CBSE Maths › Class 12 PYQs › Differential Equations

Differential Equations: 2 marks Questions (CBSE Class 12)

9 different 2 marks questions on Differential Equations from CBSE Class 12 Maths board exams 2022–2026, newest first.

1 mark (61)2 marks (9)3 marks (50)4 marks (8)5 marks (7)

Q12 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/1/1

Find the sum of the order and the degree of the differential equation :
$(x + \frac{d y}{d x})^{2} = (\frac{d y}{d x})^{2} + 1$

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Answer: 2

Expand the left side: $x^{2} + 2 x \frac{d y}{d x} + (\frac{d y}{d x})^{2} = (\frac{d y}{d x})^{2} + 1$ .
Cancel $(\frac{d y}{d x})^{2}$ : $x^{2} + 2 x \frac{d y}{d x} = 1$ .
Highest derivative is $\frac{d y}{d x}$ , so order $= 1$ ; it appears to power 1, so degree $= 1$ .
Sum $= 1 + 1 = 2$ .

Also asked in: 2022 65/1/3

Q52 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/1/2

Solve the differential equation :
$lo g (\frac{d y}{d x}) = x - y$ .

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Answer:

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

$\frac{d y}{d x} = e^{x - y} = \frac{e ^{x}}{e ^{y}}$ .
Separate: $e^{y} d y = e^{x} d x$ .
Integrate: $e^{y} = e^{x} + C$ .

Q12 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/2/1

Find the product of the order and the degree of the differential equation $[\frac{d}{d x} (x y^{2})] \cdot \frac{d y}{d x} + y = 0$ .

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Answer: 2

$\frac{d}{d x} (x y^{2}) = y^{2} + 2 x y \frac{d y}{d x}$ .
So the equation is $(y^{2} + 2 x y \frac{d y}{d x}) \frac{d y}{d x} + y = 0$ , i.e. $2 x y (\frac{d y}{d x})^{2} + y^{2} \frac{d y}{d x} + y = 0$ .
Highest derivative is $\frac{d y}{d x}$ , so order $= 1$ ; its highest power is 2, so degree $= 2$ .
Product $= 1 \times 2 = 2$ .

Also asked in: 2022 65/2/3

Q62 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/2/2

Find the particular solution of the differential equation $\frac{d y}{d x} = 2 y^{2}$ , given y = 1 when x = 1.

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Answer:

y = \frac{1}{3 - 2 x}

Separate variables: $\frac{d y}{y ^{2}} = 2 d x$ .
Integrate: $- \frac{1}{y} = 2 x + C$ .
$y = 1$ at $x = 1$ : $- 1 = 2 + C$ , so $C = - 3$ .
$- \frac{1}{y} = 2 x - 3$ , i.e. $y = \frac{1}{3 - 2 x}$ .

Q22 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/3/1

Find the general solution of the differential equation : $e^{d y / d x} = x^{2}$ .

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Answer:

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

Taking log: $\frac{d y}{d x} = lo g x^{2} = 2 lo g x$ (for $x > 0$ )
$y = 2 \int lo g x d x = 2 (x lo g x - x) + C$
So $y = 2 x lo g x - 2 x + C$

Q22 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/3/2

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

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Answer:

\frac{e ^{a x}}{a} + \frac{e ^{- b y}}{b} = C

$\frac{d y}{d x} = e^{a x + b y} = e^{a x} e^{b y}$
Separate: $e^{- b y} d y = e^{a x} d x$
Integrate: $- \frac{e ^{- b y}}{b} = \frac{e ^{a x}}{a} + C_{1}$
So $\frac{e ^{a x}}{a} + \frac{e ^{- b y}}{b} = C$ ( $a, b$
eq 0 $)$

Q32 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/3/3

Find the general solution of the differential equation : $\frac{d y}{d x} = \frac{3 e ^{2 x} + 3 e ^{4 x}}{e ^{x} + e ^{- x}}$

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Answer:

y = e^{3 x} + C

$3 e^{2 x} + 3 e^{4 x} = 3 e^{3 x} (e^{- x} + e^{x})$
So $\frac{d y}{d x} = 3 e^{3 x}$
$y = \int 3 e^{3 x} d x = e^{3 x} + C$

Q52 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/4/1

Find the general solution of the differential equation
$sec^{2} x \cdot tan y d x + sec^{2} y \cdot tan x d y = 0$ .

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Answer:

tan x tan y = C

Divide by $tan x tan y$ : $\frac{s e c ^{2} x}{t a n x} d x + \frac{s e c ^{2} y}{t a n y} d y = 0$ .
Integrate: $lo g ∣ tan x ∣ + lo g ∣ tan y ∣ = lo g ∣ C ∣$ .
So $tan x tan y = C$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q22 marksVery Short AnswerDifferential Equations CBSE 2022 · 65/5/1

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

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Answer:

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

$\frac{d y}{d x} = e^{- y} (e^{x} + x^{2})$ , so $e^{y} d y = (e^{x} + x^{2}) d x$ .
Integrating both sides: $e^{y} = e^{x} + \frac{x ^{3}}{3} + C$ .

Also asked in: 2022 65/5/2, 2022 65/5/3

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