Question

Solve the following differential equations by power series. $$ y^{\prime \prime}+x y^{\prime}+y=0 $$

Short answer

The general solution is \( y(x) = a_0 \cosh(x) + a_1 \sinh(x) \).

Step-by-step solution

Assume a Power Series Solution

Assume a solution of the form \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \].

Derive the First and Second Derivatives

The first derivative is \[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \] and the second derivative is \[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \].

Substitute the Power Series into the Differential Equation

Substitute the expressions for \(y(x)\), \(y'(x)\), and \(y''(x)\) back into the original differential equation to get: \[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x \sum_{n=1}^{\infty} n a_n x^{n-1} + \sum_{n=0}^{\infty} a_n x^n = 0 \].

Simplify and Combine the Series

Adjust the indices of the sums so that each term has the same power of \(x\). \[ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=0}^{\infty} a_n x^n = 0 \]. Combine like terms: \[ a_0 + (2a_2) + \sum_{n=1}^{\infty} [ (n+2)(n+1) a_{n+2} + (n+1)a_n ] x^n = 0 \].

Find the Recurrence Relation

Equate the coefficients of \(x^n\) to zero to find the recurrence relation: For \(n=0\): \[ a_0 + 2a_2 = 0 \]. For \(n \geq 1\): \[ (n+2)(n+1) a_{n+2} + (n+1)a_n = 0 \]. Simplify to get: \[ a_{n+2} = -\frac{a_n}{(n+2)} \].

Determine the Coefficients

Starting from the initial conditions: Assume \( a_0 \) and \( a_1 \) are arbitrary constants. Then, \[ a_2 = -\frac{a_0}{2} \] and \[ a_4 = -\frac{a_2}{4} = \frac{a_0}{4 \cdot 2} = \frac{a_0}{8} \], and so on.

Write the General Solution

The general solution in power series form is: \[ y(x) = a_0 (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots) + a_1 (x - \frac{x^3}{3!} + \ldots) \] which can also be written as \[ y(x) = a_0 \cosh(x) + a_1 \sinh(x) \] using the known series expansions for hyperbolic functions.

Key concepts

Differential Equations

A differential equation is an equation involving unknown functions and their derivatives. They are fundamental in describing various phenomena like heat conduction, wave propagation, and population dynamics. Primarily, there are two types of differential equations: ordinary differential equations (ODEs) and partial differential equations (PDEs). In this exercise, we focus on ODEs, which involve functions of a single variable and their derivatives. The given equation is \[y^{\prime \prime}+x y^{\prime}+y=0\] Solving differential equations often requires finding a specific function or set of functions that satisfy the equation. We can solve this equation using several techniques, one of which is the power series method.

Power Series

A power series is an infinite sum of the form \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \] where \(a_n\) are coefficients and \(x\) is the variable. Power series represent functions as infinite sums, useful for solving differential equations. The basic idea is to express the solution as a sum of powers of \(x\). To solve the differential equation with a power series, we first assume the solution can be written in such form. We then find its first and second derivatives: \[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \] and \[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \] Substituting these into the differential equation and collecting like terms provide us a way to design a 'recurrence relation'.

Recurrence Relation

A recurrence relation defines each term of the sequence based on previous terms. The result from the supplied power series eventually leads to a recurrence relation: For \(n = 0\): \[ a_0 + 2a_2 = 0 \] For \(n \geq 1\): \[ (n+2)(n+1) a_{n+2} + (n+1)a_n = 0 \] Simplifying gives: \[ a_{n+2} = -\frac{a_n}{(n+2)} \] This relation helps determine values of higher order terms in the series using initial terms \(a_0\) and \(a_1\). By recursively applying the relation, we can express all coefficients in terms of \(a_0\) and \(a_1\). Recurrence relations are fundamental in various areas such as numerical analysis, computer algorithms, and in our case, solving differential equations.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, rather than a circle. They are broadly used in engineering and physics. The basic hyperbolic functions are: \[\sinh(x) = \frac{e^x - e^{-x}}{2} \] and \[\cosh(x) = \frac{e^x + e^{-x}}{2} \] These have corresponding power series expansions: \[ \begin{aligned}\sinh(x) &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \cosh(x) &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \end{aligned}\] In the final step of the solution, we recognize the power series of our general solution matches that of \(\cosh(x)\) and \(\sinh(x)\). We can, therefore, capture the solution elegantly as: \[ y(x) = a_0 \cosh(x) + a_1 \sinh(x) \] This provides a neat and insightful form aligning with familiar hyperbolic functions.