
What is the solution of the differential equation \[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} + {y^2} + x{y^2} = 0\]?
A. \[\log \left( {\dfrac{x}{y}} \right) = \dfrac{1}{x} + \dfrac{1}{y} + c\]
B. \[\log \left( {\dfrac{y}{x}} \right) = \dfrac{1}{x} + \dfrac{1}{y} + c\]
C. \[\log \left( {xy} \right) = \dfrac{1}{x} + \dfrac{1}{y} + c\]
D. \[\log \left( {xy} \right) + \dfrac{1}{x} + \dfrac{1}{y} = c\]
Answer
162.6k+ views
Hint: First we separate all variables of the differential equation. Then take integration operations on both sides of the equation and by using integration formulas solve it.
Formula used:
Integration formula:
\[\int {\dfrac{1}{x}dx} = \log x + c\]
\[\int {\dfrac{1}{{{x^2}}}dx} = - \dfrac{1}{x} + c\]
Complete step by step solution:
Given differential equation is \[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} + {y^2} + x{y^2} = 0\].
Now we will separate the variable of the differential equation:
\[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} = - {y^2}\left( {1 + x} \right)\]
\[ \Rightarrow {x^2}\left( {1 - y} \right)\dfrac{{dy}}{{dx}} = - {y^2}\left( {1 + x} \right)\]
Divide both sides by \[{x^2}{y^2}\]
\[ \Rightarrow \dfrac{{{x^2}\left( {1 - y} \right)}}{{{x^2}{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{{y^2}\left( {1 + x} \right)}}{{{x^2}{y^2}}}\]
Cancel out same terms from the denominator and numerator:
\[ \Rightarrow \dfrac{{{x^2}\left( {1 - y} \right)}}{{{x^2}{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{{y^2}\left( {1 + x} \right)}}{{{x^2}{y^2}}}\]
\[ \Rightarrow \dfrac{{\left( {1 - y} \right)}}{{{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{\left( {1 + x} \right)}}{{{x^2}}}\]
\[ \Rightarrow \dfrac{{\left( {1 - y} \right)}}{{{y^2}}}dy = - \dfrac{{\left( {1 + x} \right)}}{{{x^2}}}dx\]
Break it into terms:
\[ \Rightarrow \left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{y}} \right)dy = - \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right)dx\]
\[ \Rightarrow \left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{y}} \right)dy + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right)dx = 0\]
Taking integration on both sides:
\[ \Rightarrow \int {\dfrac{1}{{{y^2}}}dy} - \int {\dfrac{1}{y}dy} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{1}{x}dx} = 0\]
Applying integration formula:
\[ \Rightarrow - \dfrac{1}{y} - \log y - \dfrac{1}{x} + \log x = c\]
Rewrite the above equation:
\[ \Rightarrow \log x - \log y = c + \dfrac{1}{x} + \dfrac{1}{y}\]
Applying quotient formula of logarithm:
\[ \Rightarrow \log \left( {\dfrac{x}{y}} \right) = c + \dfrac{1}{x} + \dfrac{1}{y}\]
Hence option A is the correct option.
Note: Students often do mistake to integrate \[\int {\dfrac{1}{{{x^2}}}dx} \].They apply the formula \[\int {\dfrac{1}{{{x^2}}}dx} = \dfrac{1}{x} + c\] which is incorrect formula. The correct formula is \[\int {\dfrac{1}{{{x^2}}}dx} = - \dfrac{1}{x} + c\].
Formula used:
Integration formula:
\[\int {\dfrac{1}{x}dx} = \log x + c\]
\[\int {\dfrac{1}{{{x^2}}}dx} = - \dfrac{1}{x} + c\]
Complete step by step solution:
Given differential equation is \[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} + {y^2} + x{y^2} = 0\].
Now we will separate the variable of the differential equation:
\[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} = - {y^2}\left( {1 + x} \right)\]
\[ \Rightarrow {x^2}\left( {1 - y} \right)\dfrac{{dy}}{{dx}} = - {y^2}\left( {1 + x} \right)\]
Divide both sides by \[{x^2}{y^2}\]
\[ \Rightarrow \dfrac{{{x^2}\left( {1 - y} \right)}}{{{x^2}{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{{y^2}\left( {1 + x} \right)}}{{{x^2}{y^2}}}\]
Cancel out same terms from the denominator and numerator:
\[ \Rightarrow \dfrac{{{x^2}\left( {1 - y} \right)}}{{{x^2}{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{{y^2}\left( {1 + x} \right)}}{{{x^2}{y^2}}}\]
\[ \Rightarrow \dfrac{{\left( {1 - y} \right)}}{{{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{\left( {1 + x} \right)}}{{{x^2}}}\]
\[ \Rightarrow \dfrac{{\left( {1 - y} \right)}}{{{y^2}}}dy = - \dfrac{{\left( {1 + x} \right)}}{{{x^2}}}dx\]
Break it into terms:
\[ \Rightarrow \left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{y}} \right)dy = - \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right)dx\]
\[ \Rightarrow \left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{y}} \right)dy + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right)dx = 0\]
Taking integration on both sides:
\[ \Rightarrow \int {\dfrac{1}{{{y^2}}}dy} - \int {\dfrac{1}{y}dy} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{1}{x}dx} = 0\]
Applying integration formula:
\[ \Rightarrow - \dfrac{1}{y} - \log y - \dfrac{1}{x} + \log x = c\]
Rewrite the above equation:
\[ \Rightarrow \log x - \log y = c + \dfrac{1}{x} + \dfrac{1}{y}\]
Applying quotient formula of logarithm:
\[ \Rightarrow \log \left( {\dfrac{x}{y}} \right) = c + \dfrac{1}{x} + \dfrac{1}{y}\]
Hence option A is the correct option.
Note: Students often do mistake to integrate \[\int {\dfrac{1}{{{x^2}}}dx} \].They apply the formula \[\int {\dfrac{1}{{{x^2}}}dx} = \dfrac{1}{x} + c\] which is incorrect formula. The correct formula is \[\int {\dfrac{1}{{{x^2}}}dx} = - \dfrac{1}{x} + c\].
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

IIIT JEE Main Cutoff 2024

IIT Full Form

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

JEE Main Cut-Off for NIT Kurukshetra: All Important Details

JEE Main Cut-Off for VNIT Nagpur 2025: Check All Rounds Cutoff Ranks

Other Pages
NEET 2025: All Major Changes in Application Process, Pattern and More

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025: Important Information and Key Updates

1 Billion in Rupees - Conversion, Solved Examples and FAQs

NEET 2025 Syllabus PDF by NTA (Released)

Important Days In June: What Do You Need To Know
