original equation.- Greatly simplifies handling non-linear differential terms by treating them in sections.This method is advantageous because if convergent, the power series gives a good approximation of \(y(x)\) for values near a given point, often around \(x = 0\).