Question

If $\frac{xdy}{dx}+2y=lnx$, then $e^{2}y\left(e\right)-y\left(1\right)=$

• A. $\frac{e^2+1}{2}$
• B. $\frac{e^2+1}{3}$
• C. $\frac{e^2+1}{4}$
• D. $e^2+1$

Answer 1

Answer: (C)

$\frac{dy}{dx}+\frac{2}{x}y=\frac{lnx}{x}$.
It is linear differential equation with $I.F.=\text{exp}\int \frac{2}{x}dx=x^2$
$\therefore$ Solution is, $y{x}^2=\int x^2\cdot \frac{lnx}{x}dx=\int xlnxdx$
$\Rightarrow y{x}^2=\frac{x^2}{2}lnx-\frac{x^2}{4}+c$
$\therefore x=e$
$\Rightarrow e^{2}y\left(e\right)=\frac{e^2}{2}-\frac{e^2}{4}+c=\frac{e^2}{4}+c$
and $x=1$
$\Rightarrow y\left(1\right)=-\frac{1}{4}+c$
So, $e^{2}y\left(e\right)-y\left(1\right)=\frac{e^2+1}{4}$.