
Find the solution of the differential equation \[9y\left( {\dfrac{{dy}}{{dx}}} \right) + 4x = 0\]
A. \[\left( {\dfrac{{{y^2}}}{9}} \right) + \left( {\dfrac{{{x^2}}}{4}} \right) = c\]
B. \[\left( {\dfrac{{{y^2}}}{4}} \right) + \left( {\dfrac{{{x^2}}}{9}} \right) = c\]
C. \[\left( {\dfrac{{{y^2}}}{9}} \right) - \left( {\dfrac{{{x^2}}}{4}} \right) = c\]
D. \[\left( {{y^2}} \right) - \left( {\dfrac{{{x^2}}}{9}} \right) = c\]
Answer
163.2k+ views
Hint: The solution of a differential equation is the relationship between the variables of a differential equation that fulfils the specified differential equation. In this question, the solutions of the differential equation can be obtained by integrating the differential equation by separating \[y\] and \[dy\] and \[x\] and \[dx\] terms.
Formula used: The following formula of integration can be used to solve this example.
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \]
Complete step-by-step solution: We know that the given differential equation is \[9y\left( {\dfrac{{dy}}{{dx}}} \right) + 4x = 0\]
Let us simplify the above differential equation.
Thus, we get
\[9y\left( {\frac{{dy}}{{dx}}} \right) = - 4x\]
\[ \Rightarrow 9ydy = - 4xdx\]
By integrating on both sides, we get
\[\int {9ydy} = \int { - 4xdx} \]
\[ \Rightarrow 9\int {ydy} = - 4\int {xdx} \]
But we know that \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \]
So, we get
\[
\Rightarrow 9\left( {\dfrac{{{y^2}}}{2}} \right) = - 4\left( {\dfrac{{{x^2}}}{2}} \right) + c' \\
\Rightarrow \left( {\dfrac{{9{y^2}}}{2}} \right) + \left( {\dfrac{{4{x^2}}}{2}} \right) = c' \\
\Rightarrow \left( {\dfrac{{9{y^2} + 4{x^2}}}{2}} \right) = c' \\
\Rightarrow \left( {9{y^2} + 4{x^2}} \right) = 2c'
\]
Here, \[c'\] is a constant of integration.
Now, take LCM (Least Common Multiple) of \[9\] and \[4\] is \[36\]
Thus, divide by \[36\] on both sides.
So, we get
\[
\left( {\dfrac{{9{y^2}}}{{36}} + \dfrac{{4{x^2}}}{{36}}} \right) = \dfrac{{2c'}}{{36}} \\
\Rightarrow \left( {\dfrac{{{y^2}}}{4} + \dfrac{{{x^2}}}{9}} \right) = \dfrac{{c'}}{{18}} \\
\]
Put \[\dfrac{{c'}}{{18}} = c\] in the above equation.
Thus, we get
\[\left( {\dfrac{{{y^2}}}{4} + \dfrac{{{x^2}}}{9}} \right) = c\]
Hence, the solution of the differential equation \[9y\left( {\dfrac{{dy}}{{dx}}} \right) + 4x = 0\] is \[\left( {\dfrac{{{y^2}}}{4} + \dfrac{{{x^2}}}{9}} \right) = c\]
Therefore, the option (B) is correct.
Additional Information: The differential equation is defined as an equation that consists of one or more functions and their derivatives. There are several methods to solve differential equations such as variable separable method, inspection method, homogeneous method, etc.
Note: There are two types of solutions to differential equations such as general solution and particular solution. A general solution of the nth order differential equation is the solution that contains vital arbitrary constants. Whereas, a particular solution of a differential equation is one generated from the general solution by assigning specific values to an arbitrary solution.
Formula used: The following formula of integration can be used to solve this example.
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \]
Complete step-by-step solution: We know that the given differential equation is \[9y\left( {\dfrac{{dy}}{{dx}}} \right) + 4x = 0\]
Let us simplify the above differential equation.
Thus, we get
\[9y\left( {\frac{{dy}}{{dx}}} \right) = - 4x\]
\[ \Rightarrow 9ydy = - 4xdx\]
By integrating on both sides, we get
\[\int {9ydy} = \int { - 4xdx} \]
\[ \Rightarrow 9\int {ydy} = - 4\int {xdx} \]
But we know that \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \]
So, we get
\[
\Rightarrow 9\left( {\dfrac{{{y^2}}}{2}} \right) = - 4\left( {\dfrac{{{x^2}}}{2}} \right) + c' \\
\Rightarrow \left( {\dfrac{{9{y^2}}}{2}} \right) + \left( {\dfrac{{4{x^2}}}{2}} \right) = c' \\
\Rightarrow \left( {\dfrac{{9{y^2} + 4{x^2}}}{2}} \right) = c' \\
\Rightarrow \left( {9{y^2} + 4{x^2}} \right) = 2c'
\]
Here, \[c'\] is a constant of integration.
Now, take LCM (Least Common Multiple) of \[9\] and \[4\] is \[36\]
Thus, divide by \[36\] on both sides.
So, we get
\[
\left( {\dfrac{{9{y^2}}}{{36}} + \dfrac{{4{x^2}}}{{36}}} \right) = \dfrac{{2c'}}{{36}} \\
\Rightarrow \left( {\dfrac{{{y^2}}}{4} + \dfrac{{{x^2}}}{9}} \right) = \dfrac{{c'}}{{18}} \\
\]
Put \[\dfrac{{c'}}{{18}} = c\] in the above equation.
Thus, we get
\[\left( {\dfrac{{{y^2}}}{4} + \dfrac{{{x^2}}}{9}} \right) = c\]
Hence, the solution of the differential equation \[9y\left( {\dfrac{{dy}}{{dx}}} \right) + 4x = 0\] is \[\left( {\dfrac{{{y^2}}}{4} + \dfrac{{{x^2}}}{9}} \right) = c\]
Therefore, the option (B) is correct.
Additional Information: The differential equation is defined as an equation that consists of one or more functions and their derivatives. There are several methods to solve differential equations such as variable separable method, inspection method, homogeneous method, etc.
Note: There are two types of solutions to differential equations such as general solution and particular solution. A general solution of the nth order differential equation is the solution that contains vital arbitrary constants. Whereas, a particular solution of a differential equation is one generated from the general solution by assigning specific values to an arbitrary solution.
Recently Updated Pages
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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Instantaneous Velocity - Formula based Examples for JEE

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

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
