
What is the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{{y^2}}}\]?
A. \[{x^3} - {y^3} = c\]
B. \[{x^3} + {y^3} = c\]
C. \[{x^2} + {y^2} = c\]
D. \[{x^2} - {y^2} = c\]
Answer
164.7k+ views
Hint: Here, the first order differential equation is given. First, simplify the given equation by rearranging the terms. Then, integrate both sides of the equation with respect to the corresponding variables. In the end, solve the integrals by using the standard integration formula and get the general solution of the differential equation.
Formula Used: \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{{y^2}}}\].
Simplify the given equation by cross multiplying the denominators.
\[{y^2}dy = {x^2}dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {{y^2}dy} = \int {{x^2}dx} \]
Apply the integration formula \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\] on both sides.
We get,
\[\dfrac{{{y^3}}}{3} = \dfrac{{{x^3}}}{3} + \dfrac{{{c^3}}}{3}\]
Simplify the equation.
Multiply both sides by 3.
\[ Rightarrow {y^3} = {x^3} + {c_1}^3\]
\[ Rightarrow {y^3} = {x^3} + c\]
\[ \Rightarrow {y^3} - {x^3} = c\] or \[{x^3} - {y^3} = c\]
Therefore, the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{{y^2}}}\] is \[{x^3} - {y^3} = c\].
Option ‘A’ is correct
Note: Students often do mistake to integrating \[\int {{x^n}} dx\]. They apply the formula \[\int {{x^n}} dx = {x^{n + 1}} + c\], which is incorrect formula. They forget to divide the term \[{x^{n + 1}}\] by \[n + 1\] The correct formula is \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\].
Formula Used: \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{{y^2}}}\].
Simplify the given equation by cross multiplying the denominators.
\[{y^2}dy = {x^2}dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {{y^2}dy} = \int {{x^2}dx} \]
Apply the integration formula \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\] on both sides.
We get,
\[\dfrac{{{y^3}}}{3} = \dfrac{{{x^3}}}{3} + \dfrac{{{c^3}}}{3}\]
Simplify the equation.
Multiply both sides by 3.
\[ Rightarrow {y^3} = {x^3} + {c_1}^3\]
\[ Rightarrow {y^3} = {x^3} + c\]
\[ \Rightarrow {y^3} - {x^3} = c\] or \[{x^3} - {y^3} = c\]
Therefore, the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{{y^2}}}\] is \[{x^3} - {y^3} = c\].
Option ‘A’ is correct
Note: Students often do mistake to integrating \[\int {{x^n}} dx\]. They apply the formula \[\int {{x^n}} dx = {x^{n + 1}} + c\], which is incorrect formula. They forget to divide the term \[{x^{n + 1}}\] by \[n + 1\] The correct formula is \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\].
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Get P Block Elements for JEE Main 2025 with clear Explanations

Sets, Relations and Functions Chapter For JEE Main Maths

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
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

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
