Question

Solution of the differential equation,
$x^{2} \frac{dy}{dx}\cdot cos \frac{1}{x}-y\,sin\, \frac{1}{x}=-1$, 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. $y=\frac{x+1}{x\,cos \frac{1}{x}}$

Answer 1

Answer: (A)

The given equation can be written as,
$\frac{dy}{dx}-\frac{y}{x^{2}}tan\left(\frac{1}{x}\right)=-sec\left(\frac{1}{x}\right)\cdot\frac{1}{x^{2}}$
$I.F.=e^{-\int\left(\frac{1}{x^{2}}\right)tan \frac{1}{x}dx}=sec\left(\frac{1}{x}\right)$
$\Rightarrow y\cdot sec\left(\frac{1}{x}\right)=-\int sec^{2}\left(\frac{1}{x}\right) \frac{1}{x^{2}}dx=tan\left(\frac{1}{x}\right)+C$
$\Rightarrow y=sin\left(\frac{1}{x}\right)+C\,cos\left(\frac{1}{x}\right)$
If $y \to-1$ and $x \to\infty$, then $C=-1$
$\Rightarrow y=sin\left(\frac{1}{x}\right)-cos\left(\frac{1}{x}\right)$