CBSE Sample Paper 2025-26 · Section A · 1 mark

Q10: Solution family of $x\,dx + y\,dy = 0$

Sample 2025-26Ch 9 · Differential Equations

The question

The solution of the differential equation $x\,dx + y\,dy = 0$ represents a family of

(A) straight lines

(B) parabolas

(D) Ellipses

Exam answer

\begin{aligned} &x\,dx + y\,dy = 0 \\ \Rightarrow\ &\int x\,dx = -\int y\,dy \end{aligned}

Why?

The variables are already separated: every $x$ is with $dx$ and every $y$ is with $dy$ ⁠. Move $y\,dy$ to the right and integrate both sides.

\begin{aligned} \Rightarrow\ &\frac{x^2}{2} = -\frac{y^2}{2} + k \\ \Rightarrow\ &x^2 + y^2 = 2k \end{aligned}

Why?

Standard integrals: $\int x\,dx = \frac{x^2}{2}$ and $\int y\,dy = \frac{y^2}{2}$ ⁠. One arbitrary constant $k$ ⁠, on one side, is enough.

Then multiply both sides by 2 and move $-y^2$ to the left.

This is a family of circles with centre at the origin ( $k$ an arbitrary constant).

Why?

A circle with centre $(0, 0)$ and radius $r$ is

x^2 + y^2 = r^2

Here $r^2 = 2k$ ⁠, with $k > 0$ ⁠. Straight lines have no squared terms, a parabola squares only one of $x$ and $y$ ⁠, and an ellipse that is not a circle has unequal coefficients of $x^2$ and $y^2$ ⁠.

Answer: (C) Circles

Where this comes from

Chapter 9, Differential Equations: Variable separable differential equations.

Question type: Variables separable

How to spot it: click to see

Type 3. Variables separable: $y$ on one side, $x$ on the other

Spot it: $\frac{dy}{dx}$ equals a part in $x$ times a part in $y$ ⁠, like $(1 + x^2)(1 + y^2)$ ⁠.

Here: $x\,dx + y\,dy = 0$ is already separated: $\int x\,dx = -\int y\,dy$ gives $x^2 + y^2 = 2k$ ⁠, circles about the origin.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Differential Equations in past papers: question 17 of 31 · see them all

Q10: Solution family of x dx+y dy=0x\,dx + y\,dy = 0xdx+ydy=0

Q10: Solution family of $x\,dx + y\,dy = 0$