Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ y^{\prime \prime}+x y^{\prime}+(3 x+2) y=0 $$

Short answer

The general power series solution for the given differential equation \(y^{\prime\prime} + xy^{\prime} + (3x + 2)y = 0\) is: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$ Where \(a_n\) can be computed recursively by the relation \(a_n = - \frac{3 a_{n-1}}{n(n+1)}\).

Step-by-step solution

Assume power series solution

Assume that there exists a power series solution of the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$

Find the first and second derivatives

Next, we need to find the first derivative \(y^{\prime}\) and the second derivative \(y^{\prime\prime}\) of \(y(x)\). For the first derivative: $$ y^{\prime}(x) = \frac{d}{dx} \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} n a_n x^{n-1} $$ For the second derivative: $$ y^{\prime\prime}(x) = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2} $$

Substitute the power series solutions into the original equation

Now, substitute \(y(x), y^{\prime}(x),\) and \(y^{\prime\prime}(x)\) into the original differential equation: $$ \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2} + x \sum_{n=0}^{\infty} n a_n x^{n-1} + (3x + 2) \sum_{n=0}^{\infty} a_n x^n = 0 $$ Now, let's simplify it: $$ \sum_{n=0}^{\infty} [n(n-1) a_n x^{n-2} + n a_n x^n + 3 a_n x^{n+1} + 2 a_n x^n] = 0 $$ In order to find the recurrence relation for the coefficients, we should have the same powers of \(x\) in all terms. Therefore, we need to shift the indices of each sum to have the same power of \(x\).

Perform index shifting and find the recurrence relation for the coefficients

Let's rewrite the equation after performing the index shifts: $$ \sum_{n=2}^{\infty} [(n(n-1) a_n + n a_n + 3 a_{n-1})x^n] + \sum_{n=0}^{\infty} (2 a_n) x^n = 0 $$ From the equation above, we can see that all the powers of \(x\) cancel out when the terms inside the summations are equal: $$ n(n-1)a_n + n a_n + 3 a_{n-1} = 0 $$ Solving this equation for \(a_n\): $$ a_n = - \frac{3 a_{n-1}}{n(n+1)} $$ Now we have a recurrence relation to compute the coefficients.

Write down the general solution in terms of power series

Using the recurrence relation found in the previous step, we can compute the coefficients and write down the general power series solution, as follows: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$ Where \(a_n\) can be computed recursively by the relation \(a_n = - \frac{3 a_{n-1}}{n(n+1)}\). By finding the particular series solutions using initial conditions or additional information, we can obtain specific solutions to the given ordinary differential equation.

Key concepts

Power Series Solutions

Power series solutions offer a fascinating way to approach differential equations, especially when traditional methods may not be applicable. Imagine a power series as an infinite sum of terms involving powers of a variable, in this case, powers of \(x\). The proposed solution takes the form:

• \(y(x) = \sum_{n=0}^{\infty} a_n x^n\)

Each term in the series has a coefficient, \(a_n\), which dictates its weight in the solution. A power series solution attempts to represent the unknown function \(y(x)\) by adjusting these coefficients.
By substituting the series representation of the function and its derivatives back into the differential equation, one can derive conditions for the coefficients. These conditions usually manifest as a recurrence relation, which you will explore further.
Ultimately, power series solutions enable us to solve complex differential equations by decomposing a problem into simpler parts. This method also assists in providing solutions around a specific point, often benefiting scenarios involving boundary conditions or initial values.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving functions and their derivatives. They appear widely in disciplines like physics, engineering, and biology, modeling phenomena from mechanical vibrations to population dynamics.
The given equation is an ODE:

• \(y^{\prime \prime}+x y^{\prime}+(3x+2) y=0\)

It seeks an unknown function \(y(x)\) that satisfies the equation. The standard method with ODEs involves finding solutions through differentiation and integration techniques.
However, not every ODE can be solved easily by traditional methods. In such cases, alternative approaches like power series solutions become invaluable. They help find approximate solutions using series expansions and can handle cases where variables are non-linear or where coefficients are complex.

Recurrence Relation

A recurrence relation is a powerful tool in mathematics, allowing us to express terms in a sequence relative to their predecessors. In the context of power series solutions, they assist in computing the coefficients \(a_n\) systematically.
In the solution given for the ODE, we derived the relation:

• \(a_n = - \frac{3 a_{n-1}}{n(n+1)}\)

This equation indicates that each coefficient depends on the previous one, effectively forming a chain of dependencies. By knowing initial terms, such as \(a_0\), all subsequent terms \(a_1, a_2, \ldots\) can be determined recursively.
Recurrence relations allow us to extend our solution as far as desired. They provide insight into the behavior of solutions around a specified point, enhancing our understanding of the differential equation's solutions.

General Solution

The general solution in differential equations refers to the overarching solution that encompasses all potential specific solutions of the equation. In constructing a general power series solution, our goal is to obtain an expression that implies all possible functions \(y(x)\) that fulfill the differential equation.
For the power series representation, the general solution can be expressed as:

• \(y(x) = \sum_{n=0}^{\infty} a_n x^n\)

The coefficients \(a_n\) are found using the recurrence relation. With these coefficients determined for each term, the general solution captures every possible function that might resolve the equation.
Specific solutions can subsequently be derived by applying initial conditions or particular constraints. Thus, while the general solution holds the key to the broader picture, initial or boundary conditions allow us to pinpoint particular solutions needed in real-world applications.