Question

If $xdy-ydx+xcoslnxdx=0$, $y(1)=1$, then $y(e)=$

• A. $e(1-cos1)$
• B. $e(1-sin1)$
• C. $e(1+cos1)$
• D. $e(1+sin1)$

Answer 1

Answer: (B)

$\frac{xdy-ydx}{x^2}+\frac{coslnx}{x}dx=0$
$\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right)+\frac{d}{dx}\left(sinlnx\right)=0$
On integration, we get $\frac{y}{x}+sinlnx=c$
$x=1$,
$y=1$
$\Rightarrow c=1$
$x=e$
$\Rightarrow y=e\left(1-sin1\right)$.