KVPY Sample Paper KVPY Stream-SX Model Paper-19

The solution of the differential equation \[{{x}^{2}}\frac{dx}{dx}.cos\left( \frac{1}{x} \right)-y\,\,\sin \left( \frac{1}{x} \right)=-\,1.\] Where \[y\to -\,1\,\,as\,x\to \infty \] is -

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

B) \[y=\frac{x+1}{x\sin (1/x)}\]

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

D) \[y=\frac{x+1}{x\cos (1/x)}\]

Correct Answer: C

Solution :

\[\because \frac{dy}{dx}-\frac{\tan \,\,(1/x)}{{{x}^{2}}}\,\,\,y=-\frac{\sec \,\,(1/x)}{{{x}^{2}}}\]		??..(1)
\[\because I.F.={{e}^{-\int{\frac{\tan \,\,(1/x)}{{{x}^{2}}}dx}}}={{e}^{\ln \,\,\sec \,\,(1/x)}}=\sec \,\,(1/x)\]
\[\therefore y\,\,\sec \,\,(1/x)=-\int{\frac{{{\sec }^{2}}\,\,(1/x)}{{{x}^{2}}}\,\,dx}\] \[\Rightarrow y\,\,\sec \,\,(1/x)=\tan \,\,(1/x)+c\]	??.(2)
If \[x\to \infty ;\,\,y\to -1\]
\[\Rightarrow (-\,1)\,\,(1)=0+c\]\[\Rightarrow c=-\,1\] put in (2)
\[y=\sin \,\,(1/x)-cos\,\,(1/x)\]