Question

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

• A. $y=lo{g}_e|\sec \frac{1}{x}+tan\frac{1}{x}|-1$.
• 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)

$y=lo{g}_e|\sec \frac{1}{x}+tan\frac{1}{x}|-1$.

Given,

$x^2\frac{dy}{dx}.cos\frac{1}{x}=-1$

$\Rightarrow dy=\frac{1}{cos\frac{1}{x}}.\frac{-1}{x^2}dx$

Let us assume $\frac{1}{x}=t$.

Therefore, $\frac{-1}{x^2}dx=dt$.

$\Rightarrow dy=\frac{1}{cost}dt$

$\Rightarrow dy=\sec tdt$

Integrating both sides, we get,

$\Rightarrow \int dy=\int \sec tdt$

$\Rightarrow y=lo{g}_e|\sec t+tant|+C$

$\Rightarrow y=lo{g}_e|\sec \frac{1}{x}+tan\frac{1}{x}|+C$

Now, given that $y\to -1asx\to \infty$

${\Rightarrow lim}_{x\to \infty }y=-1$

${\Rightarrow lim}_{x\to \infty }(lo{g}_e|\sec \frac{1}{x}+tan\frac{1}{x}|+C)=-1$

$\Rightarrow (lo{g}_e|\sec 0+tan0|+C)=-1$

$\Rightarrow C=-1$.

Therefore, the solution of given differential equation is:

$y=lo{g}_e|\sec \frac{1}{x}+tan\frac{1}{x}|-1$.