Power Series Method - Differential Equations

1. Consider the differential equation $y'' - xy' + 2y = 0$. When applying the power series method about $x = 0$, what is the recurrence relation for the coefficients $a_n$ in the series $y = \sum_{n=0}^{\infty} a_n x^n$?
   Answer: $a_{n+2} = \frac{(n+1)a_{n+1} - 2a_n}{(n+1)(n+2)}$ for $n \geq 0$ 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 $(n+1)$.
2. When applying the power series method to $y'' + \frac{1}{x}y' + y = 0$ about $x = 0$, what is the primary difficulty encountered?
   Answer: The standard power series method fails because $x = 0$ is a regular singular point, requiring the Frobenius method instead 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.
3. Consider the differential equation $(1-x^2)y'' - 2xy' + 6y = 0$. When applying the power series method about $x = 0$, what can be concluded about the radius of convergence of any power series solution?
   Answer: The radius of convergence is at most 1 because the coefficient $(1-x^2)$ has zeros at $x = \pm 1$ 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.
4. For the differential equation $y'' + x^2y' - 3y = 0$, when seeking a power series solution $y = \sum_{n=0}^{\infty} a_n x^n$ about $x = 0$, which statement about the radius of convergence is most accurate?
   Answer: The radius of convergence is infinite since all coefficient functions are polynomials with no finite singularities 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.
5. For the equation $y'' + xy' + (x^2-1)y = 0$, suppose we obtain a power series solution $y_1 = \sum_{n=0}^{\infty} a_n x^n$ where $a_0 = 1$ and $a_1 = 0$. To find a second linearly independent solution using the power series method, which approach is most appropriate?
   Answer: 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 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.
6. When solving $y'' + 2xy' + x^2y = 0$ using power series $y = \sum_{n=0}^{\infty} a_n x^n$, the recurrence relation yields $a_2 = -a_0$, $a_3 = -\frac{a_1}{3}$, and $a_4 = \frac{a_0}{12}$. Which pattern correctly describes the relationship between the even and odd coefficient subsequences?
   Answer: The even coefficients depend only on $a_0$ while the odd coefficients depend only on $a_1$, with no coupling between subsequences 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.

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