Question

Solve the given differential equation by means of a power series about the given point \(x_{0} .\) Find the recurrence relation; also find the first four terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution. \(y^{\prime \prime}+x y^{\prime}+2 y=0, \quad x_{0}=0\)

Short answer

Answer: The first four terms for the first solution (y₁(x)) are: 1 - x² The first four terms for the second solution (y₂(x)) are: x - ¹/₃x³

Step-by-step solution

Assume a power series solution for y(x)

Assume the power series solution for y(x) has the form: \(y(x) = \sum_{n=0}^{\infty} a_nx^n\) Now we need to find the first and second derivatives of y(x) with respect to x: \(y^{\prime}(x) = \sum_{n=1}^{\infty} na_nx^{n-1}\) \(y^{\prime\prime}(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\)

Substitute the power series into the given differential equation

Substitute the expressions we found for y, y', and y'' into the given differential equation: \(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + x\sum_{n=1}^{\infty} na_nx^{n-1} + 2\sum_{n=0}^{\infty} a_nx^n = 0\)

Simplify and match coefficients of like powers of x

To simplify the above equation, first shift the indices of the series to have a common power of x in all the series: \(\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n + \sum_{n=0}^{\infty} (n+1)a_{n+1}x^n + 2\sum_{n=0}^{\infty} a_nx^n = 0\) Now, combine the series: \(\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n+1)a_{n+1} + 2a_n]x^n = 0\) To satisfy the above equation, the coefficients of like powers of x on the left side must be zero: \((n+2)(n+1)a_{n+2} + (n+1)a_{n+1} + 2a_n = 0\)

Solve for the recurrence relation

From the previous equation, we have the recurrence relation: \(a_{n+2} = -\frac{(n+1)a_{n+1} + 2a_n}{(n+2)(n+1)}\)

Determine the general term and find the first four terms

Using the recurrence relation, we can find the general term and the first four terms of two linearly independent solutions. The general term can be found by applying the recurrence relation repeatedly. However, as we need only the first four terms, we can find them directly using the recurrence relation: For the first solution, let \(a_0 = 1\) and \(a_1 = 0\). Then, \(a_2 = -\frac{2a_0}{(2)(1)} = -1\) \(a_3 = -\frac{2a_1}{(3)(2)} = 0\) So the first four terms for the first solution are: \(1 - x^2\) For the second solution, let \(a_0 = 0\) and \(a_1 = 1\). Then, \(a_2 = -\frac{2a_0}{(2)(1)} = 0\) \(a_3 = -\frac{2a_1}{(3)(2)} = -\frac{1}{3}\) So the first four terms for the second solution are: \(x - \frac{1}{3}x^3\) The two linearly independent solutions are, therefore: \(y_1(x) = 1 - x^2\) \(y_2(x) = x - \frac{1}{3}x^3\) These are the first four terms of each solution.

Key concepts

Power Series Solution

When solving differential equations, a power series solution offers an approach that approximates the solution by an infinite sum of algebraic terms, each term being a multiple of a power of the variable. This method is particularly useful when finding an exact solution is complex or impossible. In the case of the given exercise, we consider a power series centered around a given point, typically the origin or another specified value, referred to as \( x_0 \) . The assumption for the solution of the differential equation is made as follows:
\[ y(x) = \sum_{n=0}^\infty a_nx^n \]
where \( a_n \) represents the series coefficients that we aim to determine. An effective power series solution should start with calculating the derivatives required by the differential equation and then substituting them into the equation. Next, by equating the coefficients of the same powers of \( x \) to zero, we establish a linkage between the coefficients in the form of a recurrence relation.

Recurrence Relation

A recurrence relation is an equation that expresses each term of a sequence as a function of the preceding terms. In the context of power series solutions to differential equations, it provides a systematic method to find the coefficients of the series. To determine a recurrence relation, we collect terms with like powers of \( x \) after substituting the series' derivatives into the differential equation. This creates a new equation where each coefficient's value depends on the values of other, often preceding, coefficients. For the given differential equation:
\[ (n+2)(n+1)a_{n+2} + (n+1)a_{n+1} + 2a_n = 0 \]
This establishes a pattern or formula that we can use recursively to find all subsequent coefficients from the initial ones, which significantly simplifies the series construction and leads toward the solution.

Linearly Independent Solutions

The concept of linearly independent solutions is crucial in the theory of differential equations. When dealing with linear differential equations, it is often the case that there are multiple solutions. For the solutions to be useful, especially in forming a general solution to the differential equation, they must be linearly independent. This independence means that no solution in the set can be written as a linear combination of the others. In our exercise, to exhibit two such solutions, we choose different initial conditions for the series coefficients \( a_0 \) and \( a_1 \) to generate distinct power series. By doing this, we ensure that our solutions—one captured by an even function set from \( a_0 \) and another by an odd function set from \( a_1 \)—are inherently dissimilar and cannot be combined through multiplication by a constant, hence satisfying the linear independence criterion.

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In practical terms, it describes a relationship involving rates of change and is a cornerstone in depicting physical, biological, economic systems, and more. The given exercise presents a second-order linear homogeneous differential equation:
\[ y''+xy'+2y=0 \]
These equations are classified by their order, determined by the highest derivative present, and linearity, which, in this case, implies the absence of terms like \( (y')^2 \) or \( y^2 \) that would make solving significantly more complicated. A 'homogeneous' equation is one where all terms are proportional to the function or its derivatives, without an extra independent term (i.e., it equals zero). The power series method applied to such equations can reveal general and particular solutions, assisting in understanding complex dynamic systems.