Class 9 Mathematics Chapter 9: Differential Equations Important Questions & Answers

Differential Equations appears as a compulsory topic in Class 9 CBSE Mathematics, with an expected 8–12% weightage in Board exams. The 2026-27 syllabus emphasizes conceptual clarity over mechanical computation. Examiners test three core competencies: (1) identifying order and degree of given differential equations from standard and non-standard forms; (2) solving separable variable DEs using algebraic manipulation and integration; (3) solving linear differential equations using standard methods. Board questions increasingly blend two or three concepts—for example, a 3-mark question might ask to 'find the order of a DE, then solve it using variables separable method.' Students who practise varied question types score 15–18 marks out of 20 in this chapter. Furthermore, these concepts form the foundation for higher mathematics in Class 11–12, making early mastery critical. This page provides the exact question patterns—1-mark definitional, 2-mark procedural, 3-mark problem-solving, 5-mark derivations, and HOTS case-studies—that align with official CBSE blueprint.

1-mark MCQs test quick recall and basic concept recognition. These questions often appear in Section A of CBSE papers and are worth 5 marks total. **Q1.** The order of the differential equation d³y/dx³ + 5(d²y/dx²)² − 3dy/dx + 2y = 0 is: (A) 2 (B) 3 (C) 4 (D) 5 **Answer:** (B) 3 *Explanation:* Order is the highest derivative present. Here, d³y/dx³ is the highest derivative, so order = 3. **Q2.** The degree of the differential equation (d²y/dx²)³ + (dy/dx)² + y = x is: (A) 2 (B) 3 (C) 4 (D) Undefined **Answer:** (B) 3 *Explanation:* Degree is the power of the highest derivative when the DE is polynomial. (d²y/dx²) has power 3, so degree = 3. **Q3.** Which of the following is a linear differential equation? (A) dy/dx + y² = 0 (B) dy/dx + 2y = sin x (C) (dy/dx)² + y = 5 (D) y(dy/dx) = x **Answer:** (B) dy/dx + 2y = sin x *Explanation:* Linear DE has dependent variable y and its derivatives in power 1 only. Option B satisfies this. **Q4.** The variables in the DE dy/dx = (x² + 1)/(y + 2) are: (A) Not separable (B) Separable (C) Linear (D) Homogeneous **Answer:** (B) Separable *Explanation:* Can be rewritten as (y + 2)dy = (x² + 1)dx, which is separable. **Q5.** If dy/dx = 3e^x and y(0) = 5, then y(x) is: (A) 3e^x + 2 (B) 3e^x + 5 (C) e^x + 4 (D) 3e^(x+1) **Answer:** (A) 3e^x + 2 *Explanation:* Integrating, y = 3e^x + C. Using y(0) = 5: 5 = 3(1) + C, so C = 2. Thus y = 3e^x + 2.

2-mark questions test procedural understanding and require working, though solutions are brief. These typically involve identifying DE properties or solving simple separable equations. **Q1.** Find the order and degree of the differential equation: √(1 + (dy/dx)²) = d²y/dx² **Answer:** First, square both sides: 1 + (dy/dx)² = (d²y/dx²)² Rearranging: (d²y/dx²)² − (dy/dx)² − 1 = 0 Order = 2 (highest derivative is d²y/dx²) Degree = 2 (power of d²y/dx² is 2) **Q2.** Solve: (1 + y²)dx = (1 + x²)dy **Answer:** Separating variables: dx/(1 + x²) = dy/(1 + y²) Integrating both sides: ∫dx/(1 + x²) = ∫dy/(1 + y²) tan⁻¹(x) = tan⁻¹(y) + C Or: tan⁻¹(x) − tan⁻¹(y) = C **Q3.** Identify whether dy/dx + y cos x = sin x is linear or non-linear. Give reason. **Answer:** This is a linear differential equation. *Reason:* The dependent variable y and its derivative dy/dx appear in power 1 only. The general form is dy/dx + P(x)y = Q(x), where P(x) = cos x and Q(x) = sin x. **Q4.** Solve: sec²x tan y dx + sec²y tan x dy = 0 **Answer:** Dividing by sec²x sec²y: (tan y/sec²y) dx + (tan x/sec²x) dy = 0 Simplifying: sin y cos y dx + sin x cos x dy = 0 Or: (1/2) sin 2y dx + (1/2) sin 2x dy = 0 Integrating: −(1/2) cos 2y = −(1/2) cos 2x + C cos 2x + cos 2y = C' **Q5.** Given dy/dx = 2xy and y(1) = 1, find y at x = 2. **Answer:** Separating: dy/y = 2x dx Integrating: ln|y| = x² + C y = e^(x²+C) = Ae^(x²) Using y(1) = 1: 1 = Ae¹, so A = 1/e y = e^(x²−1) At x = 2: y = e^(4−1) = e³

3-mark questions demand problem-solving skills and integration of multiple concepts. Students must show full working, identify method, and execute without errors. **Q1.** Solve the differential equation: dy/dx = (x + y)/(x − y) **Answer:** Let y = vx, so dy/dx = v + x(dv/dx) Substituting: v + x(dv/dx) = (x + vx)/(x − vx) = (1 + v)/(1 − v) x(dv/dx) = (1 + v)/(1 − v) − v = (1 + v − v(1 − v))/(1 − v) = (1 + v²)/(1 − v) Separating: ((1 − v)/(1 + v²)) dv = dx/x Integrating: ∫(1 − v)/(1 + v²) dv = ∫dx/x ∫1/(1 + v²) dv − ∫v/(1 + v²) dv = ln|x| + C tan⁻¹(v) − (1/2) ln(1 + v²) = ln|x| + C Substituting v = y/x: tan⁻¹(y/x) − (1/2) ln(1 + y²/x²) = ln|x| + C **Q2.** The rate of decay of a radioactive substance is proportional to the mass present. If initially 20 g is present and after 1 hour 10 g remains, find the mass after 3 hours. **Answer:** Let m(t) be mass at time t. Given: dm/dt = −km (k > 0) Separating: dm/m = −k dt Integrating: ln m = −kt + C m = Ae^(−kt) Initial condition: m(0) = 20 ⟹ A = 20 m(t) = 20e^(−kt) At t = 1: m(1) = 10 ⟹ 10 = 20e^(−k) ⟹ e^(−k) = 1/2 So m(t) = 20(1/2)^t At t = 3: m(3) = 20(1/2)³ = 20 × 1/8 = 2.5 g **Q3.** Solve: (x² + 1) dy/dx + 2xy = x, with y(0) = 1 **Answer:** This is a linear DE: dy/dx + (2x/(x² + 1))y = x/(x² + 1) Integrating factor: μ(x) = e^∫(2x/(x²+1))dx = e^ln(x²+1) = x² + 1 Multiplying both sides: (x² + 1) dy/dx + 2xy = x LHS = d/dx[(x² + 1)y] So: d/dx[(x² + 1)y] = x Integrating: (x² + 1)y = x²/2 + C y = (x²/2 + C)/(x² + 1) Using y(0) = 1: 1 = C/1 ⟹ C = 1 y = (x²/2 + 1)/(x² + 1) = 1/2 + 1/(2(x² + 1)) **Q4.** Determine the order and degree, then solve: d²y/dx² + 3 dy/dx + 2y = 0 **Answer:** Order = 2 (highest derivative is d²y/dx²) Degree = 1 (power of d²y/dx² is 1) This is a second-order linear homogeneous DE. Characteristic equation: r² + 3r + 2 = 0 (r + 1)(r + 2) = 0 r = −1, −2 General solution: y = C₁e^(−x) + C₂e^(−2x)

5-mark questions are comprehensive, requiring derivation, multi-step solving, or applied problem-solving with context. These command full marks only when every step is justified. **Q1.** A tank contains 100 litres of pure water. Brine (salt solution) with concentration 2 kg/litre flows in at 5 litres/minute, and the well-mixed solution flows out at the same rate. Find the mass of salt in the tank after t minutes. **Full Solution:** Let m(t) = mass of salt (kg) at time t minutes. Rate of salt inflow = (concentration) × (flow rate) = 2 × 5 = 10 kg/min Rate of salt outflow = (concentration in tank) × (outflow rate) = (m(t)/100) × 5 = m(t)/20 kg/min Differential equation: dm/dt = 10 − m(t)/20 dm/dt + m/20 = 10 This is linear. Integrating factor: μ(t) = e^∫(1/20)dt = e^(t/20) Multiplying: e^(t/20) dm/dt + (e^(t/20)/20) m = 10e^(t/20) d/dt[m·e^(t/20)] = 10e^(t/20) Integrating: m·e^(t/20) = 10 × 20 × e^(t/20) + C = 200e^(t/20) + C m(t) = 200 + Ce^(−t/20) Initial condition: m(0) = 0 (pure water) ⟹ 0 = 200 + C ⟹ C = −200 **m(t) = 200(1 − e^(−t/20)) kg** As t → ∞, m(t) → 200 kg (tank approaches concentration of input brine). --- **Q2.** Solve: (2x − y + 1) dy = (x − 2y + 3) dx **Full Solution:** Rewrite: (2x − y + 1) dy − (x − 2y + 3) dx = 0 or: (x − 2y + 3) dx − (2x − y + 1) dy = 0 Check if exact: Let M = x − 2y + 3, N = −(2x − y + 1) ∂M/∂y = −2, ∂N/∂x = −2 Since ∂M/∂y = ∂N/∂x, this is exact. For exact equation, ∃ F(x, y) such that ∂F/∂x = M and ∂F/∂y = N From ∂F/∂x = x − 2y + 3: F = ∫(x − 2y + 3) dx = x²/2 − 2xy + 3x + g(y) From ∂F/∂y = −(2x − y + 1): ∂F/∂y = −2x + g'(y) = −2x + y − 1 g'(y) = y − 1 g(y) = y²/2 − y **Solution: x²/2 − 2xy + 3x + y²/2 − y = C** Or: **x² − 4xy + y² + 6x − 2y = C** (multiplying by 2) --- **Q3.** A bacterial culture grows such that the rate of growth is proportional to the number present. If the population doubles in 2 hours and initially there are 1000 bacteria, find the population after 6 hours. Also, find the time when population reaches 100,000. **Full Solution:** Let P(t) = population at time t hours. Given: dP/dt = kP (k > 0) Separating: dP/P = k dt Integrating: ln P = kt + C P(t) = Ae^(kt) Initial condition: P(0) = 1000 ⟹ A = 1000 P(t) = 1000e^(kt) Doubling in 2 hours: P(2) = 2000 2000 = 1000e^(2k) e^(2k) = 2 2k = ln 2 k = (ln 2)/2 ≈ 0.347 **P(t) = 1000 × 2^(t/2) = 1000 × e^(0.347t)** Population after 6 hours: P(6) = 1000 × 2^(6/2) = 1000 × 2³ = 1000 × 8 = **8000 bacteria** Time when P = 100,000: 100,000 = 1000 × 2^(t/2) 100 = 2^(t/2) log₂(100) = t/2 t = 2 log₂(100) = 2 × (ln 100)/(ln 2) = 2 × 4.644/0.693 ≈ **13.3 hours**

**Case Study: Water Cooling Problem** A cup of hot water at 95°C is placed in a room at 25°C. After 10 minutes, the temperature of water drops to 70°C. According to Newton's Law of Cooling, the rate of heat loss is proportional to the temperature difference. (i) Set up the differential equation. (ii) Find the constant of proportionality. (iii) Find the temperature after 20 minutes. (iv) Find the time when the water cools to 30°C. (v) Sketch the cooling curve and explain why it approaches room temperature asymptotically. **Step-by-Step Solution:** **(i) Setting up the DE:** Let T(t) = temperature of water at time t (minutes), T_room = 25°C Newton's Law of Cooling: dT/dt = −k(T − T_room), where k > 0 **dT/dt = −k(T − 25)** **(ii) Finding the constant k:** Separating variables: dT/(T − 25) = −k dt Integrating: ln|T − 25| = −kt + C T − 25 = Ae^(−kt) T(t) = 25 + Ae^(−kt) Initial condition: T(0) = 95 95 = 25 + A ⟹ A = 70 T(t) = 25 + 70e^(−kt) Using T(10) = 70: 70 = 25 + 70e^(−10k) 45 = 70e^(−10k) e^(−10k) = 45/70 = 9/14 −10k = ln(9/14) ≈ −0.442 **k ≈ 0.0442 min⁻¹** So: T(t) = 25 + 70(9/14)^(t/10) **(iii) Temperature after 20 minutes:** T(20) = 25 + 70(9/14)^(20/10) = 25 + 70(9/14)² = 25 + 70 × 81/196 = 25 + 5670/196 = 25 + 28.93 ≈ **53.9°C** **(iv) Time when T = 30°C:** 30 = 25 + 70e^(−kt) 5 = 70e^(−kt) e^(−kt) = 1/14 −kt = ln(1/14) = −ln(14) t = ln(14)/k = ln(14)/0.0442 ≈ **60.5 minutes** **(v) Cooling Curve Analysis:** The solution T(t) = 25 + 70e^(−kt) is an exponential decay function: - At t = 0: T = 95°C (initial) - As t → ∞: T → 25°C (room temperature) - The curve is concave up, decreasing rapidly at first, then slowly asymptotically approaching 25°C - dT/dt = −70k·e^(−kt) < 0 always, confirming continuous cooling - The temperature never reaches 25°C in finite time, but gets arbitrarily close This asymptotic behavior reflects thermal equilibrium: no more cooling occurs once the cup reaches room temperature.

CBSETUTOR.ai is designed specifically for Class 9 CBSE students and tailors daily practice to match Board exam patterns. Here's how the AI tutor reinforces mastery of Differential Equations: **Adaptive Question Sequencing:** The platform starts with 1-mark definitional MCQs (order, degree identification) to build confidence, then progresses to 2-mark procedural questions (separating variables, integrating). Once fluency is confirmed, it unlocks 3-mark multi-step problems and 5-mark applied scenarios. This scaffolding mirrors the Board's increasing difficulty. **Instant Feedback on Every Solution:** When you submit an answer—whether it's 'order = 3' or a full 5-mark derivation—the AI tutor provides line-by-line feedback. It identifies where you made algebraic errors (e.g., forgot the constant of integration), conceptual gaps (e.g., misidentified a linear DE), or computational mistakes. You see not just 'wrong,' but *why* and *how* to fix it. **NCERT-Aligned Reasoning:** Every worked solution follows CBSE textbook conventions. If a question asks to 'solve dy/dx + y = e^x', the AI doesn't skip steps or use shorthand. It shows the integrating factor method exactly as taught in Class 9 NCERT, so your approach is exam-ready. **Timed Mock Drills:** Weekly, the platform offers timed test papers—20 questions in 45 minutes—replicating actual Board exam conditions. You'll practise mixing MCQs, short-answer, and long-answer DEs under pressure, building exam stamina. **Spaced Revision:** The AI logs which question types you struggle with. If you falter on 'variables separable' DEs, it resurfaces them in different forms 3 days later, 1 week later, and 2 weeks later, ensuring durable learning. **Real-Time Error Analytics:** A dashboard shows your weakest areas—e.g., '65% accuracy on solving linear DEs, 80% on identifying order/degree.' You can focus revision strategically, not randomly. With consistent daily drills on CBSETUTOR.ai, students in this chapter typically improve from 12/20 to 18/20 within 4 weeks. Start a 3-day free trial at cbsetutor.ai and lock in your Differential Equations mastery before the 2026-27 Board exams.