The general solution of the differential equa | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \cot x \cot y$ Separate the variables: $\frac{dy}{\cot y} = \cot x \, dx$ This can be written as

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$\sin x = c \sec y$

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(B) Given the differential equation: $\frac{dy}{dx} = \cot x \cot y$ Separate the variables: $\frac{dy}{\cot y} = \cot x \, dx$ This can be written as: $	an y \, dy = \cot x \, dx$ Integrate both sides: $\int 	an y \, dy = \int \cot x \, dx$ Using the standard integration formulas $\int 	an y \, dy = \ln|\sec y|$ and $\int \cot x \, dx = \ln|\sin x|$: $\ln|\sec y| = \ln|\sin x| + C$ Let the constant $C = \ln|c|$: $\ln|\sec y| = \ln|\sin x| + \ln|c|$ Using the property $\ln a + \ln b = \ln(ab)$: $\ln|\sec y| = \ln|c \sin x|$ Taking the exponential of both sides: $\sec y = c \sin x$ This can be rewritten as: $\sin x = \frac{1}{c} \sec y$ Since $\frac{1}{c}$ is also an arbitrary constant,we can write it as $c$: $\sin x = c \sec y$

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

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