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CBSE Class 12 · Chapter 9 · Calculus · 2026-27

Differential Equations Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

  • 15 MCQs (1 mark each)
  • 2 case studies (4 marks each)
  • No calculator needed
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Multiple-choice questions

Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceOrder and degree

Which of these differential equations has order \(2\) and degree \(3\)?

  1. (a)\(\left(\dfrac{d^2y}{dx^2}\right)^3 + \dfrac{dy}{dx} = x\)
  2. (b)\(\left(\dfrac{d^3y}{dx^3}\right)^2 + y = 0\)
  3. (c)\(\left(\dfrac{dy}{dx}\right)^3 + \dfrac{d^2y}{dx^2} = 0\)
  4. (d)\(\dfrac{d^2y}{dx^2} + \sin\left(\dfrac{dy}{dx}\right) = 0\)
Show answer
Answer: (a) \(\left(\dfrac{d^2y}{dx^2}\right)^3 + \dfrac{dy}{dx} = x\)

Why: Order = highest derivative; degree = its power, when the equation is a polynomial in the derivatives.

(a) highest derivative \(\tfrac{d^2y}{dx^2}\), power \(3\): order 2, degree 3. (b) order 3, degree 2. (c) order 2, degree 1 (the cube is on the first derivative). (d) not a polynomial in \(\tfrac{dy}{dx}\), so its degree is not defined.

Q2

·1 mark·Multiple choiceOrder and degree

The degree of the differential equation \(\left(\dfrac{dy}{dx}\right)^4 + 3y\dfrac{d^2y}{dx^2} = 0\) is

  1. (a)\(4\)
  2. (b)\(1\)
  3. (c)\(2\)
  4. (d)not defined
Show answer
Answer: (b) \(1\)

Why: Degree is the power of the highest order derivative (d²y/dx²), which is 1.

Order \(= 2\) (highest derivative \(\tfrac{d^2y}{dx^2}\)); it appears to power \(1\), so the degree is \(1\). The power \(4\) belongs to a lower-order derivative.

Q3

·1 mark·Multiple choiceOrder and degree

The order of the differential equation \(x\dfrac{d^3y}{dx^3} + \left(\dfrac{dy}{dx}\right)^5 - y = 0\) is

  1. (a)\(5\)
  2. (b)\(1\)
  3. (c)\(3\)
  4. (d)\(2\)
Show answer
Answer: (c) \(3\)

Why: Order is the order of the highest derivative present: d³y/dx³.

The highest derivative is \(\dfrac{d^3y}{dx^3}\), so the order is \(3\). (\(5\) is a power of the first derivative; it affects neither the order nor the degree, which is \(1\).)

Q4

·1 mark·Multiple choiceGeneral and particular solutions

The general solution of \(\dfrac{dy}{dx} = 2x\) is

  1. (a)\(y = 2\)
  2. (b)\(y = x^2\)
  3. (c)\(y = 2x^2 + C\)
  4. (d)\(y = x^2 + C\)
Show answer
Answer: (d) \(y = x^2 + C\)

Why: Integrate both sides: y = ∫2x dx.

\(y = \int 2x\,dx = x^2 + C\). (\(y = x^2\) is only one particular solution.)

Q5

·1 mark·Multiple choiceVariables separable

The general solution of \(\dfrac{dy}{dx} = y\) (\(y \gt 0\)) is

  1. (a)\(y = x + C\)
  2. (b)\(y = Ce^{x}\)
  3. (c)\(y = e^{x} + C\)
  4. (d)\(y = \dfrac{x^2}{2} + C\)
Show answer
Answer: (b) \(y = Ce^{x}\)

Why: Separate: dy/y = dx, so log y = x + c.

\(\dfrac{dy}{y} = dx \Rightarrow \log y = x + c \Rightarrow y = e^{x + c} = Ce^x\) (\(C = e^c \gt 0\)).

Q6

·1 mark·Multiple choiceVariables separable

The general solution of \(\dfrac{dy}{dx} = \dfrac{x}{y}\) is

  1. (a)\(x^2 + y^2 = C\)
  2. (b)\(y = Cx\)
  3. (c)\(y^2 - x^2 = C\)
  4. (d)\(xy = C\)
Show answer
Answer: (c) \(y^2 - x^2 = C\)

Why: Separate: y dy = x dx, so y²/2 = x²/2 + c.

\(y\,dy = x\,dx \Rightarrow \dfrac{y^2}{2} = \dfrac{x^2}{2} + c \Rightarrow y^2 - x^2 = C\).

Q7

·1 mark·Multiple choiceLinear equations

An integrating factor of \(\dfrac{dy}{dx} + \dfrac{2}{x}y = x\) (\(x \gt 0\)) is

  1. (a)\(x^2\)
  2. (b)\(2\log x\)
  3. (c)\(e^{2x}\)
  4. (d)\(\dfrac{1}{x^2}\)
Show answer
Answer: (a) \(x^2\)

Why: IF = e∫P dx = e2 log x = x².

\(P = \dfrac2x\), \(\int P\,dx = 2\log x = \log x^2\), so IF \(= e^{\log x^2} = x^2\).

Q8

·1 mark·Multiple choiceLinear equations

An integrating factor of \(\dfrac{dy}{dx} - y\tan x = \sin x\) for \(0 \lt x \lt \dfrac{\pi}{2}\) is

  1. (a)\(\sec x\)
  2. (b)\(\cos x\)
  3. (c)\(e^{\tan x}\)
  4. (d)\(\log(\cos x)\)
Show answer
Answer: (b) \(\cos x\)

Why: IF = e−∫tan x dx = elog cos x = cos x.

\(P = -\tan x\), \(\int P\,dx = \log(\cos x)\) (as \(\int\tan x\,dx = -\log\cos x\)), so IF \(= e^{\log\cos x} = \cos x\).

Q9

·1 mark·Multiple choiceHomogeneous equations

Which of the following differential equations is not homogeneous?

  1. (a)\(\dfrac{dy}{dx} = \dfrac{x + y}{x}\)
  2. (b)\(\dfrac{dy}{dx} = \dfrac{y}{x} + \cos\dfrac{y}{x}\)
  3. (c)\(\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}\)
  4. (d)\(\dfrac{dy}{dx} = \dfrac{x^2 + y}{x^2}\)
Show answer
Answer: (d) \(\dfrac{dy}{dx} = \dfrac{x^2 + y}{x^2}\)

Why: Scaling x and y by λ leaves the right side unchanged except for (x² + y)/x².

For a homogeneous equation \(F(\lambda x, \lambda y) = F(x, y)\). This holds for the first three (each is a function of \(\tfrac{y}{x}\)), but \(\dfrac{\lambda^2x^2 + \lambda y}{\lambda^2x^2} \ne \dfrac{x^2 + y}{x^2}\) in general.

Q10

·1 mark·Multiple choiceGeneral and particular solutions

Which of the following is a solution of \(\dfrac{d^2y}{dx^2} - 4y = 0\)?

  1. (a)\(y = e^{2x}\)
  2. (b)\(y = \sin 2x\)
  3. (c)\(y = e^{4x}\)
  4. (d)\(y = \cos 2x\)
Show answer
Answer: (a) \(y = e^{2x}\)

Why: For y = e2x, y″ = 4e2x = 4y.

\(y = e^{2x}\): \(y'' = 4e^{2x}\), so \(y'' - 4y = 0\). For \(\sin 2x\) and \(\cos 2x\), \(y'' = -4y\); for \(e^{4x}\), \(y'' = 16y\).

Q11

·1 mark·Multiple choiceGeneral and particular solutions

The particular solution of \(\dfrac{dy}{dx} = 3x^2\) with \(y(1) = 2\) is

  1. (a)\(y = x^3 + 2\)
  2. (b)\(y = 3x^3 - 1\)
  3. (c)\(y = x^3 + 1\)
  4. (d)\(y = x^3\)
Show answer
Answer: (c) \(y = x^3 + 1\)

Why: y = x³ + C and 2 = 1 + C.

\(y = x^3 + C\); \(y(1) = 2 \Rightarrow C = 1\), so \(y = x^3 + 1\).

Q12

·1 mark·Multiple choiceLinear equations

The solution of \(\dfrac{dy}{dx} + y = 2\) with \(y(0) = 0\) is

  1. (a)\(y = 2e^{-x}\)
  2. (b)\(y = 2 - 2e^{-x}\)
  3. (c)\(y = 2 + 2e^{-x}\)
  4. (d)\(y = 2x\)
Show answer
Answer: (b) \(y = 2 - 2e^{-x}\)

Why: IF = eˣ: (y eˣ)′ = 2eˣ, so y = 2 + Ce−x; y(0) = 0 gives C = −2.

IF \(= e^x\): \(\dfrac{d}{dx}(ye^x) = 2e^x \Rightarrow ye^x = 2e^x + C \Rightarrow y = 2 + Ce^{-x}\). \(y(0) = 0 \Rightarrow C = -2\): \(y = 2 - 2e^{-x}\).

Q13

·1 mark·Multiple choiceVariables separable

The solution curve of \(\dfrac{dy}{dx} = -\dfrac{y}{x}\) that passes through \((2, 3)\) also passes through

  1. (a)\((6, 6)\)
  2. (b)\((1, 3)\)
  3. (c)\((2, -3)\)
  4. (d)\((3, 2)\)
Show answer
Answer: (d) \((3, 2)\)

Why: Separate: dy/y = −dx/x gives xy = C = 6.

\(\dfrac{dy}{y} = -\dfrac{dx}{x} \Rightarrow \log|y| = -\log|x| + c \Rightarrow xy = C\). Through \((2, 3)\): \(xy = 6\), which contains \((3, 2)\).

Q14

·1 mark·Multiple choiceVariables separable

The general solution of \(\dfrac{dy}{dx} = 1 + y^2\) is

  1. (a)\(\tan^{-1}y = x + C\)
  2. (b)\(y = \tan x + C\)
  3. (c)\(\log(1 + y^2) = x + C\)
  4. (d)\(y + \dfrac{y^3}{3} = x + C\)
Show answer
Answer: (a) \(\tan^{-1}y = x + C\)

Why: Separate: dy/(1 + y²) = dx.

\(\displaystyle\int\dfrac{dy}{1 + y^2} = \int dx \Rightarrow \tan^{-1}y = x + C\). (\(y = \tan x + C\) does not satisfy the equation: its derivative is \(\sec^2 x\), not \(1 + (\tan x + C)^2\).)

Q15

·1 mark·Multiple choiceLinear equations

For \(x \gt 0\), the general solution of \(x\dfrac{dy}{dx} + y = x^2\) is

  1. (a)\(y = \dfrac{x^2}{3}\)
  2. (b)\(xy = x^3 + C\)
  3. (c)\(xy = \dfrac{x^3}{3} + C\)
  4. (d)\(y = x^2 + C\)
Show answer
Answer: (c) \(xy = \dfrac{x^3}{3} + C\)

Why: The left side is d(xy)/dx.

\(x\dfrac{dy}{dx} + y = \dfrac{d}{dx}(xy) = x^2 \Rightarrow xy = \dfrac{x^3}{3} + C\).

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.

Case study 1: Medicine in the bloodstream (4 marks)

A doctor explains that after an injection, the amount \(A\) mg of a medicine in a patient's blood decreases at a rate proportional to the amount present: \(\dfrac{dA}{dt} = -kA\), where \(t\) is in hours and \(k \gt 0\). The injection gives \(80\) mg, and after \(2\) hours \(20\) mg remains. (Here \(\log\) means \(\log_e\).)

(i) Solve the differential equation with \(A(0) = 80\). [1 mark]
Show answer
Answer: \(A = 80e^{-kt}\)
\(\dfrac{dA}{A} = -k\,dt \Rightarrow \log A = -kt + c\); \(A(0) = 80\) gives \(A = 80e^{-kt}\). A1
(ii) Find \(k\). [1 mark]
Show answer
Answer: \(k = \log 2\)
\(80e^{-2k} = 20 \Rightarrow e^{-2k} = \dfrac14 \Rightarrow 2k = \log 4 \Rightarrow k = \log 2\). A1
(iii) After how many hours will only \(5\) mg remain? [2 marks]
Show answer
Answer: \(4\) hours
\(80e^{-t\log 2} = 80 \cdot 2^{-t} = 5\) M1 \(\Rightarrow 2^{-t} = \dfrac{1}{16} \Rightarrow t = 4\) hours. A1
OR How much medicine remains after \(3\) hours? [2 marks]
Show answer
Answer: \(10\) mg
\(A(3) = 80e^{-3\log 2} = 80 \cdot 2^{-3}\) M1 \(= 10\) mg. A1

Case study 2: Drip infusion (4 marks)

A patient receives a drug through a drip. The concentration \(c\) (mg per litre) of the drug in the blood \(t\) hours after the drip starts satisfies \(\dfrac{dc}{dt} + \dfrac{c}{5} = 2\), with \(c = 0\) when \(t = 0\).

(i) Write the order and degree of the differential equation. [1 mark]
Show answer
Answer: Order \(1\), degree \(1\)
Highest derivative \(\dfrac{dc}{dt}\): order \(1\); it appears to power \(1\): degree \(1\). A1
(ii) Find the integrating factor. [1 mark]
Show answer
Answer: \(e^{t/5}\)
\(P = \dfrac15\), IF \(= e^{\int \frac15 dt} = e^{t/5}\). A1
(iii) Solve the equation to find \(c\) in terms of \(t\). [2 marks]
Show answer
Answer: \(c = 10 - 10e^{-t/5}\)
\(\dfrac{d}{dt}\left(ce^{t/5}\right) = 2e^{t/5} \Rightarrow ce^{t/5} = 10e^{t/5} + K\) M1; \(c(0) = 0 \Rightarrow K = -10\), so \(c = 10 - 10e^{-t/5}\). A1
OR Using the solution \(c = 10 - 10e^{-t/5}\), show that the concentration always stays below \(10\) mg/L and approaches \(10\) mg/L as \(t\) grows. [2 marks]
Show answer
Answer: \(c \lt 10\); \(c \to 10\)
\(10e^{-t/5} \gt 0\) for all \(t\), so \(c \lt 10\) M1; as \(t \to \infty\), \(e^{-t/5} \to 0\) and \(c \to 10\). A1

Next steps for Differential Equations

This free set is separate from the chapter's question bank. On the Differential Equations chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 40 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.

Differential Equations: notes and questionsNCERT solutions, Chapter 9What Essentials includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.