Question 51 markMCQSection A

General solution of differential equation ylog⁡y dx−x dy=0y \log y \, dx - x \, dy = 0 is :

Calculus (Differential Equations)
Differential Equations — variable separable

Options

(A)y=log⁡∣cx∣y = \log |cx|
(B)y=e∣cx∣y = e^{|cx|}
(C)y=ec+xy = e^{c + x}
(D)log⁡y=∣c+x∣\log y = |c + x|

Official answer

The correct option is B) y=e∣cx∣y = e^{|cx|} (i.e. y=ecxy = e^{cx}).


Working:


  • ylog⁡y dx−x dy=0⇒ylog⁡y dx=x dyy \log y \, dx - x \, dy = 0 \Rightarrow y \log y \, dx = x \, dy.
  • Separate variables: dxx=dyylog⁡y\frac{dx}{x} = \frac{dy}{y \log y}.
  • Integrate: ∫dxx=∫dyylog⁡y\int \frac{dx}{x} = \int \frac{dy}{y \log y}. Put u=log⁡yu = \log y, du=dyydu = \frac{dy}{y}, so the RHS = ∫duu\int \frac{du}{u}.
  • log⁡∣x∣=log⁡∣log⁡y∣\log|x| = \log|\log y| + constant ⇒ log⁡y=c⋅x\log y = c \cdot x.
  • Therefore y=ecxy = e^{cx}.
Answer keywords:variable separabledifferential equationy log ysubstitution u = log ylog y = cxy = e^{cx}general solution

Marking scheme

  1. 1.1 mark: correct separation, integration to log⁡y=cx\log y = cx, and answer y=ecxy = e^{cx} (option B).

Hint

Separate to dxx=dyylog⁡y\frac{dx}{x} = \frac{dy}{y \log y}; substitute u=log⁡yu = \log y so the right side is ∫duu\int \frac{du}{u}.

Quick oral answer

Separating gives dxx=dyylog⁡y\frac{dx}{x} = \frac{dy}{y \log y}; with u=log⁡yu = \log y this integrates to log⁡y=cx\log y = cx, so y=ecxy = e^{cx}.

Common mistakes

  1. 1.Taking the arbitrary constant additively (log⁡y=c+x\log y = c + x) instead of multiplicatively (log⁡y=cx\log y = cx).
  2. 2.Forgetting the substitution u=log⁡yu = \log y and mis-integrating ∫dyylog⁡y\int \frac{dy}{y \log y}.
  3. 3.Dropping the exponential step, leaving the answer as log⁡y\log y instead of y.

Analysis and explanation

This is a classic variable-separable differential equation with a substitution on the y-side.


Concept:


  • Separate x-terms and y-terms, then integrate each side.
  • The substitution u=log⁡yu = \log y turns ∫dyylog⁡y\int \frac{dy}{y \log y} into the standard ∫duu=log⁡∣u∣\int \frac{du}{u} = \log|u|.

Why B is correct:


  • Integration gives log⁡∣x∣=log⁡∣log⁡y∣+c1\log|x| = \log|\log y| + c_1, i.e. log⁡y=cx\log y = cx, hence y=ecxy = e^{cx}, matching option B.

Why the distractors are wrong:


  • A (y=log⁡∣cx∣y = \log|cx|): inverts the exponential relation incorrectly.
  • C (y=ec+xy = e^{c+x}): comes from treating the constant additively (log⁡y=c+x\log y = c + x) instead of multiplicatively.
  • D (log⁡y=∣c+x∣\log y = |c + x|): again uses an additive constant, which does not follow from log⁡∣x∣=log⁡∣log⁡y∣+c\log|x| = \log|\log y| + c.

Interesting facts

The integral ∫dxxlog⁡x=log⁡∣log⁡x∣\int \frac{dx}{x \log x} = \log|\log x| is a textbook example of a nested-log substitution.

Equations of the form dydx=y⋅f(x)\frac{dy}{dx} = y \cdot f(x) always yield exponential-type solutions, y=e∫f dxy = e^{\int f\,dx}.

Question details

Q5 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Calculus (Differential Equations)
Topic
Differential Equations — variable separable
Bloom's level
apply
Syllabus topic
Unit 3: Calculus — Differential Equations: variable separable method
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the constant multiplicative (cxcx) and not additive (c+xc + x)?

Integration gives log⁡∣x∣=log⁡∣log⁡y∣+c1\log|x| = \log|\log y| + c_1. Exponentiating both sides moves the additive constant into a multiplicative factor, so log⁡y=cx\log y = cx, not c+xc + x.

What substitution simplifies ∫dyylog⁡y\int \frac{dy}{y \log y}?

Let u=log⁡yu = \log y, then du=dyydu = \frac{dy}{y} and the integral becomes ∫duu=log⁡∣u∣=log⁡∣log⁡y∣\int \frac{du}{u} = \log|u| = \log|\log y|.