CELE Engineering Mathematics — Differential EquationsExam Answer Templates

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.

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

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.

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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)

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

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

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

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

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)

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

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

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

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

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

Do

• Use precise technical terms (e.g., 'order is the highest derivative present')
• State both parts of a definition if the question asks for two items
• Write formulas symbolically: μ = e^(∫P dx) not just 'integrating factor formula'

Don't

• Do not write lengthy explanations — 3 lines maximum
• Do not derive formulas unless asked — just state them
• Do not leave blank; even a partial attempt may earn the mark

Do

• Separate your answer into two visible parts that map to the two marks
• Show at least one intermediate step for calculation questions
• Label parts: 'P(x) = ...' then 'μ = ...' on separate lines

Don't

• Do not write the answer as one long run-on sentence
• Do not skip showing the integrating factor formula before evaluating it
• Do not assume the examiner will infer your reasoning — make it explicit

Do

• Number every step: Step 1, Step 2, Step 3
• State the characteristic equation or separated form explicitly before solving
• Apply the initial condition as a separate labeled step, not inline

Don't

• Do not combine all working into one block — marks cannot be awarded if steps are indistinguishable
• Do not forget the constant of integration C — it will cost you a mark
• Do not leave the answer in the form C₁e^mx + C₂e^mx for repeated roots

Do

• Write 'Given:', 'Required:', and 'Solution:' headers at the start
• State the governing DE or formula as Step 1 — this earns the first mark unconditionally
• Box or underline the final answer with units
• Add a sanity check (steady-state, special case, or dimensional analysis) for bonus credibility
• For IVP problems, show IC application as a clearly labeled separate step

Don't

• Do not skip the formula/method statement — it is 1 free mark
• Do not use only numerical values without showing the formula substitution line
• Do not round intermediate results — keep exact forms (e.g., k = ln2/1600) until the final step
• Do not write a wall of unseparated working — examiners cannot award marks they cannot locate