The general solution of the differential equa | vedclass

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Question

(C) Given differential equation is $\cos x \sin y \, dx + \sin x \cos y \, dy = 0$.Rearranging the terms,we get $\sin x \cos y \, dy = -\cos x \sin y

Vedclass · Accepted Answer

$\sin x \sin y = c$

Vedclass · Answer

(C) Given differential equation is $\cos x \sin y \, dx + \sin x \cos y \, dy = 0$.Rearranging the terms,we get $\sin x \cos y \, dy = -\cos x \sin y \, dx$.Separating the variables,we have $\frac{\cos y}{\sin y} \, dy = -\frac{\cos x}{\sin x} \, dx$.Integrating both sides,$\int \frac{\cos y}{\sin y} \, dy = -\int \frac{\cos x}{\sin x} \, dx$.Using the formula $\int \cot 	heta \, d	heta = \log |\sin 	heta| + C$,we get $\log |\sin y| = -\log |\sin x| + C_1$.Rearranging gives $\log |\sin x| + \log |\sin y| = C_1$.Using the property $\log a + \log b = \log(ab)$,we get $\log |\sin x \sin y| = C_1$.Taking the exponential on both sides,$\sin x \sin y = e^{C_1} = c$.

The general solution of the differential equation $\cos x \sin y \, dx + \sin x \cos y \, dy = 0$ is

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