Question

Solution of the differential equation,
$x^2\frac{dy}{dx}\cdot cos\frac{1}{x}-ysin\frac{1}{x}=-1$, where $y\to -1$ as $x\to \infty$, is

• A. $y=sin\frac{1}{x}-cos\frac{1}{x}$
• B. $y=\frac{x+1}{xsin\frac{1}{x}}$
• C. $y=cos\frac{1}{x}+sin\frac{1}{x}$
• D. $y=\frac{x+1}{xcos\frac{1}{x}}$

Answer 1

Answer: (A)

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