Differential Equations Quiz: Power Series Method

What this quiz covers

This quiz focuses on Power Series Method, 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. $a_{n+2} = \frac{(n-2)a_{n+1} - 2a_n}{(n+1)(n+2)}$ for $n \geq 0$
2. $a_{n+2} = \frac{(n+1)a_{n+1} - 2a_n}{(n+1)(n+2)}$ for $n \geq 0$ (correct answer)
3. $a_{n+2} = \frac{na_{n+1} - 2a_n}{(n+1)(n+2)}$ for $n \geq 0$
4. $a_{n+2} = \frac{(n+2)a_{n+1} - 2a_n}{(n+1)(n+2)}$ for $n \geq 0$

Explanation: Substituting $y = \sum_{n=0}^{\infty} a_n x^n$, $y' = \sum_{n=1}^{\infty} na_n x^{n-1}$, and $y'' = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}$ into the equation gives: $\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} - x\sum_{n=1}^{\infty} na_n x^{n-1} + 2\sum_{n=0}^{\infty} a_n x^n = 0$. Reindexing and equating coefficients of $x^n$ yields: $(n+2)(n+1)a_{n+2} - (n+1)a_{n+1} + 2a_n = 0$, which gives $a_{n+2} = \frac{(n+1)a_{n+1} - 2a_n}{(n+1)(n+2)}$. Choice A uses $(n-2)$ instead of $(n+1)$. Choice C uses $n$ instead of $(n+1)$. Choice D uses $(n+2)$ instead of $(n+1)$.

Question 2

1. The radius of convergence is exactly 1 because the coefficient functions have singularities at $x = \pm 1$
2. The radius of convergence cannot be determined without computing at least the first five terms of the series
3. The radius of convergence is 0 because the $x^2$ coefficient makes the series diverge everywhere except at $x = 0$
4. The radius of convergence is infinite since all coefficient functions are polynomials with no finite singularities (correct answer)

Explanation: When determining the radius of convergence for power series solutions of linear differential equations, you need to examine where the coefficient functions have singularities. For a second-order equation in standard form $y'' + P(x)y' + Q(x)y = 0$, the radius of convergence around $x = 0$ extends to the nearest singularity of $P(x)$ or $Q(x)$. First, rewrite the equation in standard form: $y'' + x^2y' - 3y = 0$. Here, $P(x) = x^2$ and $Q(x) = -3$. Both coefficient functions are polynomials, which are analytic everywhere in the complex plane—they have no finite singularities. Since there are no singularities to limit convergence, the power series solution converges for all finite values of $x$, giving an infinite radius of convergence. Choice A incorrectly assumes singularities exist at $x = \pm 1$, but polynomials have no finite singularities. Choice B suggests you need to compute series terms to find the radius, but the coefficient analysis immediately tells you the answer. Choice C reflects a fundamental misunderstanding—having variable coefficients like $x^2$ doesn't cause divergence; only singularities in the coefficient functions limit convergence. Study tip: For power series solutions around $x = 0$, immediately check the coefficient functions $P(x)$ and $Q(x)$ for singularities. If they're polynomials (or entire functions), the radius is infinite. If they have poles or other singularities, the radius equals the distance to the nearest one.

Question 3

1. The radius of convergence is at most 1 because the coefficient $(1-x^2)$ has zeros at $x = \pm 1$ (correct answer)
2. The radius of convergence is exactly $\frac{1}{2}$ due to the specific form of the polynomial coefficients
3. The radius of convergence is infinite since all coefficients are polynomials
4. The radius of convergence depends on the initial conditions and cannot be determined from the equation alone

Explanation: The differential equation $(1-x^2)y'' - 2xy' + 6y = 0$ can be written in standard form as $y'' - \frac{2x}{1-x^2}y' + \frac{6}{1-x^2}y = 0$. The coefficient functions $p(x) = -\frac{2x}{1-x^2}$ and $q(x) = \frac{6}{1-x^2}$ have singularities where $1-x^2 = 0$, i.e., at $x = \pm 1$. The radius of convergence of a power series solution about $x = 0$ cannot exceed the distance to the nearest singularity, which is $|\pm 1 - 0| = 1$. Choice B gives an incorrect specific value. Choice C ignores the singularities created by division. Choice D is wrong because the radius depends only on the coefficient functions, not initial conditions.

Question 4

1. Set $y_2 = \sum_{n=0}^{\infty} b_n x^n$ with $b_0 = 0$ and $b_1 = 1$, then apply the same recurrence relation (correct answer)
2. Use the method of reduction of order: $y_2 = y_1 \int \frac{e^{-\int p(x)dx}}{y_1^2} dx$ where $p(x) = x$
3. Assume $y_2 = x^r \sum_{n=0}^{\infty} c_n x^n$ and determine $r$ from the indicial equation at the regular singular point
4. The power series method automatically generates two independent solutions through the even and odd coefficient subsequences

Explanation: Since $x = 0$ is an ordinary point for this equation (all coefficients are polynomials), the power series method can be applied directly. The general solution near an ordinary point has the form $y = c_1y_1 + c_2y_2$ where both $y_1$ and $y_2$ are power series. We can find $y_1$ with initial conditions $y_1(0) = 1, y_1'(0) = 0$ and $y_2$ with $y_2(0) = 0, y_2'(0) = 1$. This corresponds to setting $a_0 = 1, a_1 = 0$ for $y_1$ and $b_0 = 0, b_1 = 1$ for $y_2$. Choice B is unnecessarily complex for an ordinary point. Choice C applies to singular points. Choice D is incorrect about automatic generation.

Question 5

1. The equation has no power series solution because the coefficient $\frac{1}{x}$ creates a logarithmic term that cannot be expressed as a power series
2. The power series converges only for $|x| < \frac{1}{2}$ due to the singular coefficient, making it impractical for most applications
3. The recurrence relation becomes undefined because division by zero occurs when computing the second coefficient
4. The standard power series method fails because $x = 0$ is a regular singular point, requiring the Frobenius method instead (correct answer)

Explanation: When encountering a differential equation with variable coefficients, you need to classify the singular points before choosing your solution method. The equation $y'' + \frac{1}{x}y' + y = 0$ has a coefficient that's undefined at $x = 0$, making it a singular point. To determine if $x = 0$ is regular or irregular, rewrite the equation in standard form and check if $xP(x) = x \cdot \frac{1}{x} = 1$ and $x^2Q(x) = x^2 \cdot 1 = x^2$ are analytic at $x = 0$. Since both expressions are polynomials (hence analytic) at $x = 0$, this confirms $x = 0$ is a regular singular point. At regular singular points, the standard power series method $y = \sum a_n x^n$ fails because the solution typically involves fractional or logarithmic terms that can't be captured by integer powers alone. The Frobenius method, using $y = x^r \sum a_n x^n$, is designed specifically for this situation. Option A is wrong because regular singular points do have solutions—they just require the Frobenius method, not standard power series. Option B incorrectly focuses on convergence radius, which isn't the primary issue here. Option C suggests computational problems with coefficients, but the recurrence relation itself isn't the fundamental obstacle. Option D correctly identifies that the standard power series method fails at regular singular points. Study tip: Always classify singular points first. If $x = x_0$ makes coefficients undefined, check if $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are analytic there to distinguish regular from irregular singular points.

Question 6

1. The even coefficients $\{a_0, a_2, a_4, \ldots\}$ and odd coefficients $\{a_1, a_3, a_5, \ldots\}$ form two independent subsequences with separate recurrence relations
2. All coefficients are determined by $a_0$ and $a_1$, but the even and odd subsequences are coupled through cross-terms in the recurrence relation
3. The even coefficients depend only on $a_0$ while the odd coefficients depend only on $a_1$, with no coupling between subsequences (correct answer)
4. The recurrence relation alternates between even and odd terms, so each coefficient depends on the previous coefficient regardless of parity

Explanation: From the given values: $a_2 = -a_0$ (even, depends on $a_0$), $a_3 = -\frac{a_1}{3}$ (odd, depends on $a_1$), and $a_4 = \frac{a_0}{12}$ (even, depends on $a_0$). The pattern shows that even-indexed coefficients depend only on $a_0$ and other even coefficients, while odd-indexed coefficients depend only on $a_1$ and other odd coefficients. This decoupling is characteristic of equations where the recurrence relation has the form $a_{n+2} = f(n)a_n + g(n)a_{n-2} + \cdots$, connecting terms of the same parity. Choice A is partially correct but misses the independence aspect. Choice B incorrectly suggests coupling. Choice D incorrectly describes alternating dependence.