Question

The solution of the differential equation $x^2\frac{dy}{dx}\cos \left(\frac{1}{x}\right)-y\sin \left(\frac{1}{x}\right)=-1;wherey\to -1asx\to \infty$ is

• A. $y=\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)$
• B. $y=\frac{x+1}{x\sin \left(\frac{1}{x}\right)}$
• C. $y=\cos \left(\frac{1}{x}\right)+\sin \left(\frac{1}{x}\right)$
• D. $y=\frac{x+1}{x\cos \left(\frac{1}{x}\right)}$

Answer 1

Answer: (A)

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

$x^2\frac{dy}{dx}cos\left(\frac{1}{x}\right)-ysin\left(\frac{1}{x}\right)=-1$

$\frac{dy}{dx}-\frac{y}{x^2}tan\left(\frac{1}{x}\right)=-\frac{1}{x^2}cos\left(\frac{1}{x}\right)$

$\frac{dy}{dx}+\left(\frac{-1}{x^2}tan\left(\frac{1}{x}\right)\right)y=\frac{-1}{x^{2}cos\left(\frac{1}{x}\right)}$

$I.F=e^{\int -\frac{1}{x^2}tan\left(\frac{1}{x}\right)dx}$

$\frac{1}{x}=t$

$-\frac{dx}{x^2}=dt$

$=e^{\int tan\left(t\right)dt}$

$=e^{ln\left(sect\right)}$

$=sec\left(\frac{1}{x}\right)$

$sec\left(\frac{1}{x}\right).y=\int \frac{1}{x^{2}cos\left(\frac{1}{x}\right)}.sec\left(\frac{1}{x}\right)$

$=-\int \frac{se{c}^2\left(\frac{1}{x}\right)}{x^2}$

Let $\frac{1}{x}=t$

$\frac{-dx}{x^2}=dt$

$=\int se{c}^2\left(t\right)dt$

$sec\left(\frac{1}{x}\right).y=tan\left(\frac{1}{x}\right)+C$

$1.-1=0+C$ $C=-1$

$sec\left(\frac{1}{x}\right).y=tan\left(\frac{1}{x}\right)-1,$

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