The general solution of the differential equation $\cos x \cdot \sin y d x$ $+\sin x \cdot \cos y d y=0$ is

The general solution of the differential equation $\cos x \cdot \sin y d x$ $+\sin x \cdot \cos y d y=0$ is
  1. $\sin x+\sin y=c$
  2. $\cos x+\cos y=c$
  3. $\sin x \cdot \sin y=c$
  4. $\cos x \cdot \cos y=c$

Solution

$\begin{aligned} & \cos \mathrm{x} \sin \mathrm{ydx}+\sin \mathrm{x} \cos \mathrm{y} d \mathrm{y}=0 \\ & \therefore \cos \mathrm{x} \sin \mathrm{ydx}=-\sin \mathrm{x} \cos \mathrm{ydy} \\ & \therefore \int \frac{\cos \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}=-\int \frac{\cos \mathrm{y}}{\sin \mathrm{y}} \mathrm{dy} \\ & \therefore \log |\sin \mathrm{x}|=-\log |\sin \mathrm{y}|+\mathrm{c}_1 \\ & \therefore \log |\sin \mathrm{x}|+\log |\sin \mathrm{y}|=\mathrm{c}_1 \Rightarrow \log [\sin \mathrm{x} \sin \mathrm{y}]=\mathrm{c}_1 \\ & \therefore \sin \mathrm{x} \sin \mathrm{y}=\mathrm{e}^{c_1}=\mathrm{c} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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