Differential Equations: CBSE Class 12 Previous Year Questions

37 different questions from Differential Equations (NCERT Chapter 9) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions18 questions 3 marks questions17 questions 5 marks questions2 questions

Most asked Differential Equations questions

Q111 markMCQDifferential Equations CBSE 2026 · 65/1/1

The integrating factor of differential equation $R \frac{d x}{d y} + P x = Q$ where P, Q, R are functions of y is

(A) $e^{\int \frac{P}{Q} d y}$
(B) $e^{\int P d y}$
(C) $e^{\int \frac{P}{R} d y}$
(D) $e^{\int \frac{P}{R} d x}$

Show answer & solution

Answer: (C)

e^{\int \frac{P}{R} d y}

Divide by R: $\frac{d x}{d y} + \frac{P}{R} x = \frac{Q}{R}$ , a linear equation in $x$ .
Integrating factor $= e^{\int \frac{P}{R} d y}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q121 markMCQDifferential Equations CBSE 2026 · 65/2/1

The integrating factor of the differential equation $2 x \frac{d y}{d x} - y = 3$ is

(A) $x$
(B) $\frac{1}{x}$
(C) $e^{x}$
(D) $e^{- x}$

Show answer & solution

Answer: (B)

\frac{1}{x}

Standard form: $\frac{d y}{d x} - \frac{1}{2 x} y = \frac{3}{2 x}$ , so $P = - \frac{1}{2 x}$ .
I.F. $= e^{\int - \frac{1}{2 x} d x} = e^{- \frac{1}{2} l o g x} = \frac{1}{x}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q131 markMCQDifferential Equations CBSE 2026 · 65/3/1

$\frac{d y}{d x} = F (x, y)$ will be a homogeneous differential equation for which of the following functions ?
(i) $F (x, y) = 3 x + 2 y$
(ii) $F (x, y) = sin \frac{y}{x} + lo g y - lo g x$
(iii) $F (x, y) = e^{y / x} + 1$
(iv) $F (x, y) = x^{2} + y^{2} - y$

(A)(i) and (ii)
(B)(i), (ii) and (iii)
(C)(ii), (iii) and (iv)
(D)(ii) and (iii)

Show answer & solution

Answer: (D) (ii) and (iii)

$\frac{d y}{d x} = F (x, y)$ is homogeneous when F is homogeneous of degree zero, i.e. $F (λ x, λ y) = F (x, y)$ .
(i) $3 x + 2 y$ has degree 1: not homogeneous of degree zero.
(ii) $sin \frac{y}{x} + lo g \frac{y}{x}$ : degree zero.
(iii) $e^{y / x} + 1$ : degree zero.
(iv) $x^{2} + y^{2} - y$ has degree 1.
So (ii) and (iii).

Also asked in: 2026 65/3/2, 2026 65/3/3

Q191 markAssertion–ReasonDifferential Equations CBSE 2026 · 65/3/1

Assertion (A) : One of the particular solutions of the differential equation $\frac{d y}{d x} = e^{x + y}$ can be $e^{x} + e^{- y} = - 2$ .
Reason (R) : $e^{x} + e^{- y} = C$ is the general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ .

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (D) Assertion (A) is false, but Reason (R) is true.

$e^{- y} d y = e^{x} d x$ gives $- e^{- y} = e^{x} + k$ , i.e. $e^{x} + e^{- y} = C$ . So R is true.
$e^{x} + e^{- y} > 0$ for all real x, y, so it can never equal $- 2$ . A is false.

Also asked in: 2026 65/3/2, 2026 65/3/3

Q131 markMCQDifferential Equations CBSE 2026 · 65/4/1

Which of the following is not a Linear Differential Equation ?

(A) $(1 + x^{2}) d y + 2 x y d x = cot x d x$
(B) $y + \frac{d}{d x} (x y) = x (sin x + lo g x)$
(C) $x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0$
(D) $y d x - (x + 3 y^{2}) d y = 0$

Show answer & solution

Answer: (C)

x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0

(A): $\frac{d y}{d x} + \frac{2 x}{1 + x ^{2}} y = \frac{c o t x}{1 + x ^{2}}$ , linear in y.
(B): $y + x \frac{d y}{d x} + y = x (sin x + lo g x)$ , linear in y.
(D): $\frac{d x}{d y} - \frac{x}{y} = 3 y$ , linear in x.
(C): $\frac{d y}{d x} = \frac{x ( 1 + y ^{2} )}{y ( 1 + x ^{2} )}$ contains $y^{2}$ , and $\frac{d x}{d y}$ contains $x^{2}$ , so it is not linear.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q141 markMCQDifferential Equations CBSE 2026 · 65/5/1

The general solution for the differential equation $\frac{d y}{d x} = e^{3 x - y}$ is :

(A) $3 e^{y} = e^{3 x} + C$
(B) $lo g (3 x - y) = C$
(C) $e^{3 x - y} = C$
(D) $- e^{y} + 3 e^{3 x} = C$

Show answer & solution

Answer: (A)

3 e^{y} = e^{3 x} + C

$\frac{d y}{d x} = e^{3 x} e^{- y} \Rightarrow e^{y} d y = e^{3 x} d x$ .
Integrating: $e^{y} = \frac{e ^{3 x}}{3} + C_{1}$ , i.e. $3 e^{y} = e^{3 x} + C$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

Q293 marksShort AnswerDifferential Equations CBSE 2026 · 65/2/1

Solve the following differential equation :
$x \frac{d y}{d x} = y - x sin^{2} (\frac{y}{x})$ , given that $y (1) = \frac{π}{6}$

Show answer & solution

Answer:

cot (\frac{y}{x}) = lo g ∣ x ∣ + 3

$\frac{d y}{d x} = \frac{y}{x} - sin^{2} \frac{y}{x}$ (homogeneous). Put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$v + x \frac{d v}{d x} = v - sin^{2} v \Rightarrow \frac{d v}{s i n ^{2} v} = - \frac{d x}{x}$ .
Integrating: $- cot v = - lo g ∣ x ∣ + C_{1}$ , i.e. $cot \frac{y}{x} = lo g ∣ x ∣ + C$ .
$y (1) = \frac{π}{6}$ : $cot \frac{π}{6} = 3 = 0 + C \Rightarrow C = 3$ .
Solution: $cot (\frac{y}{x}) = lo g ∣ x ∣ + 3$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q29 (OR) (OR)3 marksShort AnswerDifferential Equations CBSE 2026 · 65/2/1

Find the general solution of the differential equation : $y lo g y \frac{d x}{d y} + x = \frac{2}{y}$ .

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Answer:

x lo g y = - \frac{2}{y} + C

$\frac{d x}{d y} + \frac{1}{y l o g y} x = \frac{2}{y ^{2} l o g y}$ , linear in x.
I.F. $= e^{\int \frac{d y}{y l o g y}} = e^{l o g (l o g y)} = lo g y$ .
$x lo g y = \int \frac{2}{y ^{2} l o g y} lo g y d y = \int \frac{2}{y ^{2}} d y = - \frac{2}{y} + C$ .
General solution: $x lo g y = - \frac{2}{y} + C$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q345 marksLong AnswerDifferential Equations CBSE 2026 · 65/4/1

Solve the differential equation $y e^{y} d x = (y^{3} + 2 x e^{y}) d y$ , when $y (0) = 1$ .

Show answer & solution

Answer:

x = y^{2} (e^{- 1} - e^{- y})

, i.e.

x = \frac{y ^{2}}{e} - y^{2} e^{- y}

Write as $\frac{d x}{d y} = \frac{y ^{3} + 2 x e ^{y}}{y e ^{y}}$ , i.e. $\frac{d x}{d y} - \frac{2}{y} x = y^{2} e^{- y}$ , linear in x.
I.F. $= e^{\int - \frac{2}{y} d y} = e^{- 2 l o g y} = \frac{1}{y ^{2}}$ .
$\frac{x}{y ^{2}} = \int y^{2} e^{- y} \cdot \frac{1}{y ^{2}} d y = \int e^{- y} d y = - e^{- y} + C$ .
$y (0) = 1$ : $x = 0$ , $y = 1$ gives $0 = - e^{- 1} + C$ , so $C = \frac{1}{e}$ .
Solution: $x = y^{2} (\frac{1}{e} - e^{- y})$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q34 (OR) (OR)5 marksLong AnswerDifferential Equations CBSE 2026 · 65/4/1

Find the general solution of the differential equation
$(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ .

Show answer & solution

Answer:

x^{2} - y^{2} = C (x^{2} + y^{2})^{2}

$\frac{d y}{d x} = \frac{x ^{3} - 3 x y ^{2}}{y ^{3} - 3 x ^{2} y}$ is homogeneous. Put $y = v x$ : $v + x \frac{d v}{d x} = \frac{1 - 3 v ^{2}}{v ^{3} - 3 v}$ .
$x \frac{d v}{d x} = \frac{1 - 3 v ^{2} - v ^{4} + 3 v ^{2}}{v ^{3} - 3 v} = \frac{1 - v ^{4}}{v ^{3} - 3 v}$ , so $\int \frac{v ^{3} - 3 v}{1 - v ^{4}} d v = \int \frac{d x}{x}$ .
Put $v^{2} = t$ : $\frac{1}{2} \int \frac{t - 3}{( 1 - t ) ( 1 + t )} d t = \frac{1}{2} \int (\frac{- 1}{1 - t} - \frac{2}{1 + t}) d t = \frac{1}{2} lo g ∣1 - t ∣ - lo g (1 + t)$ .
So $\frac{1}{2} lo g ∣1 - v^{2} ∣ - lo g (1 + v^{2}) = lo g ∣ x ∣ + c$ , i.e. $\frac{∣1 - v ^{2} ∣}{( 1 + v ^{2} ) ^{2} x ^{2}} =$ constant.
With $v = \frac{y}{x}$ : $\frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = C$ , i.e. $x^{2} - y^{2} = C (x^{2} + y^{2})^{2}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

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