The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation: $\frac{dy}{dx} = \cot x \cdot \cot y$. Separating the variables,we get: $\frac{dy}{\cot y} = \cot x \cdot dx$. Th

Vedclass · Accepted Answer

$\sin x = c \cos y$,where $c$ is the constant of integration.

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \cot x \cdot \cot y$. Separating the variables,we get: $\frac{dy}{\cot y} = \cot x \cdot dx$. This can be written as: $	an y \cdot dy = \cot x \cdot dx$. Integrating both sides: $\int 	an y \cdot dy = \int \cot x \cdot dx$. Using the standard integration formulas $\int 	an y \cdot dy = \ln|\sec y|$ and $\int \cot x \cdot dx = \ln|\sin x|$,we get: $\ln|\sec y| = \ln|\sin x| + \ln|c|$. Using the property $\ln|a| + \ln|b| = \ln|ab|$,we have: $\ln|\sec y| = \ln|c \sin x|$. Taking the exponential of both sides: $\sec y = c \sin x$. Since $\sec y = \frac{1}{\cos y}$,we have $\frac{1}{\cos y} = c \sin x$,which implies $\sin x = \frac{1}{c} \cos y$. Letting $k = \frac{1}{c}$,we get $\sin x = k \cos y$. Comparing this with the given options,option $C$ is the correct form.

The general solution of the differential equation $\frac{dy}{dx} = \cot x \cdot \cot y$ is

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