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Q.
If $x \,dy -y \,dx + x\,cos\,ln\,x\,dx = 0$, $y(1) = 1$, then $y(e) =$
Differential Equations
Solution:
$\frac{xdy-ydx}{x^{2}}+\frac{cos\,ln\,x}{x}dx=0$
$\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right)+\frac{d}{dx}\left(sin\,ln\,x\right)=0$
On integration, we get $\frac{y}{x}+sin\,ln\,x=c$
$x=1$,
$y=1$
$\Rightarrow c=1$
$x=e$
$\Rightarrow y=e\left(1-sin\,1\right)$.