What is the solution of the differential equation \[\cos x \cos y \dfrac{{dy}}{{dx}} = - \sin x \sin y \]?
A. \[\sin y + \cos x = c\]
B. \[\sin y - \cos x = c\]
C. \[\sin y \cos x = c\]
D. \[\sin y = c \cos x\]
Answer
262.2k+ views
Hint: Here, the first order differential equation is given. First, simplify the given equation. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve both integrals by using the standard integral rule. In the end, apply the properties of a logarithm to get the solution of the differential equation.
Formula used:
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Integration rule: \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \left[ {\log f\left( x \right)} \right] + c\] , where \[c\] is an integration constant.
Complete step by step solution:
The given differential equation is \[\cos x \cos y \dfrac{{dy}}{{dx}} = - \sin x \sin y \].
Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[\dfrac{{\cos y}}{{\sin y}}dy = - \dfrac{{\sin x}}{{\cos x}} dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{\cos y}}{{\sin y}}dy} = \int { - \dfrac{{\sin x}}{{\cos x}} dx} \]
We know the integration rule \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \left[ {\log f\left( x \right)} \right] + c\].
Apply this rule to solve both integrals.
\[\log\left( {\sin y} \right) = \log\left( {\cos x} \right) + \log c\]
Apply the logarithmic property of addition \[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\].
We get,
\[\log\left( {\sin y} \right) = \log\left( {c \cos x} \right)\]
Equating both sides, we get
\[\sin y = c \cos x\]
Hence the correct option is D.
Note: Students often make mistakes when integrating \[ \int {- \dfrac{{\sin x}}{{\cos x}} dx}\]. They take that \[\sin x\] is the derivative of \[\cos x\]. For this reason they get \[\sin y \cos x = c\] as an answer. But the correct answer is \[\sin y = c \cos x\].
Formula used:
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Integration rule: \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \left[ {\log f\left( x \right)} \right] + c\] , where \[c\] is an integration constant.
Complete step by step solution:
The given differential equation is \[\cos x \cos y \dfrac{{dy}}{{dx}} = - \sin x \sin y \].
Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[\dfrac{{\cos y}}{{\sin y}}dy = - \dfrac{{\sin x}}{{\cos x}} dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{\cos y}}{{\sin y}}dy} = \int { - \dfrac{{\sin x}}{{\cos x}} dx} \]
We know the integration rule \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \left[ {\log f\left( x \right)} \right] + c\].
Apply this rule to solve both integrals.
\[\log\left( {\sin y} \right) = \log\left( {\cos x} \right) + \log c\]
Apply the logarithmic property of addition \[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\].
We get,
\[\log\left( {\sin y} \right) = \log\left( {c \cos x} \right)\]
Equating both sides, we get
\[\sin y = c \cos x\]
Hence the correct option is D.
Note: Students often make mistakes when integrating \[ \int {- \dfrac{{\sin x}}{{\cos x}} dx}\]. They take that \[\sin x\] is the derivative of \[\cos x\]. For this reason they get \[\sin y \cos x = c\] as an answer. But the correct answer is \[\sin y = c \cos x\].
Recently Updated Pages
JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

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: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

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

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Advanced Marks vs Rank 2025 - Predict Your IIT Rank Based on Score

