
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
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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.
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