
Select the general solution of the differential equation \[ydx - xdy = {x^2}ydx\]from the following options.
A \[y{e^{{x^2}}} = c{x^2}\]
B \[y{e^{ - {x^2}}} = c{x^2}\]
C \[{y^2}{e^{{x^2}}} = c{x^2}\]
D \[{y^2}{e^{ - {x^2}}} = c{x^2}\]
Answer
216.3k+ views
Hint: The general solution of the differential equation can be found by using the variable separation method. Simplify the equation to separate the x containing terms and y containing terms and integrate it to find a general solution.
Formula Used: Integrating formula
\[\begin{array}{l}\int {\dfrac{1}{x}dx} = \log x + c\\\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\end{array}\]
Here, c is an arbitrary constant and n can not be zero.
Complete step by step solution: The given differential equation is \[ydx - xdy = {x^2}ydx\].
Simplify the equation by subtracting \[ydx\]on both sides of the equation.
\[ - xdy = {x^2}ydx - ydx\]
Now, Multiply the equation by -1 on both sides of the equation.
\[xdy = ydx - {x^2}ydx\]
Take \[ydx\]common from the right side of the equation.
\[xdy = ydx\left( {1 - {x^2}} \right)\]
Now divide by x on both sides of the equation.
\[dy = \dfrac{y}{x}dx\left( {1 - {x^2}} \right)\]
Divide both sides of the equation by y to separate the x containing terms and y containing terms.
\[\dfrac{{dy}}{y} = \left( {\dfrac{{1 - {x^2}}}{x}} \right)dx\]
Integrate both sides of the equation.
\[\begin{array}{l}\int {\dfrac{{dy}}{y}} = \int {\dfrac{{1 - {x^2}}}{x}dx} \\\log y = \int {\dfrac{1}{x}dx} - \int {xdx} \\\log y = \log x - \dfrac{{{x^2}}}{2} + \log c\end{array}\]
Multiply both sides of the equation by \[2\]
\[2\log y = 2\log x - {x^2} + 2\log c\]
Now add \[{x^2}\]on both sides of the equation.
\[2\log y + {x^2} = 2\log x + 2\log c\]
Subtract \[2\log y\]on both sides of the equation.
\[\begin{array}{l}{x^2} = 2\log x + 2\log c - 2\log y\\{x^2} = \log {x^2} + \log c - \log {y^2}\\{x^2} = \log \dfrac{{c{x^2}}}{{{y^2}}}\end{array}\]
Apply exponential function on both sides of the equation.
\[{e^{{x^2}}} = \dfrac{{c{x^2}}}{{{y^2}}}\]
Multiply both sides of the equation by \[{y^2}\]
\[{y^2}{e^{{x^2}}} = c{x^2}\]
The solution of the equation is \[{y^2}{e^{{x^2}}} = c{x^2}\].
Option ‘C’ is correct
Note: The common mistake in this type of question is first taking direct integration of both sides without separating variables. This leads to the wrong solution.
Formula Used: Integrating formula
\[\begin{array}{l}\int {\dfrac{1}{x}dx} = \log x + c\\\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\end{array}\]
Here, c is an arbitrary constant and n can not be zero.
Complete step by step solution: The given differential equation is \[ydx - xdy = {x^2}ydx\].
Simplify the equation by subtracting \[ydx\]on both sides of the equation.
\[ - xdy = {x^2}ydx - ydx\]
Now, Multiply the equation by -1 on both sides of the equation.
\[xdy = ydx - {x^2}ydx\]
Take \[ydx\]common from the right side of the equation.
\[xdy = ydx\left( {1 - {x^2}} \right)\]
Now divide by x on both sides of the equation.
\[dy = \dfrac{y}{x}dx\left( {1 - {x^2}} \right)\]
Divide both sides of the equation by y to separate the x containing terms and y containing terms.
\[\dfrac{{dy}}{y} = \left( {\dfrac{{1 - {x^2}}}{x}} \right)dx\]
Integrate both sides of the equation.
\[\begin{array}{l}\int {\dfrac{{dy}}{y}} = \int {\dfrac{{1 - {x^2}}}{x}dx} \\\log y = \int {\dfrac{1}{x}dx} - \int {xdx} \\\log y = \log x - \dfrac{{{x^2}}}{2} + \log c\end{array}\]
Multiply both sides of the equation by \[2\]
\[2\log y = 2\log x - {x^2} + 2\log c\]
Now add \[{x^2}\]on both sides of the equation.
\[2\log y + {x^2} = 2\log x + 2\log c\]
Subtract \[2\log y\]on both sides of the equation.
\[\begin{array}{l}{x^2} = 2\log x + 2\log c - 2\log y\\{x^2} = \log {x^2} + \log c - \log {y^2}\\{x^2} = \log \dfrac{{c{x^2}}}{{{y^2}}}\end{array}\]
Apply exponential function on both sides of the equation.
\[{e^{{x^2}}} = \dfrac{{c{x^2}}}{{{y^2}}}\]
Multiply both sides of the equation by \[{y^2}\]
\[{y^2}{e^{{x^2}}} = c{x^2}\]
The solution of the equation is \[{y^2}{e^{{x^2}}} = c{x^2}\].
Option ‘C’ is correct
Note: The common mistake in this type of question is first taking direct integration of both sides without separating variables. This leads to the wrong solution.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

Understanding Atomic Structure for Beginners

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

