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Question

(D) Given the differential equation: $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ Separate the variables $x$ and $y$: $\frac{\cos y}{\sin y} dy = -

Vedclass · Accepted Answer

$\sin y = c \cos x$

Vedclass · Answer

(D) Given the differential equation: $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ Separate the variables $x$ and $y$: $\frac{\cos y}{\sin y} dy = - \frac{\sin x}{\cos x} dx$ This simplifies to: $\cot y \, dy = - 	an x \, dx$ Integrating both sides: $\int \cot y \, dy = - \int 	an x \, dx$ Using the standard integrals $\int \cot y \, dy = \log |\sin y|$ and $\int 	an x \, dx = \log |\sec x| = - \log |\cos x|$,we get: $\log |\sin y| = - (- \log |\cos x|) + \log |c|$ $\log |\sin y| = \log |\cos x| + \log |c|$ Using the property $\log a + \log b = \log(ab)$: $\log |\sin y| = \log |c \cos x|$ Taking the exponential of both sides: $\sin y = c \cos x$

The solution of the differential equation $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ is

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