Differential Equations Quiz: Series Solutions Ordinary Points

What this quiz covers

This quiz focuses on Series Solutions Ordinary Points, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $x = 1$ is an ordinary point since the original coefficients are polynomials at this point
2. $x = 1$ is a regular singular point because the equation has polynomial coefficients
3. $x = 1$ is an ordinary point because $\frac{x}{x^2-4}$ and $\frac{x+1}{x^2-4}$ are both analytic at $x = 1$ (correct answer)
4. $x = 1$ is an irregular singular point due to the behavior of $Q(x)$ near $x = 1$

Explanation: 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.

Question 2

1. $a_{n+2} = -\frac{(n+1)a_{n+1} + (n-1)a_{n-1}}{(n+2)(n+1)}$ for $n \geq 1$ (correct answer)
2. $a_{n+2} = -\frac{(n+1)a_{n+1} + na_n}{(n+2)(n+1)}$ for $n \geq 0$
3. $a_{n+2} = -\frac{na_{n+1} + (n-1)a_{n-1}}{(n+2)(n+1)}$ for $n \geq 1$
4. $a_{n+2} = -\frac{(n+1)a_{n+1} + (n+1)a_{n-1}}{(n+2)(n+1)}$ for $n \geq 1$

Explanation: 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 $a_{n-1}$.

Question 3

1. The radius of convergence is infinite since all coefficients are polynomials (correct answer)
2. The radius of convergence is determined by the nearest singularity of the rewritten standard form
3. The radius of convergence is 1 due to the coefficient of the $y$ term
4. The radius of convergence cannot be determined without solving the recurrence relation

Explanation: 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.

Question 4

1. The coefficient functions are not polynomial, so the standard power series method does not apply
2. The series for $\sin x$ and $\cos x$ converge too slowly to provide accurate coefficient calculations
3. The recurrence relations will be non-linear in the coefficients $a_n$, making them unsolvable
4. The recurrence relation will be infinite-order, relating each coefficient to infinitely many previous coefficients (correct answer)

Explanation: 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.

Question 5

1. $(n+2)(n+1)a_{n+2} + na_{n-1} + a_n - a_{n-2} = 0$ (correct answer)
2. $(n+2)(n+1)a_{n+2} + (n-1)a_{n-1} + a_n - a_{n-2} = 0$
3. $(n+2)(n+1)a_{n+2} + na_{n-1} + a_n = 0$
4. $(n+2)(n+1)a_{n+2} + (n-1)a_{n-1} + a_n = 0$

Explanation: 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.

Question 6

1. The series will converge globally since the coefficients are rational functions with real poles
2. The series will converge only for $|x| < 1$ due to the poles of the coefficient functions at $x = \pm i$ (correct answer)
3. The series will converge only for $|x| < \sqrt{2}$ based on the denominator $1+x^2$
4. The convergence radius cannot be determined without computing the series coefficients explicitly

Explanation: 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.

Question 7

1. A second solution can be found, but it will involve a logarithmic term and cannot be a pure power series
2. A second solution exists but requires the Frobenius method since the given solution suggests a logarithmic term
3. No second power series solution exists because the coefficient pattern suggests a unique solution structure
4. A second power series solution exists and can be found using the same method since $x = 0$ is an ordinary point (correct answer)

Explanation: 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.