Question:
The general solution of the differential equation \(\frac{d y}{d x}=\cot x \cdot \cot y\) is
The general solution of the differential equation \(\frac{d y}{d x}=\cot x \cdot \cot y\) is
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Separate the variables and integrate both sides.
Updated On: Nov 13, 2026
\(\mathrm{cos}x=\mathrm{ccosec}y\), where c is the constant of integration.
\(\mathrm{sin}x=\mathrm{csec}y\), where c is the constant of integration.
\(\mathrm{sin}x=x\mathrm{cos}y\), where c is the constant of integration.
\(\mathrm{cos}x=\mathrm{csin}y\), where c is the constant of integration.
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The equation \(\dfrac{dy}{dx}=\cot x\cot y\) has separable variables.
Step 2: Separate the variables
\(\tan y\,dy=\cot x\,dx\).
Step 3: Integrate
\(\int\tan y\,dy=-\log|\cos y|\) and \(\int\cot x\,dx=\log|\sin x|+C_1\).
So \(\log|\sin x|+\log|\cos y|=\log c\), which gives \(\sin x\cos y=c\).
Step 4: Compare the options
Rewrite as \(\sin x=\dfrac{c}{\cos y}=c\sec y\). This matches option (B).
Final Answer:
The general solution is sin x equals c times sec y. \[ \boxed{\text{(B) }\sin x=c\sec y} \]
The equation \(\dfrac{dy}{dx}=\cot x\cot y\) has separable variables.
Step 2: Separate the variables
\(\tan y\,dy=\cot x\,dx\).
Step 3: Integrate
\(\int\tan y\,dy=-\log|\cos y|\) and \(\int\cot x\,dx=\log|\sin x|+C_1\).
So \(\log|\sin x|+\log|\cos y|=\log c\), which gives \(\sin x\cos y=c\).
Step 4: Compare the options
Rewrite as \(\sin x=\dfrac{c}{\cos y}=c\sec y\). This matches option (B).
Final Answer:
The general solution is sin x equals c times sec y. \[ \boxed{\text{(B) }\sin x=c\sec y} \]
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