MathsCBSE Class 12 · Board exam 2023, set 65/5/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2023: question 38

Set in the CBSE Class 12 board exam of 2023, paper 65/5/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedDifferential Equations

Case Study-III

An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form $\frac{d y}{d x} = F (x, y)$ is said to be homogeneous if $F (x, y)$ is a homogeneous function of degree zero, whereas a function $F (x, y)$ is a homogenous function of degree n if $F (λ x, λ y) = λ^{n} F (x, y)$ . To solve a homogeneous differential equation of the type $\frac{d y}{d x} = F (x, y) = g (\frac{y}{x})$ , we make the substitution $y = v x$ and then separate the variables.

Based on the above, answer the following questions :

(I)
Show that $(x^{2} - y^{2}) d x + 2 x y d y = 0$ is a differential equation of the type $\frac{d y}{d x} = g (\frac{y}{x})$ .
2 marks
(II)
Solve the above equation to find its general solution.
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 65/5/1.

Hints

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

Hint 1

Make dy/dx the subject, then divide the top and bottom by x².

Hint 2

Substitute y = vx and dy/dx = v + x dv/dx, then simplify.

Hint 3

After separating, notice that 2v/(1+v²) is the derivative of 1+v² divided by 1+v², so it integrates to a logarithm.

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.

(I) From $(x^{2} - y^{2}) d x + 2 x y d y = 0$ :
$\frac{d y}{d x} = - \frac{x ^{2} - y ^{2}}{2 x y} = \frac{y ^{2} - x ^{2}}{2 x y}$
1 mark
Divide the numerator and denominator by $x^{2}$ :
$\frac{d y}{d x} = \frac{( \frac{y}{x} ) ^{2} - 1}{2 ( \frac{y}{x} )} = g (\frac{y}{x})$
So the equation is of the type $\frac{d y}{d x} = g (\frac{y}{x})$ . It is homogeneous, with $F (λ x, λ y) = λ^{0} F (x, y)$ .
1 mark
(II) Put $y = v x$ , so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Then
$v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} \Rightarrow x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v = \frac{- 1 - v ^{2}}{2 v}$
½ mark
Separate the variables:
$\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{d x}{x}$
1 mark
Integrate:
$lo g ∣1 + v^{2} ∣ + lo g ∣ x ∣ = lo g C \Rightarrow x (1 + \frac{y ^{2}}{x ^{2}}) = C$
Hence the general solution is $x^{2} + y^{2} = C x$ .
½ mark

Answer: (I) dy/dx = ((y/x)² − 1)/(2(y/x)) = g(y/x); (II) x² + y² = Cx

Where the marks are

(I) Writes dy/dx = (y² − x²)/(2xy) (1 mark)
(I) Expresses it as a function of y/x (1 mark)
(II) Substitutes y = vx, dy/dx = v + x dv/dx (½ mark)
(II) Separates variables: 2v/(1+v²) dv = −dx/x (1 mark)
(II) Integrates to get x² + y² = Cx (½ mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Differential Equations, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 65/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2023: question 38

Hints

Step-by-step solution

Revise the chapter

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