
What is the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \cot x \cot y\]?
A. \[\cos x = c \csc y\]
B. \[\sin x = c \sec y\]
C. \[\sin x = c \cos y\]
D. \[\cos x = c \sin y\]
Answer
161.4k+ views
Hint: Here, the first order differential equation is given. First, simplify the given equation by using the trigonometric ratios. Then, integrate both sides of the equation with respect to the corresponding variables. In the end, solve the integrals by using logarithmic properties to find the general solution of the differential equation.
Formula Used: \[\dfrac{1}{{\cot x}} = \tan x\]
\[\int {\tan x dx} = log\left( {\sec x} \right) + c\]
\[\int {\cot x dx} = log\left( {\sin x} \right) + c\]
Addition Property of logarithm: \[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = \cot x \cot y\].
Simplify the given equation.
\[\dfrac{{dy}}{{\cot y}} = \cot x dx\]
Apply the trigonometric ratio \[\dfrac{1}{{\cot x}} = \tan x\] and simplify the equation.
\[\tan y dy = \cot x dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\tan y dy} = \int {\cot x dx} \]
Solve the above integrals by applying the integral formulas \[\int {\tan x dx} = log\left( {\sec x} \right) + c\], and \[\int {\cot x dx} = log\left( {\sin x} \right) + c\].
We get,
\[\log\left( {\sec y} \right) + log c = \log\left( {\sin x} \right)\]
Apply the logarithmic property of addition on the right-hand side.
\[\log\left( {c \sec y} \right) = \log\left( {\sin x} \right)\]
Equate both sides.
\[c \sec y = \sin x\]
Therefore, the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \cot x \cot y\] is \[c \sec y = \sin x\].
Option ‘B’ is correct
Note: 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.
If all terms of the solution are in the form of a logarithm, then the integration constant is also in the form of a logarithm.
If all terms of the solution follow a similar pattern, then we can write the integration constant in that pattern. Because it makes calculations so much easier.
Formula Used: \[\dfrac{1}{{\cot x}} = \tan x\]
\[\int {\tan x dx} = log\left( {\sec x} \right) + c\]
\[\int {\cot x dx} = log\left( {\sin x} \right) + c\]
Addition Property of logarithm: \[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\]
Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = \cot x \cot y\].
Simplify the given equation.
\[\dfrac{{dy}}{{\cot y}} = \cot x dx\]
Apply the trigonometric ratio \[\dfrac{1}{{\cot x}} = \tan x\] and simplify the equation.
\[\tan y dy = \cot x dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\tan y dy} = \int {\cot x dx} \]
Solve the above integrals by applying the integral formulas \[\int {\tan x dx} = log\left( {\sec x} \right) + c\], and \[\int {\cot x dx} = log\left( {\sin x} \right) + c\].
We get,
\[\log\left( {\sec y} \right) + log c = \log\left( {\sin x} \right)\]
Apply the logarithmic property of addition on the right-hand side.
\[\log\left( {c \sec y} \right) = \log\left( {\sin x} \right)\]
Equate both sides.
\[c \sec y = \sin x\]
Therefore, the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \cot x \cot y\] is \[c \sec y = \sin x\].
Option ‘B’ is correct
Note: 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.
If all terms of the solution are in the form of a logarithm, then the integration constant is also in the form of a logarithm.
If all terms of the solution follow a similar pattern, then we can write the integration constant in that pattern. Because it makes calculations so much easier.
Recently Updated Pages
Crack JEE Advanced 2025 with Vedantu's Live Classes

JEE Advanced Maths Revision Notes

JEE Advanced Chemistry Revision Notes

Download Free JEE Advanced Revision Notes PDF Online for 2025

The students S1 S2 S10 are to be divided into 3 groups class 11 maths JEE_Advanced

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 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

JEE Advanced Cut Off 2024

JEE Advanced Exam Pattern 2025

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Displacement-Time Graph and Velocity-Time Graph for JEE

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

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
