
What is the solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\]?
A. \[y + \log x + c = 0\]
B. \[y = \log x + c\]
C. \[{y^{\log x}} + c = 0\]
D. None of these
Answer
232.8k+ views
Hint: Here, the first order differential equation is given. First, simplify the given equation by rearranging the terms or cross multiplication method. Then, integrate both sides of the equation with respect to the corresponding variables. In the end, solve the integrals using the standard integration formulas \[\int {ndx} = x + c\] and \[\int {\dfrac{1}{x}dx} = \log x + c\] to get the solution of the differential equation.
Formula Used: \[\int {ndx} = x + c\], where \[n\] is a constant.
\[\int {\dfrac{1}{x}dx} = \log x + c\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\].
Simplify the given equation.
\[dy = \dfrac{{dx}}{x}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {dy} = \int {\dfrac{{dx}}{x}} \]
Apply the integration formulas \[\int {ndx} = x + c\], and \[\int {\dfrac{1}{x}dx} = \log x + c\].
We get,
\[y = log x + c\]
Therefore, the solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\] is \[y = \log x + c\].
Option ‘B’ is correct
Note: Students often apply a wrong formula to integrate \[\dfrac {1}{x}\]. They integrate it by using the power formula of integration. But the correct formula is \[\int {\dfrac{1}{x}dx} = \log x + c\].
It is necessary to use an integration constant as soon as integration is performed if we solve a first-order differential equation by a variable method.
Formula Used: \[\int {ndx} = x + c\], where \[n\] is a constant.
\[\int {\dfrac{1}{x}dx} = \log x + c\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\].
Simplify the given equation.
\[dy = \dfrac{{dx}}{x}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {dy} = \int {\dfrac{{dx}}{x}} \]
Apply the integration formulas \[\int {ndx} = x + c\], and \[\int {\dfrac{1}{x}dx} = \log x + c\].
We get,
\[y = log x + c\]
Therefore, the solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\] is \[y = \log x + c\].
Option ‘B’ is correct
Note: Students often apply a wrong formula to integrate \[\dfrac {1}{x}\]. They integrate it by using the power formula of integration. But the correct formula is \[\int {\dfrac{1}{x}dx} = \log x + c\].
It is necessary to use an integration constant as soon as integration is performed if we solve a first-order differential equation by a variable method.
Recently Updated Pages
JEE Advanced 2026 Revision Notes for Vectors - Free PDF Download

JEE Advanced 2026 Revision Notes for Trigonometry - Free PDF Download

JEE Advanced 2026 Surface Chemistry Revision Notes - Free PDF Download

JEE Advanced Study Plan 2026: Expert Tips and Preparation Guide

JEE Advanced 2026 Revision Notes for Chemistry Solutions - Free PDF Download

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Difference Between Exothermic and Endothermic Reactions Explained

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

IIT Fees Structure 2025

Top IIT Colleges in India 2025

Other Pages
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

