CBSE Maths › Class 12 PYQs › Differential Equations

Differential Equations: 3 marks Questions (CBSE Class 12)

17 different 3 marks questions on Differential Equations from CBSE Class 12 Maths board exams 2026, newest first.

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

Find the general solution of the following differential equation :
$x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2}$

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Answer:

tan^{- 1} \frac{y}{x} = lo g ∣ x ∣ + C

$\frac{d y}{d x} = 1 + \frac{y}{x} + \frac{y ^{2}}{x ^{2}}$ , a homogeneous equation.
Put $y = v x$ : $v + x \frac{d v}{d x} = 1 + v + v^{2} \Rightarrow x \frac{d v}{d x} = 1 + v^{2}$ .
$\int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x} \Rightarrow tan^{- 1} v = lo g ∣ x ∣ + C$ .
General solution: $tan^{- 1} \frac{y}{x} = lo g ∣ x ∣ + C$ .

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

Find the particular solution of the differential equation
$x y \frac{d y}{d x} = (x + 2) (y + 2)$ , given that $y (1) = - 1$ .

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Answer:

y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ - 2

Separate: $\frac{y}{y + 2} d y = \frac{x + 2}{x} d x$ , i.e. $(1 - \frac{2}{y + 2}) d y = (1 + \frac{2}{x}) d x$ .
Integrate: $y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ + C$ .
At $x = 1, y = - 1$ : $- 1 - 2 lo g 1 = 1 + 0 + C \Rightarrow C = - 2$ .
Particular solution: $y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ - 2$ .

Q273 marksShort AnswerDifferential Equations CBSE 2026 · 65/1/2

Find the general solution of the differential equation :
$y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$

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Answer:

tan^{- 1} \frac{x}{y} + lo g ∣ y ∣ = C

$\frac{d x}{d y} = - \frac{x ^{2} - x y + y ^{2}}{y ^{2}}$ , homogeneous in $x$ and $y$ .
Put $x = v y$ : $v + y \frac{d v}{d y} = - (v^{2} - v + 1) \Rightarrow y \frac{d v}{d y} = - (1 + v^{2})$ .
$\int \frac{d v}{1 + v ^{2}} = - \int \frac{d y}{y} \Rightarrow tan^{- 1} v = - lo g ∣ y ∣ + C$ .
General solution: $tan^{- 1} \frac{x}{y} + lo g ∣ y ∣ = C$ .

Q27 (OR) (OR)3 marksShort AnswerDifferential Equations CBSE 2026 · 65/1/2

Find the particular solution of the differential equation $\frac{d y}{d x} = y tan x$ , given that y = 2 if $x = 0$ .

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Answer:

y = 2 sec x

Separate: $\frac{d y}{y} = tan x d x$ .
Integrate: $lo g ∣ y ∣ = lo g ∣ sec x ∣ + lo g ∣ C ∣ \Rightarrow y = C sec x$ .
$y = 2$ at $x = 0$ : $2 = C sec 0 = C$ .
Particular solution: $y = 2 sec x$ .

Q303 marksShort AnswerDifferential Equations CBSE 2026 · 65/1/3

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

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Answer:

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

$\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ , a homogeneous equation.
Put $y = v x$ : $v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} \Rightarrow x \frac{d v}{d x} = - \frac{1 + v ^{2}}{2 v}$ .
$\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{d x}{x} \Rightarrow lo g (1 + v^{2}) = - lo g ∣ x ∣ + lo g ∣ C ∣$ .
So $x (1 + v^{2}) = C$ , i.e. $x (1 + \frac{y ^{2}}{x ^{2}}) = C$ .
General solution: $x^{2} + y^{2} = C x$ .

Q30 (OR) (OR)3 marksShort AnswerDifferential Equations CBSE 2026 · 65/1/3

Find the particular solution of the differential equation
$(1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0$ , given that y(1) = 0.

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Answer:

tan^{- 1} y + tan^{- 1} (e^{x}) = tan^{- 1} e

Separate: $\frac{d y}{1 + y ^{2}} = - \frac{e ^{x}}{1 + e ^{2 x}} d x$ .
Put $t = e^{x}$ on the right: $\int \frac{e ^{x} d x}{1 + e ^{2 x}} = \int \frac{d t}{1 + t ^{2}} = tan^{- 1} (e^{x})$ .
So $tan^{- 1} y = - tan^{- 1} (e^{x}) + C$ .
$y (1) = 0$ : $0 = - tan^{- 1} e + C \Rightarrow C = tan^{- 1} e$ .
Particular solution: $tan^{- 1} y + tan^{- 1} (e^{x}) = tan^{- 1} e$ .

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 $y (1) = \frac{π}{6}$

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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

Q293 marksShort AnswerDifferential Equations CBSE 2026 · 65/3/1

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

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Answer:

y = \frac{- 2 x}{l o g ∣ x ∣ + C}

(equivalently

2 x + y lo g ∣ x ∣ + C y = 0

)

$\frac{d y}{d x} = \frac{y ^{2} + 2 x y}{2 x ^{2}}$ is homogeneous. Put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$v + x \frac{d v}{d x} = \frac{v ^{2}}{2} + v$ , so $x \frac{d v}{d x} = \frac{v ^{2}}{2}$ .
$\frac{2 d v}{v ^{2}} = \frac{d x}{x}$ , giving $- \frac{2}{v} = lo g ∣ x ∣ + C$ .
With $v = \frac{y}{x}$ : $- \frac{2 x}{y} = lo g ∣ x ∣ + C$ , i.e. $y = \frac{- 2 x}{l o g ∣ x ∣ + C}$ .

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

Find a particular solution of the differential equation $(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1$ , given that $y = 0$ when $x = 0$ .

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Answer:

y = lo g \frac{2 x + 1}{x + 1}

(i.e.

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

)

$2 e^{- y} - 1 = \frac{2 - e ^{y}}{e ^{y}}$ , so $\frac{e ^{y} d y}{2 - e ^{y}} = \frac{d x}{x + 1}$ .
Integrating: $- lo g ∣2 - e^{y} ∣ = lo g ∣ x + 1∣ + C$ .
At $x = 0$ , $y = 0$ : $- lo g 1 = lo g 1 + C$ , so $C = 0$ .
Hence $(2 - e^{y}) (x + 1) = 1$ , i.e. $e^{y} = 2 - \frac{1}{x + 1} = \frac{2 x + 1}{x + 1}$ .
$y = lo g \frac{2 x + 1}{x + 1}$ .

Q273 marksShort AnswerDifferential Equations CBSE 2026 · 65/3/2

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

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Answer:

x^{2} = 2 y^{2} (lo g ∣ y ∣ + C)

, i.e.

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

Write $\frac{d x}{d y} = \frac{x ^{2} + y ^{2}}{x y} = \frac{x}{y} + \frac{y}{x}$ , homogeneous in x and y.
Put $x = v y$ , $\frac{d x}{d y} = v + y \frac{d v}{d y}$ : $v + y \frac{d v}{d y} = v + \frac{1}{v}$ .
$v d v = \frac{d y}{y}$ , so $\frac{v ^{2}}{2} = lo g ∣ y ∣ + C$ .
With $v = \frac{x}{y}$ : $\frac{x ^{2}}{2 y ^{2}} = lo g ∣ y ∣ + C$ , i.e. $x^{2} = 2 y^{2} (lo g ∣ y ∣ + C)$ .

Q27 (OR) (OR)3 marksShort AnswerDifferential Equations CBSE 2026 · 65/3/2

Find the particular solution of the differential equation $\frac{d y}{d x} - 3 y cot x = sin 2 x$ , given that $y = 2$ when $x = \frac{π}{2}$ .

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Answer:

y = 4 sin^{3} x - 2 sin^{2} x

Linear with $P = - 3 cot x$ , $Q = sin 2 x$ .
I.F. $= e^{- 3 \int c o t x d x} = e^{- 3 l o g ∣ s i n x ∣} = \frac{1}{s i n ^{3} x}$ .
$\frac{y}{s i n ^{3} x} = \int \frac{2 s i n x c o s x}{s i n ^{3} x} d x = \int \frac{2 c o s x}{s i n ^{2} x} d x = - \frac{2}{s i n x} + C$ .
So $y = - 2 sin^{2} x + C sin^{3} x$ .
At $x = \frac{π}{2}$ , $y = 2$ : $2 = - 2 + C$ , so $C = 4$ .
$y = 4 sin^{3} x - 2 sin^{2} x$ .

Q303 marksShort AnswerDifferential Equations CBSE 2026 · 65/3/3

Find the general solution of the differential equation $(x^{2} - y^{2}) d x + 2 x y d y = 0$ .

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Answer:

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

$\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ , homogeneous. Put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v}$ , so $x \frac{d v}{d x} = - \frac{1 + v ^{2}}{2 v}$ .
$\frac{2 v d v}{1 + v ^{2}} = - \frac{d x}{x}$ , giving $lo g (1 + v^{2}) = - lo g ∣ x ∣ + lo g ∣ C ∣$ .
$x (1 + v^{2}) = C$ ; with $v = \frac{y}{x}$ : $x^{2} + y^{2} = C x$ .

Q30 (OR) (OR)3 marksShort AnswerDifferential Equations CBSE 2026 · 65/3/3

Solve the differential equation $sin x cos y d x + cos x sin y d y = 0$ , given that $y = \frac{π}{4}$ when $x = 0$ .

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Answer:

cos x cos y = \frac{1}{2}

(i.e.

2 cos x cos y = 1

)

Divide by $cos x cos y$ : $tan x d x + tan y d y = 0$ .
Integrating: $- lo g ∣ cos x ∣ - lo g ∣ cos y ∣ = - lo g ∣ C ∣$ , i.e. $cos x cos y = C$ .
At $x = 0$ , $y = \frac{π}{4}$ : $C = 1 \cdot \frac{1}{2} = \frac{1}{2}$ .
Particular solution: $cos x cos y = \frac{1}{2}$ .

Q283 marksShort AnswerDifferential Equations CBSE 2026 · 65/5/1

Solve the differential equation $(x + 2 y^{3}) d y = y d x$ .

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Answer:

x = y^{3} + C y

Write as $\frac{d x}{d y} = \frac{x + 2 y ^{3}}{y}$ , i.e. $\frac{d x}{d y} - \frac{1}{y} x = 2 y^{2}$ (linear in x).
I.F. $= e^{- \int \frac{1}{y} d y} = \frac{1}{y}$ .
$\frac{x}{y} = \int 2 y^{2} \cdot \frac{1}{y} d y = y^{2} + C$ .
So $x = y^{3} + C y$ .

Q283 marksShort AnswerDifferential Equations CBSE 2026 · 65/5/2

Solve the differential equation $y d x + (x - y^{3}) d y = 0$ .

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Answer:

x y = \frac{y ^{4}}{4} + C

Write as $\frac{d x}{d y} = \frac{y ^{3} - x}{y}$ , i.e. $\frac{d x}{d y} + \frac{1}{y} x = y^{2}$ (linear in x).
I.F. $= e^{\int \frac{1}{y} d y} = y$ .
$x y = \int y^{2} \cdot y d y = \frac{y ^{4}}{4} + C$ .
So the solution is $x y = \frac{y ^{4}}{4} + C$ .

Q283 marksShort AnswerDifferential Equations CBSE 2026 · 65/5/3

Solve the differential equation $(x - sin y) d y + tan y d x = 0$ .

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Answer:

x sin y = \frac{s i n ^{2} y}{2} + C

Write as $\frac{d x}{d y} = \frac{s i n y - x}{t a n y}$ , i.e. $\frac{d x}{d y} + x cot y = cos y$ (linear in x).
I.F. $= e^{\int c o t y d y} = sin y$ .
$x sin y = \int sin y cos y d y = \frac{s i n ^{2} y}{2} + C$ .
So the solution is $x sin y = \frac{s i n ^{2} y}{2} + C$ .

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