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)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. 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 $3$ ?

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Answer: (a)

{(\frac{d^{2} y}{d x^{2}})}^{3} + \frac{d y}{d x} = x

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

(a) highest derivative $\frac{d^{2} y}{d x^{2}}$ , 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 $\frac{d y}{d x}$ , so its degree is not defined.

Q2

·1 mark·Multiple choiceOrder and degree

The degree of the differential equation ${(\frac{d y}{d x})}^{4} + 3 y \frac{d^{2} y}{d x^{2}} = 0$ is

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Answer: (b)

1

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

Order $= 2$ (highest derivative $\frac{d^{2} y}{d x^{2}}$ ); 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 $x \frac{d^{3} y}{d x^{3}} + {(\frac{d y}{d x})}^{5} - y = 0$ is

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Answer: (c)

3

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

The highest derivative is $\frac{d^{3} y}{d x^{3}}$ , 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 $\frac{d y}{d x} = 2 x$ is

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Answer: (d)

y = x^{2} + C

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

$y = \int 2 x d x = x^{2} + C$ . ( $y = x^{2}$ is only one particular solution.)

Q5

·1 mark·Multiple choiceVariables separable

The general solution of $\frac{d y}{d x} = y$ ( $y > 0$ ) is

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Answer: (b)

y = C e^{x}

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

$\frac{d y}{y} = d x \Rightarrow \log y = x + c \Rightarrow y = e^{x + c} = C e^{x}$ ( $C = e^{c} > 0$ ).

Q6

·1 mark·Multiple choiceVariables separable

The general solution of $\frac{d y}{d x} = \frac{x}{y}$ is

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Answer: (c)

y^{2} - x^{2} = C

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

$y d y = x d x \Rightarrow \frac{y^{2}}{2} = \frac{x^{2}}{2} + c \Rightarrow y^{2} - x^{2} = C$ .

Q7

·1 mark·Multiple choiceLinear equations

An integrating factor of $\frac{d y}{d x} + \frac{2}{x} y = x$ ( $x > 0$ ) is

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Answer: (a)

x^{2}

Why: IF = e^{∫P dx} = e^{2 log x} = x².

$P = \frac{2}{x}$ , $\int P d x = 2 \log x = \log x^{2}$ , so IF $= e^{\log x^{2}} = x^{2}$ .

Q8

·1 mark·Multiple choiceLinear equations

An integrating factor of $\frac{d y}{d x} - y \tan x = \sin x$ for $0 < x < \frac{π}{2}$ is

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Answer: (b)

\cos x

Why: IF = e^{−∫tan x dx} = e^{log cos x} = cos x.

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

Q9

·1 mark·Multiple choiceHomogeneous equations

Which of the following differential equations is not homogeneous?

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Answer: (d)

\frac{d y}{d x} = \frac{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 $F (λ x, λ y) = F (x, y)$ . This holds for the first three (each is a function of $\frac{y}{x}$ ), but $\frac{λ^{2} x^{2} + λ y}{λ^{2} x^{2}} \neq \frac{x^{2} + y}{x^{2}}$ in general.

Q10

·1 mark·Multiple choiceGeneral and particular solutions

Which of the following is a solution of $\frac{d^{2} y}{d x^{2}} - 4 y = 0$ ?

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Answer: (a)

y = e^{2 x}

Why: For y = e^2x, y″ = 4e^2x = 4y.

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

Q11

·1 mark·Multiple choiceGeneral and particular solutions

The particular solution of $\frac{d y}{d x} = 3 x^{2}$ with $y (1) = 2$ is

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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 $\frac{d y}{d x} + y = 2$ with $y (0) = 0$ is

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Answer: (b)

y = 2 - 2 e^{- x}

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

IF $= e^{x}$ : $\frac{d}{d x} (y e^{x}) = 2 e^{x} \Rightarrow y e^{x} = 2 e^{x} + C \Rightarrow y = 2 + C e^{- x}$ . $y (0) = 0 \Rightarrow C = - 2$ : $y = 2 - 2 e^{- x}$ .

Q13

·1 mark·Multiple choiceVariables separable

The solution curve of $\frac{d y}{d x} = - \frac{y}{x}$ that passes through $(2, 3)$ also passes through

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Answer: (d)

(3, 2)

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

$\frac{d y}{y} = - \frac{d x}{x} \Rightarrow \log | y | = - \log | x | + c \Rightarrow x y = C$ . Through $(2, 3)$ : $x y = 6$ , which contains $(3, 2)$ .

Q14

·1 mark·Multiple choiceVariables separable

The general solution of $\frac{d y}{d x} = 1 + y^{2}$ is

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Answer: (a)

\tan^{- 1} y = x + C

Why: Separate: dy/(1 + y²) = dx.

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

Q15

·1 mark·Multiple choiceLinear equations

For $x > 0$ , the general solution of $x \frac{d y}{d x} + y = x^{2}$ is

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Answer: (c)

x y = \frac{x^{3}}{3} + C

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

$x \frac{d y}{d x} + y = \frac{d}{d x} (x y) = x^{2} \Rightarrow x y = \frac{x^{3}}{3} + C$ .

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. All Class 12 case study questions by chapter.

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: $\frac{d A}{d t} = - k A$ , where $t$ is in hours and $k > 0$ . 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]

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Answer:

A = 80 e^{- k t}

\frac{d A}{A} = - k d t \Rightarrow \log A = - k t + c

;

A (0) = 80

gives

A = 80 e^{- k t}

. A1

(ii) Find

k

. [1 mark]

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Answer:

k = \log 2

80 e^{- 2 k} = 20 \Rightarrow e^{- 2 k} = \frac{1}{4} \Rightarrow 2 k = \log 4 \Rightarrow k = \log 2

. A1

(iii) After how many hours will only

5

mg remain? [2 marks]

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Answer:

4

hours

80 e^{- t \log 2} = 80 \cdot 2^{- t} = 5

\Rightarrow 2^{- t} = \frac{1}{16} \Rightarrow t = 4

hours. A1

OR How much medicine remains after

3

hours? [2 marks]

Show answer

Answer:

10

A (3) = 80 e^{- 3 \log 2} = 80 \cdot 2^{- 3}

= 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 $\frac{d c}{d t} + \frac{c}{5} = 2$ , with $c = 0$ when $t = 0$ .

(i) Write the order and degree of the differential equation. [1 mark]

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Answer: Order

1

, degree

1

Highest derivative

\frac{d c}{d t}

: order

1

; it appears to power

1

: degree

1

. A1

(ii) Find the integrating factor. [1 mark]

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Answer:

e^{t / 5}

P = \frac{1}{5}

, IF

= e^{\int \frac{1}{5} d t} = e^{t / 5}

. A1

(iii) Solve the equation to find

c

in terms of

t

. [2 marks]

Show answer

Answer:

c = 10 - 10 e^{- t / 5}

\frac{d}{d t} (c e^{t / 5}) = 2 e^{t / 5} \Rightarrow c e^{t / 5} = 10 e^{t / 5} + K

M1;

c (0) = 0 \Rightarrow K = - 10

, so

c = 10 - 10 e^{- t / 5}

. A1

OR Using the solution

c = 10 - 10 e^{- 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 < 10

;

c \to 10

10 e^{- t / 5} > 0

for all

t

, so

c < 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 Pro adds all 50 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, the full common-mistakes library and AI marking of your written working.

Differential Equations: important questions What Pro 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 checked twice, the second time 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.