The solution of the differential equation x2dydxcos⁡1x−ysin1x=−1 When y→−1 as x→∞ is

dydx−yx2tan⁡1x=−sec⁡1x⋅1x2 I. F=e∫-1x2tan1xdx=eln sec1x=sec⁡1x Solution of the D.E is ysec⁡1x=−∫sec2⁡1x⋅1x2dxysec⁡1x=tan⁡1x+c........(1) Given y→−1 and x→∞-sec1∞=tan1∞+c -sec0=tan0+c c=−1substituting in (1)ysec⁡1x=tan⁡1x-1y=sin⁡1x−cos⁡1x

The solution of the differential equation x2dydxcos⁡1x−ysin1x=−1 When y→−1 as x→∞ is

dydx−yx2tan⁡1x=−sec⁡1x⋅1x2 I. F=e∫-1x2tan1xdx=eln sec1x=sec⁡1x Solution of the D.E is ysec⁡1x=−∫sec2⁡1x⋅1x2dxysec⁡1x=tan⁡1x+c........(1) Given y→−1 and x→∞-sec1∞=tan1∞+c -sec0=tan0+c c=−1substituting in (1)ysec⁡1x=tan⁡1x-1y=sin⁡1x−cos⁡1x

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$The solution of the differential equation x^{2} \frac{dy}{dx} \cos (\frac{1}{x}) - ysin (\frac{1}{x}) = - 1$ $When y \to - 1 as x \to \infty is$

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$y=sin(\frac{1}{x}) - \cos (\frac{1}{x})$

$y= \frac{x+1}{xsin(\frac{1}{x})}$

$y=cos(\frac{1}{x}) + \sin (\frac{1}{x})$

$y= \frac{x+1}{xcos(\frac{1}{x})}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{dy}{dx} - \frac{y}{x^{2}} \tan (\frac{1}{x}) = - \sec (\frac{1}{x}) \cdot \frac{1}{x^{2}} \\ I. F = e^{\int - \frac{1}{x^{2}} \tan \frac{1}{x} d x} = e^{\ln s e c (\frac{1}{x})} = \sec (\frac{1}{x}) \\ Solution of the D.E is \\ ysec (\frac{1}{x}) = - \int \sec^{2} (\frac{1}{x}) \cdot \frac{1}{x^{2}} d x \\ ysec (\frac{1}{x}) = \tan (\frac{1}{x}) + c . . . . . . . . (1) \\ Given y \to - 1 a n d x \to \infty \\ - s e c (\frac{1}{\infty}) = \tan (\frac{1}{\infty}) + c \\ - s e c 0 = \tan 0 + c \\ c = - 1 \\ s u b s t i t u t i n g i n (1) \\ ysec (\frac{1}{x}) = \tan (\frac{1}{x}) - 1 \\ y = \sin (\frac{1}{x}) - \cos (\frac{1}{x}) \end{array}$