Problem 18 - Advanced Engineering Mathematics

Question

Find two power series solutions of the given differential equation about the ordinary point \(x=0\). $$ y^{\prime \prime}+x^{2} y=0 $$

Step-by-Step Solution

Answer

The two power series solutions are obtained by choosing initial conditions and using the recurrence relation. They represent solutions mainly in even and odd powers of \( x \).

Step 1: Identify the Standard Form

The given differential equation is \( y'' + x^2 y = 0 \). We recognize this as a second-order linear ordinary differential equation and it's already in its standard form since it includes \( y'' \), \( y \), and a function of \( x \). This allows for power series solutions around the ordinary point \( x = 0 \).

Step 2: Assume a Power Series Solution

Assume a power series solution of the form \( y(x) = \sum_{n=0}^{\infty} a_n x^n \). The power series representation provides a way to solve the differential equation by finding coefficients \( a_n \) that satisfy the equation.

Step 3: Find the Derivatives

Compute the necessary derivatives of the assumed solution: \( y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} \). This shifting in the power of \( x \) is essential as we need to equate to \( x^2 y \).

Step 4: Substitute into the Differential Equation

Substitute \( y(x) \), \( y''(x) \), and the term \( x^2 y(x) = \sum_{n=0}^{\infty} a_n x^{n+2} \) back into the differential equation: \( \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + \sum_{n=0}^{\infty} a_n x^{n+2} = 0 \). This gives us a single power series expression.

Step 5: Align Powers of x

To combine these into one power series, make a change of index so both sums have the same power of \( x \). Adjust the second sum: let \( m = n + 2 \), hence \( n = m - 2 \). Thus, the terms align as \( \sum_{m=0}^{\infty} (m+2)(m+1)a_{m+2} x^m + \sum_{m=2}^{\infty} a_{m-2} x^m = 0 \).

Step 6: Find a Recurrence Relation

Equate the coefficients of \( x^m \) from both series: \( (m+2)(m+1)a_{m+2} + a_{m-2} = 0 \). This gives a recurrence relation: \( a_{m+2} = -\frac{a_{m-2}}{(m+2)(m+1)} \).

Step 7: Determine the Initial Conditions

To find particular solutions, choose initial conditions: Assume \( a_0 = 1 \) and \( a_1 = 0 \) for the first solution; assume \( a_0 = 0 \) and \( a_1 = 1 \) for the second. Apply these to the recurrence relation to determine all other coefficients.

Step 8: Construct the Series Solutions

Using the initial conditions from the previous step: for \( a_0 = 1 \); \( a_4 = -\frac{a_0}{4!} \), \( a_8 = \frac{a_0}{8!} \), leading to the series solution for even powers. For \( a_1 = 1 \); \( a_5 = -\frac{a_1}{5!} \), \( a_9 = \frac{a_1}{9!} \), leading to the series solution for odd powers.

Key Concepts

Power Series Solution

A power series solution is a method used to solve differential equations by expressing the solution as an infinite sum of terms. In many cases, this provides a way to construct solutions around a specific point, often referred to as an ordinary point. For an equation such as \( y'' + x^2 y = 0 \), we assume a solution in the form of a power series:

• \( y(x) = \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) are the series coefficients to be determined.
• This series form turns the differential equation into an algebraic problem of finding the values of \( a_n \).

Power series are particularly useful for their flexibility, as they allow you to approximate the behavior of functions near the chosen point by considering sufficiently many terms.

Initial Conditions

Initial conditions are crucial in finding specific solutions to differential equations. They specify the value of the function and its derivatives at a certain point, which helps to pin down one precise solution from the general form that might have many fits. For our problem:

• We have two sets of initial conditions: \( a_0 = 1 \), \( a_1 = 0 \) for the first solution, and \( a_0 = 0 \), \( a_1 = 1 \) for the second one.
• These choices are arbitrary initial conditions that help generate two linearly independent solutions, which in turn span the solution space of the differential equation.

By applying these initial conditions to the recurrence relations, you can derive all the subsequent coefficients of the series.

Recurrence Relation

A recurrence relation arises when you substitute the power series and its derivatives into the differential equation. It effectively gives a rule to find higher order coefficients from known lower order ones. For the equation \( y'' + x^2 y = 0 \), we establish:

• The recurrence relation: \( a_{m+2} = -\frac{a_{m-2}}{(m+2)(m+1)} \).
• This relation helps calculate each coefficient \( a_n \) once the initial conditions provide the first terms.

Recurrence relations simplify the process of finding series solutions by breaking the problem into manageable steps. It's like setting up a chain reaction: start with initial conditions, then use the rule repeatedly to get all necessary coefficients.

Second-Order Linear Differential Equation

Differential equations classify by the order, which is determined by the highest derivative present in the equation. A second-order linear differential equation, like \( y'' + x^2 y = 0 \), is characterized by:

• The highest derivative is the second derivative \( y'' \).
• It involves linearly combined terms of the function and its derivatives.
• The coefficients of each term could be constants or functions of the independent variable \( x \).

Such equations often model a wide range of physical phenomena, like oscillations and waves, and can frequently be approached using power series especially when they present no straightforward solutions.