Differential Equations Quiz: Frobenius Method

What this quiz covers

This quiz focuses on Frobenius 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. The second solution has the form $y_2 = y_1 \ln x + x^{-1}\sum_{n=0}^{\infty} b_n x^n$ where $y_1$ is the first solution
2. The second solution has the form $y_2 = x^{-1}\sum_{n=0}^{\infty} b_n x^n$ and can be found directly using the second root $r_2 = -1$
3. The second solution requires the method of reduction of order and has the form $y_2 = y_1 \ln x + \sum_{n=1}^{\infty} b_n x^n$ (correct answer)
4. The second solution cannot be determined by the Frobenius method alone due to the integer difference in roots and requires alternative techniques

Explanation: The correct answer is C. The indicial equation for this equation gives roots $r_1 = 0$ and $r_2 = -1$, with difference 1 (an integer). When the roots differ by an integer, the second solution typically has the form $y_2 = cy_1 \ln x + x^{r_2}\sum_{n=0}^{\infty} b_n x^n$ where $c$ may be zero. For this specific equation, the second solution is $y_2 = y_1 \ln x + \sum_{n=1}^{\infty} b_n x^n$ (starting from $n=1$ due to the specific structure). Choice A has incorrect powers of $x$. Choice B ignores the logarithmic term that's needed. Choice D is incorrect because Frobenius method can handle this case, though it requires the generalized form with logarithmic terms.

Question 2

1. The modified equation yields exponentially growing and decaying solutions rather than oscillatory Bessel functions, but both maintain the same $x^{±1/2}$ behavior near $x = 0$ (correct answer)
2. The modified equation requires logarithmic terms in both solutions due to the negative coefficient, while the standard equation has pure power series solutions
3. Both equations yield identical solutions since the sign change only affects higher-order terms in the series expansion
4. The modified equation produces polynomial solutions when $ν = 1/2$, while the standard equation yields infinite series solutions

Explanation: The correct answer is A. The key difference is the sign of the $x^2$ term. Both equations have the same indicial equation giving roots $r = ±1/2$, so both have the same $x^{±1/2}$ behavior near $x = 0$. However, the modified Bessel equation leads to modified Bessel functions $I_{1/2}(x)$ and $K_{1/2}(x)$, which have exponential rather than oscillatory character for large $x$. The recurrence relations differ due to the sign change, leading to fundamentally different asymptotic behavior. Choice B is wrong because logarithmic terms don't automatically appear due to sign changes. Choice C is incorrect because the sign change fundamentally alters the solution character. Choice D is wrong because neither equation produces polynomial solutions for $ν = 1/2$.

Question 3

1. No additional constraint is needed since the roots differ by 2, which ensures two independent power series solutions exist
2. The coefficient $a_2$ in the first solution must satisfy $a_2 = 0$ to prevent resonance in the second solution's recurrence relation
3. The second solution automatically requires logarithmic terms because the difference $r_1 - r_2 = 2$ is an even integer
4. We must verify that the recurrence relation for the second solution doesn't lead to division by zero at the critical step corresponding to the root difference (correct answer)

Explanation: The correct answer is D. When the roots of the indicial equation differ by an integer (here, $r_1 - r_2 = 2$), the second solution may or may not exist in standard Frobenius form. The critical issue is whether the recurrence relation for the second solution leads to division by zero at the step where $n = r_1 - r_2 = 2$. At this step, the coefficient of $a_2$ in the recurrence becomes zero, potentially making $a_2$ indeterminate or forcing logarithmic terms. We must check if the numerator is also zero at this step. Choice A is wrong because integer difference doesn't guarantee two standard solutions. Choice B incorrectly identifies which coefficient matters. Choice C is wrong because even integers don't automatically require logarithmic terms.

Question 4

1. Both solutions behave like $x^{3/2}$ and $x^{-3/2}$ respectively, with the second requiring logarithmic terms due to resonance
2. Both solutions behave like $x^{3/2}$ and $x^{-3/2}$ respectively, and both can be expressed as pure power series without logarithmic terms (correct answer)
3. One solution behaves like $x^{3/2}$ while the second behaves like $x^{-3/2} \ln x$ due to the half-integer difference in roots
4. Both solutions require logarithmic terms because $ν = \frac{3}{2}$ creates a resonance condition in the recurrence relations

Explanation: The correct answer is B. This is Bessel's equation with $ν = 3/2$. The indicial equation gives roots $r_1 = 3/2$ and $r_2 = -3/2$, with difference 3 (an integer). However, since $ν = 3/2$ is not an integer, both solutions $J_{3/2}(x)$ and $J_{-3/2}(x)$ exist as linearly independent power series solutions. The key insight is that logarithmic terms only appear when the parameter $ν$ itself is an integer, not when the difference of roots is an integer. Choice A incorrectly mentions logarithmic terms. Choice C incorrectly assumes logarithmic behavior due to half-integer difference. Choice D incorrectly claims resonance occurs for non-integer $ν$.

Question 5

1. Logarithmic terms are always required when the indicial equation has a double root, regardless of the equation parameters
2. The presence of logarithmic terms depends on whether $\gamma = 0$, since this affects the lowest-order terms in the recurrence
3. The coefficient $\beta$ determines the need for logarithmic terms, as it controls the $x^2$ term that creates resonance with the double root
4. Whether logarithmic terms appear depends on the solvability condition in the recurrence relation at the step where the coefficient determinant vanishes (correct answer)

Explanation: The correct answer is D. When the indicial equation has a double root, the second solution may or may not require logarithmic terms depending on the specific equation. The determining factor is whether the recurrence relation for the second solution leads to a consistency condition (solvability condition) at the step where the coefficient determinant becomes zero. If this condition is satisfied, a second power series solution exists; if not, logarithmic terms are required. Choice A is incorrect because double roots don't automatically require logarithmic terms. Choice B oversimplifies the relationship to just one parameter. Choice C incorrectly identifies $\beta$ as the sole determining factor. The actual determination requires careful analysis of the full recurrence system.

Question 6

1. The indicial equation becomes a quartic rather than quadratic, requiring numerical methods to find the roots
2. The radius of convergence of the Frobenius series is limited by the distance to the next singular point at $x = 1$ (correct answer)
3. Both singular points must be regular for the Frobenius method to work, requiring simultaneous analysis at both points
4. The recurrence relations become coupled due to interference from the second singular point, preventing closed-form solutions

Explanation: The correct answer is B. When applying the Frobenius method near $x = 0$, the presence of another singular point at $x = 1$ limits the radius of convergence of the power series to at most the distance between singular points, which is 1. This is a fundamental limitation of series solutions - they can only converge in a neighborhood that doesn't contain other singular points. The indicial equation remains quadratic (choice A is wrong). The regularity of both points is important for global analysis but doesn't prevent local Frobenius analysis (choice C is wrong). The recurrence relations are determined locally and aren't directly coupled by distant singular points (choice D is wrong). This radius limitation is a crucial practical consideration when using Frobenius series.