Question

Find the power series in \(x\) for the general solution. $$ \left(1+x^{2}\right) y^{\prime \prime}+6 x y^{\prime}+6 y=0 $$

Short answer

Question: Determine the general solution of the given second-order linear differential equation using the power series method. Given Differential Equation: $$(1+x^{2})y''+6xy'+6y=0$$ Answer: The general solution of the given differential equation is given by: $$ y = a_0 \sum_{n=0}^{\infty} a_n x^n + a_1 \sum_{n=1}^{\infty} a_n x^n $$ where $a_0$ and $a_1$ are constants of integration, and the coefficients $a_n$ are given by the recursion relation: $$ a_{n+2} = \frac{-n(n+1)a_n-6(n+1)a_{n+1}-6a_n}{n(n-1)} $$

Step-by-step solution

Write down the general power series representation of \(y\), \(y'\), and \(y''\).

Assume that the solution to the given differential equation has the form of a power series: $$ y = \sum_{n=0}^{\infty} a_n x^n $$ Differentiate \(y\) with respect to \(x\) to find \(y'\), and then differentiate again to find \(y''\). The derivatives are: $$ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} $$ $$ y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} $$

Substitute the power series representations into the given differential equation.

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

Simplify and regroup terms.

Now, rearrange the equation so that terms with the same power of \(x\) are grouped together: $$ \sum_{n=0}^{\infty}(n(n-1)a_{n+2}+n(n+1)a_n+6(n+1)a_{n+1}+6a_n)x^n = 0 $$

Equate coefficients of corresponding powers of \(x\).

Since the terms on the right side of the equation are all zeros, equate coefficients of corresponding powers of \(x\) on the left side of the equation to zero: $$ n(n-1)a_{n+2}+n(n+1)a_n+6(n+1)a_{n+1}+6a_n = 0 $$ This is a recursion relation for the coefficients \(a_n\).

Solve the recursion relation.

Solve the above recursion relation for \(a_{n+2}\): $$ a_{n+2} = \frac{-n(n+1)a_n-6(n+1)a_{n+1}-6a_n}{n(n-1)} $$ Plug in the initial values of \(a_0\) and \(a_1\) as constants and use the above recursion relation to find the values of the other coefficients, \(a_2, a_3, a_4, \dots .\)

Write down the general solution using power series.

Now that we have a recursion relation for the coefficients, we can write down the general solution representing the power series: $$ y = a_0 \sum_{n=0}^{\infty} a_n x^n + a_1 \sum_{n=1}^{\infty} a_n x^n $$ This general solution represents the linearly independent solutions of the given differential equation, where \(a_0\) and \(a_1\) are the constants of integration.

Key concepts

Differential Equations

A differential equation is a type of equation that involves a function and its derivatives. It tells us how a particular quantity changes with respect to changes in another quantity, often time or space. In the context of our problem, we deal with a second-order linear differential equation given by \((1+x^2) y'' + 6x y' + 6y = 0\). This equation describes how the function \(y(x)\) changes based on its own value and the values of its first and second derivatives.
To solve differential equations, different methods can be employed, such as separation of variables, integrating factors, or, as in this exercise, power series solutions.
Power series solutions are useful when the equation does not have a simple solution or when solving around a particular point where the function is analytic. We express the solution as an infinite series, which can be easier to handle than directly solving complex differential equations.

Recurrence Relations

In the context of differential equations, a recurrence relation gives us a way to calculate the terms of a sequence based on the previous terms. When we express a function as a power series, its coefficients must satisfy certain conditions. These conditions form the recurrence relation.
In the given exercise, once we substitute the power series into the differential equation and simplify, we obtain the recurrence relation: \(a_{n+2} = \frac{-n(n+1)a_n - 6(n+1)a_{n+1} - 6a_n}{n(n-1)}\). This relation allows us to compute the coefficients \(a_2, a_3, a_4, \ldots\) based on the initial values of \(a_0\) and \(a_1\), which are constants.

• These coefficients help in constructing the series that form the solution to the differential equation.
• Being able to determine these terms is key in obtaining the power series solution and understanding the behavior of the function \(y(x)\) around the point of expansion.

Linear Independence

Linear independence is a principle from linear algebra that applies to functions as well as to vectors. When we consider solutions to differential equations, particularly in a power series form, we aim to find solutions that are linearly independent.
The reason we look for linear independence is to ensure that each solution we find provides new information about the system rather than being a mere derivative or combination of other solutions.
In our exercise, the final power series solution includes constants \(a_0\) and \(a_1\). By adjusting these constants, we obtain two linearly independent solutions, which together form a general solution to the differential equation.

• Each solution contributes differently to the behavior of the differential system.
• This means that they span a solution space, capturing all possible behaviors of the system described by the differential equation.

Linear independence is crucial for understanding the complete picture of the system's dynamics.