CELE Engineering Mathematics — Differential EquationsExam Answer Templates

Marks

1

Topic

Classification of Differential Equations

Difficulty

easy

Template Id

T1

Examiner Tip

One-mark questions reward precision. Two correct keyword concepts (order = highest derivative; degree = power of that derivative) are sufficient. Do not over-explain.

Model Answer

A differential equation is an equation relating a function and one or more of its derivatives. The order is the highest derivative present; the degree is the power of that highest-order derivative once the equation is expressed as a polynomial in its derivatives.

Question Type

very_short_answer

Answer Structure

• Line 1: Define differential equation (relation between function and its derivatives) [½ mark]
• Line 2: Define order (highest derivative) and degree (power of that derivative) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Confusing order with degree — they are not the same
• Stating degree without the condition that the equation must be polynomial (radical-free) in derivatives
• Incomplete definition that omits mention of derivatives

Key Phrases To Include

• highest derivative
• order
• degree
• polynomial in its derivatives

Marks

2

Topic

Classification of Differential Equations

Difficulty

easy

Template Id

T2

Examiner Tip

Always give the reason for linearity/nonlinearity. An answer that says 'nonlinear' without explanation earns only half the available mark.

Model Answer

Given: (d²y/dx²)³ + 5(dy/dx) = x² Order = 2 (highest derivative is the second derivative) Degree = 3 (the second derivative is raised to the power 3) Type: Nonlinear — because the highest-order derivative (d²y/dx²) appears with a power greater than 1. ∴ The ODE is of order 2, degree 3, and is nonlinear.

Question Type

short_answer

Answer Structure

• Step 1: Identify highest derivative → order = 2 [½ mark]
• Step 2: Identify power of highest derivative → degree = 3 [½ mark]
• Step 3: Assess linearity — power > 1 on d²y/dx² → nonlinear [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing order (2) with degree (3) — the most common error on this question type
• Calling it linear because x² is linear in x — linearity refers to the dependent variable y and its derivatives, not x
• Omitting the reason for the linearity classification

Key Phrases To Include

• order 2
• degree 3
• nonlinear
• highest-order derivative raised to power greater than 1

Marks

3

Topic

First-Order Separable ODEs

Difficulty

easy

Template Id

T3

Examiner Tip

Three-mark separable ODE questions almost always follow: separate → integrate → apply IC. Each step earns exactly one mark. Never skip writing the separated form — it is an explicit mark.

Model Answer

Given: dy/dx = 2xy, y(0) = 3 Required: y(x) Solution: Step 1 — Separate variables: dy/y = 2x dx Step 2 — Integrate both sides: ∫ dy/y = ∫ 2x dx ln|y| = x² + C₁ Step 3 — Exponentiate: |y| = e^(x² + C₁) = e^(C₁) · e^(x²) y = C e^(x²) where C = ±e^(C₁) Step 4 — Apply initial condition y(0) = 3: 3 = C · e^(0) = C ∴ C = 3 ∴ y = 3e^(x²) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Separate variables — dy/y = 2x dx [1 mark]
• Step 2: Integrate both sides correctly — ln|y| = x² + C [1 mark]
• Step 3: Apply initial condition to find C = 3 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the constant of integration C after integrating
• Not applying the initial condition — leaving the answer as y = Ce^(x²) without finding C
• Incorrect integration: writing ∫dy/y = y² / 2 instead of ln|y|
• Missing absolute value in ln|y| (minor, but noted by strict examiners)

Key Phrases To Include

• separate variables
• dy/y
• integrate both sides
• ln|y|
• apply initial condition
• C = 3

Marks

2

Topic

First-Order Linear ODEs

Difficulty

easy

Template Id

T4

Examiner Tip

Always write μ = e^(∫P dx) as the formula before substituting. This formula line earns the method mark independent of the arithmetic that follows.

Model Answer

Given: dy/dx + (2/x)y = x³ → Standard form: dy/dx + P(x)y = Q(x) Step 1 — Identify P(x): P(x) = 2/x Step 2 — Compute integrating factor: μ = e^(∫P dx) = e^(∫(2/x) dx) = e^(2 ln|x|) = e^(ln x²) = x² ∴ Integrating factor μ = x²

Question Type

short_answer

Answer Structure

• Step 1: Identify standard form and state P(x) = 2/x [1 mark]
• Step 2: Correctly compute μ = e^(∫P dx) = x² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not rewriting in standard form first — P(x) and Q(x) must be clearly separated
• Errors in integrating 2/x: writing x²/2 instead of 2 ln|x|
• Failing to simplify e^(2 ln x) to x²

Key Phrases To Include

• standard form
• P(x) = 2/x
• integrating factor
• μ = e^(∫P dx)
• x²

Marks

5

Topic

First-Order Linear ODEs

Difficulty

medium

Template Id

T5

Examiner Tip

In a 5-mark linear ODE, each logical step is worth exactly 1 mark. Never skip the 'd/dx[μy]' recognition — it is the pivot of the entire method and earns a dedicated mark.

Model Answer

Given: dy/dx + (2/x)y = x³, y(1) = 1 Required: y(x) ─── Step 1: Standard form ─── dy/dx + P(x)y = Q(x) where P(x) = 2/x, Q(x) = x³ ─── Step 2: Integrating factor ─── μ = e^(∫(2/x) dx) = e^(2 ln x) = x² ─── Step 3: Multiply both sides by μ ─── x² dy/dx + 2xy = x⁵ d/dx [x² y] = x⁵ ─── Step 4: Integrate both sides ─── x² y = ∫ x⁵ dx = x⁶/6 + C ─── Step 5: Solve for y ─── y = x⁴/6 + C/x² ─── Step 6: Apply initial condition y(1) = 1 ─── 1 = (1)⁴/6 + C/(1)² 1 = 1/6 + C C = 1 − 1/6 = 5/6 ∴ y = x⁴/6 + 5/(6x²) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Write standard form, identify P(x) and Q(x) [1 mark]
• Step 2: Compute integrating factor μ = x² [1 mark]
• Step 3: Multiply through and recognize d/dx[μy] = μQ(x) [1 mark]
• Step 4: Integrate and include constant C [1 mark]
• Step 5: Apply IC y(1) = 1 to find C = 5/6 and write final answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Multiplying only the left side by μ and forgetting to multiply Q(x) — loses Step 3 mark
• Not writing the LHS as a derivative: d/dx[μy] — misses the key recognition step
• Forgetting the constant C after integration — then no IC application possible
• Arithmetic error in applying IC, but correct method still earns 4/5 marks

Key Phrases To Include

• integrating factor
• μ = x²
• d/dx[x²y]
• multiply both sides
• x²y = x⁶/6 + C
• apply initial condition
• C = 5/6

Marks

5

Topic

Exact Differential Equations

Difficulty

hard

Template Id

T6

Examiner Tip

Show the partial derivatives in full, not just the final values. Examiners need to see the differentiation process to award the exactness mark.

Model Answer

Given: M dx + N dy = 0 where M = 2xy + y², N = x² + 2xy Required: Verify exactness; find solution F(x, y) = C ─── Step 1: Check exactness ─── ∂M/∂y = 2x + 2y ∂N/∂x = 2x + 2y Since ∂M/∂y = ∂N/∂x, the equation IS exact. ✓ ─── Step 2: Find F(x, y) from ∂F/∂x = M ─── F = ∫M dx = ∫(2xy + y²) dx = x²y + xy² + g(y) ─── Step 3: Use ∂F/∂y = N to find g(y) ─── ∂F/∂y = x² + 2xy + g'(y) = x² + 2xy (= N) g'(y) = 0 → g(y) = constant (absorbed into C) ─── Step 4: Write general solution ─── ∴ F(x, y) = x²y + xy² = C ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Compute ∂M/∂y and ∂N/∂x; confirm they are equal → exactness established [1 mark]
• Step 2: Integrate M w.r.t. x to get F = x²y + xy² + g(y) [1 mark]
• Step 3: Differentiate F w.r.t. y, equate to N, solve for g'(y) = 0 [1 mark]
• Step 4: State g(y) = const and write final solution x²y + xy² = C [1 mark]
• Step 5: Complete logical flow from exactness check to final implicit solution [1 mark]

Scoring Breakdown

Common Mark Deductions

• Failing to state the exactness condition ∂M/∂y = ∂N/∂x before checking
• Integrating N dy first instead of M dx — valid, but must be consistent; mixing approaches loses marks
• Dropping the arbitrary function g(y) when integrating M w.r.t. x
• Forgetting to equate ∂F/∂y to N — skipping this step loses 2 consecutive marks

Key Phrases To Include

• ∂M/∂y = ∂N/∂x
• exact equation
• ∂F/∂x = M
• ∂F/∂y = N
• g'(y) = 0
• x²y + xy² = C

Marks

3

Topic

Higher-Order Linear ODEs — Distinct Real Roots

Difficulty

easy

Template Id

T7

Examiner Tip

The characteristic equation approach is worth 1 mark by itself. Even if you cannot factor the quadratic, using the quadratic formula to get correct m values still earns the roots mark.

Model Answer

Given: y'' − 5y' + 6y = 0 Required: General solution y(x) Step 1 — Write the characteristic equation: m² − 5m + 6 = 0 Step 2 — Solve (factor): (m − 2)(m − 3) = 0 m₁ = 2, m₂ = 3 (distinct real roots) Step 3 — Write general solution: y = C₁e^(2x) + C₂e^(3x) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Write characteristic equation m² − 5m + 6 = 0 [1 mark]
• Step 2: Find roots m = 2, m = 3 [1 mark]
• Step 3: State general solution y = C₁e^(2x) + C₂e^(3x) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing the characteristic equation as m² − 5m + 6 = y (including y) instead of = 0
• Correct roots but wrong solution form — e.g., writing y = C₁e^(2x) only (one term)
• Using C₁ and C₂ without superscript: writing e2x instead of e^(2x) — penalized in strict marking

Key Phrases To Include

• characteristic equation
• m² − 5m + 6 = 0
• distinct real roots
• m = 2, m = 3
• C₁e^(2x) + C₂e^(3x)

Marks

3

Topic

Higher-Order Linear ODEs — Repeated Roots

Difficulty

medium

Template Id

T8

Examiner Tip

The repeated-root form (C₁ + C₂x)e^(mx) is tested almost every board exam cycle. Memorize it and its derivation rationale (reduction of order).

Model Answer

Given: y'' − 6y' + 9y = 0 Required: General solution y(x) Step 1 — Characteristic equation: m² − 6m + 9 = 0 Step 2 — Solve: (m − 3)² = 0 m = 3 (repeated root, multiplicity 2) Step 3 — General solution for repeated root: y = (C₁ + C₂x)e^(3x) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Form characteristic equation m² − 6m + 9 = 0 [1 mark]
• Step 2: Identify repeated root m = 3 [1 mark]
• Step 3: Write correct solution form y = (C₁ + C₂x)e^(3x) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing y = C₁e^(3x) + C₂e^(3x) — this is the distinct-root form applied to a repeated root; loses the solution mark
• Factoring incorrectly and getting m = 3 and m = −3
• Omitting the x factor in C₂x — the most common error on repeated-root problems

Key Phrases To Include

• repeated root
• m = 3 twice
• (C₁ + C₂x)e^(3x)
• multiplicity 2

Marks

3

Topic

Higher-Order Linear ODEs — Complex Roots

Difficulty

medium

Template Id

T9

Examiner Tip

For purely imaginary roots (no real part), e^(0·x) = 1, so the exponential term vanishes. Write this simplification explicitly to earn full marks.

Model Answer

Given: y'' + 4y = 0 Required: General solution y(x) Step 1 — Characteristic equation: m² + 4 = 0 m² = −4 m = ±2i (purely imaginary, i.e., α = 0, β = 2) Step 2 — Complex root solution form: y = e^(αx)[C₁cos(βx) + C₂sin(βx)] = e^(0)[C₁cos(2x) + C₂sin(2x)] Step 3 — Final answer: y = C₁cos(2x) + C₂sin(2x) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Form characteristic equation, solve to get m = ±2i [1 mark]
• Step 2: Identify α = 0, β = 2 [1 mark]
• Step 3: Write general solution y = C₁cos(2x) + C₂sin(2x) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing y = e^(2x)(C₁cos + C₂sin) — incorrectly including e^(2x) when α = 0
• Giving only one trigonometric term in the solution
• Confusing β = 2 with β = 4 (from m² = −4, students sometimes take β = 4)

Key Phrases To Include

• complex roots
• m = ±2i
• α = 0
• β = 2
• C₁cos(2x) + C₂sin(2x)

Marks

5

Topic

Growth and Decay Applications

Difficulty

medium

Template Id

T10

Examiner Tip

Always verify: at t = 6 h (two 3-hour periods), the population should double twice: 500 × 2 × 2 × 2 × 2 = no — actually, 8³ happens for each 3-hour interval — but since it triples every period by factor 8, y(6) = 500 × 8² = 32 000. Use this as a quick sanity check.

Model Answer

Given: y₀ = 500, y(3) = 4000 Required: (a) k; (b) y(6) Governing Equation: dy/dt = ky → y = y₀ e^(kt) ─── Part (a): Find k ─── At t = 3 h: y(3) = 4000 4000 = 500 e^(3k) e^(3k) = 4000/500 = 8 3k = ln 8 = 3 ln 2 k = ln 2 = 0.6931 hr⁻¹ ─── Part (b): Find y(6) ─── y(6) = 500 e^(0.6931 × 6) = 500 e^(4.1589) = 500 × 64 = 32 000 cells ∴ (a) k = ln 2 ≈ 0.6931 hr⁻¹ (b) y(6) = 32 000 cells ← Final Answers

Question Type

numerical

Answer Structure

• Step 1: State governing equation y = y₀e^(kt) [1 mark]
• Step 2: Substitute y₀ = 500, y(3) = 4000 to set up equation for k [1 mark]
• Step 3: Solve k = ln 2 / 1 = 0.6931 hr⁻¹ [1 mark]
• Step 4: Substitute into y(6) with found k [1 mark]
• Step 5: Evaluate y(6) = 32 000 cells with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to state the governing differential equation — loses method mark
• Leaving k as ln 8 / 3 without simplifying — loses partial credit for k
• Using k from part (a) in a rounded form causing compounded error in part (b)
• Omitting units: k must have units of hr⁻¹, and y(6) must have units of cells

Key Phrases To Include

• y = y₀e^(kt)
• k = ln 2
• k = 0.6931 hr⁻¹
• y(6) = 32 000 cells
• growth constant

Marks

5

Topic

Newton's Law of Cooling

Difficulty

medium

Template Id

T11

Examiner Tip

Notice that T − T_s at t = 5 is 40 (= 60 × 2/3), at t = 10 is 60 × (2/3)² = 26.67, and at t = 15 is 60 × (2/3)³ = 17.78. Recognizing this geometric pattern allows a fast check.

Model Answer

Given: T₀ = 80°C, T_s = 20°C (surroundings), T(5) = 60°C Required: T(15) Governing Equation (Newton's Law of Cooling): dT/dt = −k(T − T_s) → T(t) = T_s + (T₀ − T_s) e^(−kt) T(t) = 20 + 60 e^(−kt) ─── Step 1: Find k from T(5) = 60 ─── 60 = 20 + 60 e^(−5k) 40 = 60 e^(−5k) e^(−5k) = 40/60 = 2/3 −5k = ln(2/3) k = −ln(2/3)/5 = ln(3/2)/5 = 0.08109 min⁻¹ ─── Step 2: Find T(15) ─── T(15) = 20 + 60 e^(−0.08109 × 15) = 20 + 60 e^(−1.2164) = 20 + 60 × (2/3)³ = 20 + 60 × 8/27 = 20 + 17.78 = 37.78°C ≈ 37.8°C ∴ T(15) ≈ 37.8°C ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: State Newton's Law of Cooling equation and general solution form [1 mark]
• Step 2: Substitute T_s, T₀ to get T(t) = 20 + 60e^(−kt) [1 mark]
• Step 3: Apply T(5) = 60 and solve for k = ln(3/2)/5 [1 mark]
• Step 4: Substitute k and t = 15 into T(t) [1 mark]
• Step 5: Evaluate T(15) ≈ 37.8°C with unit °C [1 mark]

Scoring Breakdown

Common Mark Deductions

• Positive sign in dT/dt = +k(T − T_s) — sign error means the body heats up instead of cools
• Using T₀ − T_s = 80 − 20 = 60 incorrectly as T_s = 0 — must keep T_s in the formula
• Not substituting t = 15 (three 5-minute periods) without recognizing the geometric progression shortcut
• Rounding k too early (e.g., k = 0.08) causing significant error in T(15)

Key Phrases To Include

• Newton's Law of Cooling
• dT/dt = −k(T − T_s)
• T(t) = T_s + (T₀ − T_s)e^(−kt)
• k = ln(3/2)/5
• T(15) ≈ 37.8°C

Marks

3

Topic

Radioactive Decay — Applications

Difficulty

medium

Template Id

T12

Examiner Tip

After finding k, substitute as k = ln2/1600 symbolically before reaching for a calculator. This avoids rounding errors and shows the examiner your mastery of the method.

Model Answer

Given: Half-life t₁/₂ = 1600 yr, t = 4000 yr Required: Fraction remaining y/y₀ Step 1 — Find decay constant k: y = y₀ e^(−kt) At t = 1600: y = y₀/2 y₀/2 = y₀ e^(−1600k) 1/2 = e^(−1600k) k = ln 2 / 1600 = 4.332 × 10⁻⁴ yr⁻¹ Step 2 — Fraction at t = 4000 yr: y/y₀ = e^(−kt) = e^(−4000 × ln2/1600) = e^(−(4000/1600) ln2) = e^(−2.5 ln2) = 2^(−2.5) = 1 / (2^2.5) = 1 / (4√2) ≈ 0.1768 ∴ Fraction remaining ≈ 0.177 (17.7%) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Use half-life condition to find k = ln2/1600 [1 mark]
• Step 2: Compute y/y₀ = e^(−4000k) using exact k [1 mark]
• Step 3: Simplify to 2^(−2.5) = 1/(4√2) ≈ 0.177 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using k = −ln2/1600 as a positive constant then placing a positive exponent — double-negative error
• Computing 4000/1600 = 2.5 but then evaluating 2^(2.5) instead of 2^(−2.5)
• Forgetting to express the answer as a fraction or percentage — 0.177 or 17.7% required

Key Phrases To Include

• half-life
• k = ln2/t₁/₂
• y/y₀ = e^(−kt)
• 2^(−2.5)
• 0.177

Marks

5

Topic

Mixing Tank Applications

Difficulty

hard

Template Id

T13

Examiner Tip

The mixing problem is a near-guaranteed 5-mark question in PRC CE board exams. The key phrase 'rate in − rate out' always starts the model answer. Steady-state value (Q → 50 kg) is an excellent self-check.

Model Answer

Given: V = 100 L (constant), Q₀ = 10 kg, inflow: 0.5 kg/L × 5 L/min = 2.5 kg/min; outflow concentration = Q/100 kg/L × 5 L/min = Q/20 kg/min Required: Q(t) ─── Step 1: Set up the DE ─── dQ/dt = Rate in − Rate out dQ/dt = 2.5 − Q/20 ─── Step 2: Rewrite in standard linear form ─── dQ/dt + (1/20)Q = 2.5 ─── Step 3: Integrating factor ─── μ = e^(∫(1/20) dt) = e^(t/20) ─── Step 4: Multiply and integrate ─── d/dt [Q e^(t/20)] = 2.5 e^(t/20) Q e^(t/20) = ∫ 2.5 e^(t/20) dt = 2.5 × 20 e^(t/20) + C = 50 e^(t/20) + C Q(t) = 50 + C e^(−t/20) ─── Step 5: Apply IC Q(0) = 10 ─── 10 = 50 + C C = −40 ∴ Q(t) = 50 − 40 e^(−t/20) [kg] ← Final Answer Steady-state check: as t → ∞, Q → 50 kg ✓ (= 0.5 kg/L × 100 L)

Question Type

numerical

Answer Structure

• Step 1: Formulate dQ/dt = Rate in − Rate out = 2.5 − Q/20 [1 mark]
• Step 2: Rewrite in standard linear form and find integrating factor e^(t/20) [1 mark]
• Step 3: Multiply through and write as d/dt[Qe^(t/20)] [1 mark]
• Step 4: Integrate to get Q(t) = 50 + Ce^(−t/20) [1 mark]
• Step 5: Apply Q(0) = 10 → C = −40; state Q(t) = 50 − 40e^(−t/20) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using outflow rate as 0.5 kg/L × 5 L/min instead of Q/100 × 5 — incorrect because concentration changes with Q
• Forgetting that volume is constant — if inflow ≠ outflow rate, volume changes and the DE becomes more complex
• Sign error: writing dQ/dt = Rate out − Rate in gives a decreasing-only solution — physically wrong here
• Omitting the steady-state check or units for the final answer

Key Phrases To Include

• rate in − rate out
• dQ/dt + Q/20 = 2.5
• integrating factor e^(t/20)
• Q(t) = 50 − 40e^(−t/20)
• steady state

Marks

5

Topic

Higher-Order Linear ODEs — Initial Value Problem

Difficulty

hard

Template Id

T14

Examiner Tip

When differentiating y = (C₁ + C₂x)e^(−3x), explicitly write 'by product rule' before differentiating. This labels the step and earns the differentiation mark even if there is a minor slip.

Model Answer

Given: y'' + 6y' + 9y = 0, y(0) = 2, y'(0) = −1 Required: Particular solution y(x) ─── Step 1: Characteristic equation ─── m² + 6m + 9 = 0 (m + 3)² = 0 m = −3 (repeated root) ─── Step 2: General solution ─── y = (C₁ + C₂x) e^(−3x) ─── Step 3: Apply y(0) = 2 ─── 2 = (C₁ + 0) e^0 = C₁ C₁ = 2 ─── Step 4: Find y' ─── y' = C₂ e^(−3x) + (C₁ + C₂x)(−3) e^(−3x) = [C₂ − 3(C₁ + C₂x)] e^(−3x) ─── Step 5: Apply y'(0) = −1 ─── −1 = [C₂ − 3C₁] e^0 = C₂ − 3(2) = C₂ − 6 C₂ = 5 ∴ y = (2 + 5x) e^(−3x) ← Final Answer

Question Type

numerical

Answer Structure

• Step 1: Characteristic equation → repeated root m = −3 [1 mark]
• Step 2: General solution y = (C₁ + C₂x)e^(−3x) [1 mark]
• Step 3: Apply y(0) = 2 → C₁ = 2 [1 mark]
• Step 4: Differentiate y correctly using product rule [1 mark]
• Step 5: Apply y'(0) = −1 → C₂ = 5; state final particular solution [1 mark]

Scoring Breakdown

Common Mark Deductions

• Differentiating y' = C₂ e^(−3x) without the product rule — forgetting to differentiate (C₁ + C₂x)
• Writing the wrong general solution form C₁e^(−3x) + C₂e^(−3x) — these are linearly dependent
• Applying both ICs simultaneously without first finding y' — disorganized approach loses marks
• Final answer not simplified; leaving C₁ and C₂ symbolic

Key Phrases To Include

• repeated root
• m = −3
• (C₁ + C₂x)e^(−3x)
• product rule
• C₁ = 2
• C₂ = 5
• particular solution

Marks

2

Topic

Laplace Transform

Difficulty

easy

Template Id

T15

Examiner Tip

Board exams frequently test Laplace transform pairs as 2-mark questions. Memorize the table: e^(at) → 1/(s−a); sin(bt) → b/(s²+b²); cos(bt) → s/(s²+b²); t^n → n!/s^(n+1).

Model Answer

The Laplace transform of a function f(t) is defined as: L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt, s > a (for convergence) Standard transforms: L{e^(at)} = 1/(s − a), s > a L{sin(bt)} = b/(s² + b²), s > 0

Question Type

short_answer

Answer Structure

• Line 1: State the integral definition of the Laplace transform [1 mark]
• Line 2: State both L{e^(at)} = 1/(s−a) and L{sin(bt)} = b/(s²+b²) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing L{e^(at)} = 1/(s + a) — sign error in denominator
• Writing L{sin(bt)} = 1/(s² + b²) without the b numerator
• Omitting lower limit of integration (should be 0 for one-sided Laplace transform)

Key Phrases To Include

• ∫₀^∞ e^(−st) f(t) dt
• 1/(s − a)
• b/(s² + b²)
• convergence condition s > a