Question

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

Short answer

The general solution is \( y = a_0 \cos x + a_1 \sin x \).

Step-by-step solution

Identify the Differential Equation

Given the differential equation: \( y'' + xy = 0 \)

Assume a Power Series Solution

Assume a solution of the form: \( y = \sum_{n=0}^{\infty} a_n x^n \)then, the first and second 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 Series into the Differential Equation

Substitute \( y \) and \( y'' \) into the equation \( \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x \sum_{n=0}^{\infty} a_n x^n = 0 \)

Adjust Index to Combine Series

Adjust the index of the first series: \( \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + \sum_{n=0}^{\infty} a_n x^{n+1} = 0 \)Combine into one series: \( \sum_{n=0}^{\infty} [(n+2)(n+1) a_{n+2} + a_{n-1}] x^n = 0 \)

Find the Recurrence Relation

For the series to be zero for all \( x \), the coefficients of each power of \( x \) must be zero: \( (n+2)(n+1) a_{n+2} + a_{n-1} = 0 \)Rearrange the equation to form the recurrence relation: \( a_{n+2} = -\frac{a_{n-1}}{(n+2)(n+1)} \)

Determine the First Few Coefficients

To find the first few coefficients, start with initial conditions: Let \( a_0 \) and \( a_1 \) be constants. Then: \( a_2 = -\frac{a_0}{2} \), \( a_3 = -\frac{a_1}{6} \), \( a_4 = \frac{a_0}{24} \), \( a_5 = \frac{a_1}{120} \)

Write the General Solution

Combine the terms to form the general solution: \( y = a_0 \bigg[ 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \bigg] + a_1 \bigg[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \bigg] \)Recognize the series as expansions of cosine and sine: \( y = a_0 \cos x + a_1 \sin x \)

Key concepts

Power Series Solutions

When tackling differential equations, one useful approach is to assume that the solution can be expressed as a power series. This means we write our solution in the form of an infinite sum:

• Assume the solution: \( y = \sum_{n=0}^{\infty} a_n x^n \).


• Then calculate the necessary derivatives of the power series, like\( y' \) and \( y'' \).

To solve a differential equation using this method, we first substitute this series into the given equation. This helps us transform the differential equation into a series-based form.
This method is particularly effective for solving linear differential equations with variable coefficients, especially when they cannot be solved using simpler methods. Using a power series allows us to approximate the solution as closely as necessary by including more terms in the series.

Recurrence Relations

Once the power series and its derivatives are substituted into the differential equation, the next step usually involves combining and simplifying the resulting series.
By aligning the series so that all the terms have the same powers of \( x \), you can identify the recurrence relation. This relation ties together the coefficients of different powers of \( x \).
For our exercise, the recurrence relation comes from the condition that the series must sum to zero for all values of \( x \). Thus, the recurrence relation for our example is derived as:

• \( a_{n+2} = -\frac{a_{n-1}}{(n+2)(n+1)}. \)

This recurrence relation helps us determine the coefficients of the power series step-by-step, starting from the initial conditions given (like \( a_0 \) and \( a_1 \)).

Second-Order Differential Equations

Second-order differential equations involve the second derivative, and they often appear in physics and engineering. In our example, the equation is:

• \( y'' + xy = 0 \)

When dealing with such equations, power series methods become invaluable, especially when typical analytical methods are not feasible. Here’s how it works in our example: The equation includes both the second derivative and a first-degree term in \( x \). We expand both the second derivative and the original series, substituting them back into the original differential equation. This allows us to equate coefficients from the same power of \( x \), and hence, solve for the unknown coefficients.

Series Expansion

Series expansion is a technique where functions are written as the sum of their series representations. For our differential equation, once we identify the recurrence relation, we can determine the initial coefficients.
In this example:

• Let's pick the initial constants: \( a_0 \) and \( a_1 \).
• First few terms are: \( a_2 = -\frac{a_0}{2} \), \( a_3 = -\frac{a_1}{6} \), \( a_4 = \frac{a_0}{24} \), \( a_5 = \frac{a_1}{120} \).

Putting these back into the series, we form the general solution:

• \( y = a_0 \left[ 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \right] + a_1 \left[ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right] \).

Recognizing these series expansions as forms of cosine and sine, we end up with the compact form:
\( y = a_0 \cos x + a_1 \sin x \).