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Q.
If $\frac{xdy}{dx}+2y=ln\,x$, then $e^{2}\,y\left(e\right)-y\left(1\right)=$
Differential Equations
Solution:
$\frac{dy}{dx}+\frac{2}{x}y=\frac{ln\,x}{x}$.
It is linear differential equation with $I.F.=\text{exp} \int \frac{2}{x}dx=x^{2}$
$\therefore $ Solution is, $yx^{2}=\int x^{2}\cdot\frac{ln\,x}{x}dx=\int x\,ln\,x\,dx$
$\Rightarrow yx^{2}=\frac{x^{2}}{2}ln\,x-\frac{x^{2}}{4}+c$
$\therefore x=e$
$\Rightarrow e^{2}y\left(e\right)=\frac{e^{2}}{2}-\frac{e^{2}}{4}+c=\frac{e^{2}}{4}+c$
and $x=1$
$\Rightarrow y\left(1\right)=-\frac{1}{4}+c$
So, $e^{2}y\left(e\right)-y\left(1\right)=\frac{e^{2}+1}{4}$.