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) \[y=\frac{x+1}{x\cos 1/x}\]
Correct Answer: A
Solution :
[a] \[{{x}^{2}}\frac{dy}{dx}\cos \frac{1}{x}-y\sin \frac{1}{x}=-1\] Or \[\frac{dy}{dx}-\frac{y}{{{x}^{2}}}\tan \frac{1}{x}=-\sec \frac{1}{x}\frac{1}{{{x}^{2}}}(linear)\] I.F. \[={{e}^{-\int{\frac{1}{{{x}^{2}}}\tan \frac{1}{x}dx}}}=\sec \frac{1}{x}\] Thus, solution is \[y\sec \frac{1}{x}=-\int{{{\sec }^{2}}\left( \frac{1}{x} \right)\frac{1}{{{x}^{2}}}dx=\tan \frac{1}{x}+c}\] Given \[y\to -1,x\to \infty .\,\,\,\text{Thus,}\,C=-1.\] Hence, equation of curve is\[y=\sin \frac{1}{x}-\cos \frac{1}{x}\].
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