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  • the solution of the differential equation \cos x \sin y \;dx+\sin x \cos y\; dy=0 is (a)\frac{\sin x}{\sin y}=c\\ (b)\sin x \sin y = c\\ (c)\sin x +\sin y = c\\ (d)\cos x \cos y = c

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

\\(a)\frac{\sin x}{\sin y}=c\\ (b)\sin x \sin y = c\\ (c)\sin x +\sin y = c\\ (d)\cos x \cos y = c

Answers (1)

\\\cos x \sin y dx +\sin x \cos y dy=0\\ \Rightarrow \sin x \cos y dy=-\cos x \sin y dx\\ \Rightarrow \frac{\cos y}{sin y}dy=-\frac{\cos x}{sin x}dx\\ \Rightarrow \cot y dy =-\cot x dx

Upon integration of both sides,

\Rightarrow \int \cot y dy =-\int \cot x dx\\ \Rightarrow \log \left | \sin y \right |=-\log \left | \sin x \right |+\log c\\ \Rightarrow \log \left | \sin y \right |+\log \left | \sin x \right |=\log c\\ \Rightarrow \log \left | \sin y . \sin x \right |=\log c\\ \Rightarrow \sin y \sin x=c

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