Series Solutions: Ordinary Points - Differential Equations

1. The differential equation $y'' + P(x)y' + Q(x)y = 0$ has an ordinary point at $x = x_0$ if both $P(x)$ and $Q(x)$ are analytic at $x_0$. For the equation $(x^2 - 4)y'' + xy' + (x+1)y = 0$, which statement about the point $x = 1$ is correct?
   Answer: $x = 1$ is an ordinary point because $\frac{x}{x^2-4}$ and $\frac{x+1}{x^2-4}$ are both analytic at $x = 1$ To determine the nature of $x = 1$, we must write the equation in standard form: $y'' + \frac{x}{x^2-4}y' + \frac{x+1}{x^2-4}y = 0$. Since $x^2-4 = (x-2)(x+2)$, at $x = 1$ we have $x^2-4 = -3 \neq 0$. Therefore both $P(x) = \frac{x}{x^2-4}$ and $Q(x) = \frac{x+1}{x^2-4}$ are analytic at $x = 1$, making it an ordinary point. Choice A incorrectly focuses on the original form; Choice B misclassifies the point type; Choice D incorrectly identifies it as irregular singular.
2. For the equation $y'' + (\sin x)y' + (\cos x)y = 0$, a student wants to find a power series solution about $x = 0$. After expanding $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$ and $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$, what is the most significant computational challenge?
   Answer: The recurrence relation will be infinite-order, relating each coefficient to infinitely many previous coefficients When you encounter a differential equation with non-polynomial coefficients like $\sin x$ and $\cos x$, you can still use power series methods by substituting the known series expansions for these functions. The key insight is understanding what happens when you multiply infinite series together. Let's say you assume a solution $y = \sum_{n=0}^{\infty} a_n x^n$. When you substitute this into the differential equation along with the series for $\sin x$ and $\cos x$, you must multiply the series $y' = \sum_{n=1}^{\infty} na_n x^{n-1}$ by $\sin x = x - \frac{x^3}{6} + \cdots$, and $y = \sum_{n=0}^{\infty} a_n x^n$ by $\cos x = 1 - \frac{x^2}{2} + \cdots$. The crucial problem emerges here: when you multiply two infinite series, each coefficient in the resulting series depends on infinitely many terms from the original series. For example, to find the coefficient of $x^n$ in the product $(\sin x) \cdot y'$, you need terms from every power in the $\sin x$ expansion paired with the appropriate term from $y'$. This creates recurrence relations where each $a_n$ depends on $a_0, a_1, a_2, \ldots, a_{n-1}$ and potentially more. Choice A is wrong because power series methods do work with analytic coefficient functions. Choice B misunderstands the issue—convergence speed doesn't affect the method's validity. Choice C is incorrect because the recurrence relations remain linear in the coefficients $a_n$. Study tip: Remember that non-polynomial coefficients in differential equations lead to infinite-order recurrence relations, making hand calculations extremely tedious beyond the first few terms.
3. Consider the differential equation $y'' - xy' + (x^2 - 1)y = 0$. When seeking a series solution about $x = 0$, what is the most restrictive condition on the radius of convergence of the power series solution?
   Answer: The radius of convergence is infinite since all coefficients are polynomials Since $x = 0$ is an ordinary point (all coefficients $P(x) = -x$, $Q(x) = x^2 - 1$ are analytic at $x = 0$), and both coefficient functions are polynomials (hence entire functions), the series solution converges everywhere. The radius of convergence is infinite. Choice B incorrectly assumes singularities exist; Choice C misinterprets the role of the $y$ coefficient; Choice D incorrectly suggests the recurrence relation is needed to determine convergence radius.
4. A student seeks a series solution to $y'' + \frac{x}{1+x^2}y' + \frac{1}{1+x^2}y = 0$ about $x = 0$. Before applying the power series method, what should the student verify about the convergence properties?
   Answer: The series will converge only for $|x| < 1$ due to the poles of the coefficient functions at $x = \pm i$ Since $x = 0$ is an ordinary point, the series solution converges in a disk centered at $x = 0$ extending to the nearest singularity. The coefficient functions $P(x) = \frac{x}{1+x^2}$ and $Q(x) = \frac{1}{1+x^2}$ have poles where $1+x^2 = 0$, i.e., at $x = \pm i$. The distance from $x = 0$ to either $x = i$ or $x = -i$ is $|\pm i| = 1$, so the radius of convergence is 1. Choice A ignores the complex singularities; Choice C miscalculates the distance; Choice D incorrectly suggests coefficient computation is needed.
5. For the differential equation $y'' + (1 + x^2)y' + xy = 0$, suppose we seek a power series solution $y = \sum_{n=0}^{\infty} a_n x^n$ about $x = 0$. After substituting and collecting terms, which recurrence relation correctly relates the coefficients?
   Answer: $a_{n+2} = -\frac{(n+1)a_{n+1} + (n-1)a_{n-1}}{(n+2)(n+1)}$ for $n \geq 1$ Substituting $y = \sum a_n x^n$ gives: $\sum (n+2)(n+1)a_{n+2}x^n + \sum (n+1)a_{n+1}x^n + \sum na_nx^{n+2} + \sum na_nx^n = 0$. Shifting indices and collecting coefficients of $x^n$ for $n \geq 1$: $(n+2)(n+1)a_{n+2} + (n+1)a_{n+1} + (n-1)a_{n-1} = 0$. Choice B incorrectly uses $na_n$ instead of $(n-1)a_{n-1}$; Choice C has wrong coefficient on $a_{n+1}$; Choice D has wrong coefficient on $a_{n-1}$.
6. For the differential equation $y'' + \frac{2x}{1-x^2}y' + \frac{2}{1-x^2}y = 0$, suppose we know that one solution can be written as $y_1(x) = 1 + x^2 + \frac{x^4}{3} + \frac{x^6}{5} + \cdots$. What can be concluded about finding a second linearly independent solution using power series methods about $x = 0$?
   Answer: A second power series solution exists and can be found using the same method since $x = 0$ is an ordinary point When you encounter a second-order linear differential equation and need to find a second linearly independent solution, the key question is whether the point in question is ordinary or singular. This determines which methods will work and what form the solutions will take. To determine the nature of $x = 0$, examine the equation in standard form. The coefficients are $P(x) = \frac{2x}{1-x^2}$ and $Q(x) = \frac{2}{1-x^2}$. At $x = 0$, we have $P(0) = 0$ and $Q(0) = 2$, so both coefficients are analytic (expressible as convergent power series) at $x = 0$. This makes $x = 0$ an ordinary point. At an ordinary point, the fundamental theorem guarantees that both linearly independent solutions can be expressed as convergent power series. Since we already have one power series solution $y_1(x)$, we can find a second power series solution $y_2(x)$ using standard power series methods—no special techniques required. Option A is wrong because logarithmic terms only appear when dealing with singular points or repeated roots in the indicial equation. Option B incorrectly suggests using the Frobenius method, which is necessary only at regular singular points, not ordinary points. Option C is incorrect because linear second-order equations always have exactly two linearly independent solutions. The correct answer is D: a second power series solution exists and can be found using standard power series substitution since $x = 0$ is an ordinary point. Study tip: Always check if coefficients are analytic at your point of interest—this immediately tells you whether standard power series methods will work for both solutions.
7. For the differential equation $y'' + xy' + (1-x^2)y = 0$, a student attempts to find a power series solution $y = \sum_{n=0}^{\infty} a_n x^n$ about $x = 0$. After substitution and simplification, which equation must be satisfied by the coefficients for $n \geq 2$?
   Answer: $(n+2)(n+1)a_{n+2} + na_{n-1} + a_n - a_{n-2} = 0$ Substituting $y = \sum a_n x^n$ into the equation: $\sum (n+2)(n+1)a_{n+2}x^n + \sum na_nx^n + \sum a_nx^n - \sum a_nx^{n+2} = 0$. Shifting the last sum and collecting terms for $n \geq 2$: $(n+2)(n+1)a_{n+2} + na_{n-1} + a_n - a_{n-2} = 0$. The key is recognizing that $xy'$ contributes $na_{n-1}x^n$ after index shifting, and $(1-x^2)y$ gives both $+a_n$ and $-a_{n-2}$ terms. Choice B has wrong coefficient; Choices C and D omit the $-a_{n-2}$ term.

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