The solution of the differential equation \[{{x}^{2}}\frac{dy}{dx}\cos \frac{1}{x}-y\sin \frac{1}{x}=-1\], where \[y\to -1\,\,as\,\,x\to \infty \]is

\[y=\frac{x+1}{x\sin \frac{1}{x}}\]

\[y=\sin \frac{1}{x}-\cos \frac{1}{x}\]

\[y=\cos \frac{1}{x}+sin\frac{1}{x}\]

The solution of the differential equation \[{{x}^{2}}\frac{dy}{dx}\cos

1.	The solution of the differential equation \[{{x}^{2}}\frac{dy}{dx}\cos \frac{1}{x}-y\sin \frac{1}{x}=-1\], where \[y\to -1\,\,as\,\,x\to \infty \]is
A.	\[y=\sin \frac{1}{x}-\cos \frac{1}{x}\]
B.	\[y=\frac{x+1}{x\sin \frac{1}{x}}\]
C.	\[y=\cos \frac{1}{x}+sin\frac{1}{x}\]
D.	\[y=\frac{x+1}{x\cos 1/x}\]
Answer» B. \[y=\frac{x+1}{x\sin \frac{1}{x}}\]

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The solution of the differential equation \[{{x}^{2}}\frac{dy}{dx}\cos \frac{1}{x}-y\sin \frac{1}{x}=-1\], where \[y\to -1\,\,as\,\,x\to \infty \]is

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