MathsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2025-26: question 34(B)

From the official CBSE sample question paper for 2025-26, section D, question 34(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(B) · 5 marks · Long answerDifferential Equations

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

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Look at $e^{x / y}$ : the equation is easier to handle as $\frac{d x}{d y}$ rather than $\frac{d y}{d x}$ .

Hint 2

Since the equation is homogeneous in $x / y$ , substitute $x = v y$ .

Hint 3

After simplifying, the $v$ terms separate as $e^{v} d v$ ; then use $x = 0$ , $y = 1$ to find the constant.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Rearrange the equation:
$\frac{d x}{d y} = \frac{2 x e ^{x / y} - y}{2 y e ^{x / y}} = \frac{2 \frac{x}{y} e ^{x / y} - 1}{2 e ^{x / y}} .$
The right side is a function of $\frac{x}{y}$ only, so the equation is homogeneous.
1 mark
Put $x = v y$ , so that $\frac{d x}{d y} = v + y \frac{d v}{d y}$ .
Then $v + y \frac{d v}{d y} = \frac{2 v e ^{v} - 1}{2 e ^{v}}$ .
1 mark
Simplify:
$y \frac{d v}{d y} = \frac{2 v e ^{v} - 1 - 2 v e ^{v}}{2 e ^{v}} = \frac{- 1}{2 e ^{v}}$ .
Separate the variables:
$2 e^{v} d v = - \frac{d y}{y}$ .
1 mark
Integrate both sides: $2 e^{v} = - lo g ∣ y ∣ + C$ .
Substitute back $v = \frac{x}{y}$ :
$2 e^{x / y} + lo g ∣ y ∣ = C$ .
1 mark
Use $y (0) = 1$ : $2 e^{0} + lo g 1 = C \Rightarrow C = 2$ .
The particular solution is
$2 e^{x / y} + lo g ∣ y ∣ = 2.$
1 mark

Answer: $2 e^{x / y} + lo g ∣ y ∣ = 2$

Where the marks are

Writes $\frac{d x}{d y}$ as a function of $x / y$ and recognises the equation as homogeneous (1 mark)
Substitutes $x = v y$ with $\frac{d x}{d y} = v + y \frac{d v}{d y}$ (1 mark)
Simplifies to the separable form $2 e^{v} d v = - \frac{d y}{y}$ (1 mark)
Integrates to get $2 e^{x / y} + lo g ∣ y ∣ = C$ (1 mark)
Uses $y (0) = 1$ to get $C = 2$ and states the particular solution (1 mark)

Revise the chapter

Chapter: Differential Equations (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Differential Equations, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Mathematics sample paper 2025-26: question 34(B)

Hints

Step-by-step solution

Revise the chapter

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