Laplace Transform for GATE MA: What the Syllabus and Papers Ask

The Laplace transform of a function f(t) defined for t ≥ 0 is the function F(s) = ∫₀^∞ e^(−st) f(t) dt. It converts differentiation into multiplication by s. That is why a linear differential equation with constant coefficients becomes an algebra problem in s.

The integral is not magic, and the derivative rule is one integration by parts:

∫₀^∞ e^(−st) f′(t) dt = [e^(−st) f(t)]₀^∞ + s∫₀^∞ e^(−st) f(t) dt = sF(s) − f(0)

This holds provided e^(−st) f(t) → 0 as t → ∞. That proviso is why s must be large enough. For example, the transform of e^(at) is 1/(s − a) only when s > a. The initial value f(0) arrives inside the rule, so you use the initial conditions while transforming, not at the end.

L{1} = 1/s is the same integral: ∫₀^∞ e^(−st) dt = 1/s for s > 0. The result is a function of s, not a number. Each value of s gives a different weighted total of f. For complex s = σ + iω, the transform is the Fourier transform of f(t)e^(−σt) (taking f as zero for t < 0). The damping factor e^(−σt) is what lets the transform handle functions that grow.

The GATE MA syllabus lists it in two places, and the CSIR HRDG syllabus files contain no Laplace transform. The table compares the documents' wording.

Your cycle's CSIR information bulletin is the final word, so read its Unit 3 before deciding.

Our recommendation:

• GATE MA: learn the transform now. It is a short topic, and the same ODE section also lists series solutions, the Frobenius method and Bessel functions (see Frobenius Method for GATE MA).
• CSIR NET only: give it no dedicated block until your bulletin lists it. For the full GATE topic list, see our GATE MA syllabus page.

In the GATE MA papers from 2022 to 2026, exactly one question was on the Laplace transform: Q.12 in 2025, worth one mark.

Count: questions in each paper that use the Laplace transform, from the linked papers. Five papers is a small sample, so read it as a rough rate, not a trend or a weightage.

The question. Y(s) is the Laplace transform of the Bessel function J₀(x). Which equation does Y satisfy? The official answer key gives option D: dY/ds + sY/(s²+1) = 0, s > 0. Here is the derivation:

1. J₀ solves x y″ + y′ + x y = 0, with y(0) = 1 and y′(0) = 0.
2. L{y} = Y, L{y′} = sY − 1 and L{y″} = s²Y − s.
3. The multiply-by-x rule is L{x·f} = −F′(s). So L{x y″} = −d/ds(s²Y − s) = −2sY − s²Y′ + 1, and L{x y} = −Y′.
4. Add the three transforms: (−2sY − s²Y′ + 1) + (sY − 1) − Y′ = −(s²+1)Y′ − sY = 0.
5. Therefore Y′ + sY/(s²+1) = 0, which is option D.

Eliminating options without the derivation. This is our own reasoning. For a function that is 1 at t = 0, sY(s) → 1 as s → ∞, so Y must decay like 1/s.

• Option A gives Y = C/(s²+1), which decays like 1/s², so it fails.
• Options B and C give Y = C(s²+1) and Y = C√(s²+1), which grow with s, so they fail.
• Option D gives Y = C/√(s²+1), with C = 1.

Transform every term with its initial conditions, solve for Y(s), simplify, then invert and check. An initial value problem (IVP) is an ODE together with the values of y and its derivatives at t = 0. These four rules cover most of what you need. They and the standard pairs (1 → 1/s, tⁿ → n!/sⁿ⁺¹, sin at → a/(s²+a²)) are in Lamar University's Laplace transform notes:

Each step has a pass/fail test:

1. Transform each term and insert the initial conditions. Pass: a second-order equation contributes two initial-value terms. Fail: your equation for Y has no constants even though y(0) or y′(0) is non-zero.
2. Solve for Y(s). Pass: the coefficient of Y equals the ODE's characteristic polynomial (s² + 3s + 2 for y″ + 3y′ + 2y). Fail: a different coefficient means a sign slipped in a derivative rule.
3. Cancel common factors, then use partial fractions. Pass: no factor appears in both numerator and denominator, and a repeated factor (s+1)² gets both A/(s+1) and B/(s+1)². Fail: your partial fractions include a factor that also appears in the numerator, or you cannot show that the coefficients of the cancelled factor are zero.
4. Invert and substitute back. Pass: y(0), y′(0) and the ODE are all satisfied. Fail: any one is not.

Worked check. Take y″ + 3y′ + 2y = e^(−t), y(0) = 0, y′(0) = 1.

• Step 1: (s²Y − 1) + 3sY + 2Y = 1/(s+1).
• Step 2: (s²+3s+2)Y = 1 + 1/(s+1) = (s+2)/(s+1).
• Step 3: Y = (s+2)/[(s+1)²(s+2)] = 1/(s+1)² after cancelling (s+2).
• Step 4: y = t·e^(−t).

Substituting back: y′ = e^(−t)(1−t) and y″ = e^(−t)(t−2). Then y″ + 3y′ + 2y = e^(−t)(t − 2 + 3 − 3t + 2t) = e^(−t), with y(0) = 0 and y′(0) = 1. The factor t appears because the forcing e^(−t) sits on a root of the characteristic polynomial.

Four slips are common in Laplace-transform work (our view from the worked problems above), and each has a quick test.

• Dropping f(0) or f′(0). Count the initial-condition terms in step 1 against the order of the equation.
• Reading the Laplace equation as the transform. "Laplace equation" (∂²u/∂x² + ∂²u/∂y² = 0) is a PDE solved by separation of variables, as in GATE 2022 Q.61. No integral is involved.
• Matching poles to exponentials. A pole of Y(s) at s = a gives a term e^(at), and a double pole gives t·e^(at). A pole at s = −1 must give e^(−t).
• Ignoring where the transform exists. The transform of e^(at) needs s > a. Q.12 states s > 0 because the transform of J₀ is only needed for positive s.

Our recommendation: about 6 to 8 hours over three or four sittings for GATE MA, and none for CSIR NET until your bulletin lists it. The block covers the definition, the four rules above, five or six standard pairs, partial fractions and a handful of IVPs. Then work Q.12-style questions that apply the rules to a non-standard function. This is our recommendation from the syllabus wording and the five papers above.

Before paying for anything, our free mathematics study content and PDF Drive give you notes and previous-year questions at no cost. Our guide to the CSIR NET test series explains what to check before you pay for one.

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