
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
217.8k+ 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
Arithmetic, Geometric & Harmonic Progressions Explained

Cartesian Form of Vector Explained: Formula, Examples & Uses

Apparent Frequency Explained: Formula, Uses & Examples

Calorimetry: Definition, Principles & Calculations

Centrifugal Force Explained: Definition, Formula & Examples

Charge in a Magnetic Field: Definition, Formula & Examples

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

Understanding Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

