Frobenius Method for GATE MA: Triage Before the Series

Only GATE MA lists the Frobenius method. As of November 2026, the GATE 2027 MA syllabus names "series solutions (power series, Frobenius method)" in its ordinary differential equations (ODE) section. The CSIR NET, IIT JAM and CUET PG syllabi do not include it.

The December 2026 CSIR-UGC NET bulletin from the National Testing Agency directs candidates to CSIR's own site for the syllabus (Information Bulletin, October 2026). CSIR's Mathematical Sciences page carries the syllabus file in the table, dated December 2020.

If you sit CSIR NET, you can leave this topic and spend the time on the CSIR NET exam pattern. If you sit GATE MA, the topic is on your list, and our GATE MA syllabus page shows where it sits among the other sections.

The 2025 GATE MA paper had no question on the Frobenius method itself. Its related questions covered the Laplace transform of the Bessel function J₀ (Q12), a Legendre polynomial (Q28) and a further Legendre-polynomial question (Q45) (GATE 2025 MA paper, set by IIT Roorkee).

The paper contains no question on the Frobenius method, indicial equations or singular points. That is one paper, and the syllabus does list the method, so a question can still come.

Our inference: since the syllabus lists the method, learn the opening steps first, because they are fast. Learn the Bessel and Legendre properties the syllabus names beside the method, and build full series last. To see how marks spread across sections, read GATE Mathematics previous year papers by section.

Run these five checks in order, and stop as soon as the question is answered. Write the equation as y″ + P(x)y′ + Q(x)y = 0 and take the point as x = 0. For a point x = a, put t = x − a, rewrite the equation in t, and use tP and t²Q in step 1; the solutions are then series in (x − a) times (x − a)^r. "Analytic" means the function has a convergent power series around the point.

A log term means a term of the form (first solution) × ln x added to the second solution.

No. Equal roots force a log term, a non-integer gap never needs one, and a positive integer gap N sometimes does, because the constant multiplying the log term can be zero (Boyce and DiPrima, Elementary Differential Equations, chapter on series solutions near a regular singular point). One test settles each case.

Write the equation as x²y″ + x·p(x)·y′ + q(x)·y = 0, with p and q analytic, and substitute y = Σ aₙ x^(n+r). Let F(r) = r(r − 1) + p₀r + q₀ = (r − r₁)(r − r₂). The coefficients then satisfy:

F(r + n)·aₙ = −(a sum of the earlier coefficients a₀ … aₙ₋₁, each multiplied by constants from p and q)

• Larger root, r = r₁: F(r₁ + n) = n(n + N), which is positive for every n ≥ 1. Every coefficient is defined, so the larger root always gives a series.
• Smaller root, r = r₂: F(r₂ + n) = n(n − N), which is zero at n = N. At that step the equation reads 0 · a_N = (right side).

The step-N test: compute the right side at n = N using r = r₂.

• If it is 0, then a_N is free and there is no log term. Pass.
• If it is not 0, the equation is impossible without a log term. The second solution is C·y₁·ln x plus a series in x^(r₂+n).

The roots were checked with sympy; the log-term verdicts come from the step-N test worked by hand.

Compare rows 3 and 4. Both have a gap of 1, yet Bessel's equation of order 1/2 has no log term and xy″ + y = 0 does. For Bessel order 1/2, p(x) = 1 and q(x) = x² − 1/4 have no x¹ term, so at n = 1 the right side is 0 · a₀ = 0. For xy″ + y = 0, the x⁰ coefficient of a plain power series is a₀, which must vanish, so the smaller root r = 0 gives no series on its own. That forces the log term. In the last two rows the equation has constant p and q, so no earlier coefficients feed the right side, and it is always 0.

Take 4(ln x)·y″ + 3y′ + y = 0 for x > 1, and ask for |r₁ − r₂| at the point x = 1. The answer is 0.25. This is our own practice example, worked below.

1. Divide through by 4 ln x: P = 3/(4 ln x) and Q = 1/(4 ln x).
2. Shift the point: put t = x − 1. Near t = 0, ln x = t − t²/2 + …, so t/ln x tends to 1.
3. Step 1: tP tends to 3/4, and t²Q = t · (t/(4 ln x)) tends to 0. Both are analytic at t = 0, so x = 1 is a regular singular point.
4. Step 2: p₀ = 3/4 and q₀ = 0.
5. Step 3: r(r − 1) + (3/4)r = r² − r/4 = r(r − 1/4), so the roots are 0 and 1/4.
6. Step 4: the gap N = 1/4 is not an integer, so there are two Frobenius series and no log term.
7. Step 5: the question asks only for the gap, so |r₁ − r₂| = 0.25. No series is built.

The whole answer needs a few lines of algebra, and the series is never opened.

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