Question

Let $y=y\left(x\right)$ be the solution of the differential equation, $x\frac{dy}{dx}+y=x{\log }_{e}x,(x>1).$ If $2y\left(2\right)={\log }_{e}4-1,$ then $y\left(e\right)$ is equal to:

• A. $\frac{e}{4}$
• B. $\frac{e^2}{4}$
• C. $-\frac{e}{2}$
• D. $-\frac{e^2}{2}$

Answer 1

The correct option is A $\frac{e}{4}$

$x\frac{dy}{dx}+y=x\ln x$
$\Rightarrow \frac{dy}{dx}+\frac{1}{x}.y=\ln x$
$I.F.=e^{\int \frac{1}{x}dx}=x$
$\Rightarrow x.y=\int x.\ln xdx\Rightarrow x.y=\ln x.\frac{x^2}{2}-\int \frac{1}{x}.\frac{x^2}{2}dx+C=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$
at $x=2,2y={\log }_{e}4-1$
So,
$\ln 4-1=\frac{4}{2}\ln 2-\frac{4}{4}+C$
$\Rightarrow C=0$
Therefore,
$x.y=\frac{x^2}{2}\ln x-\frac{x^2}{4}$
at $x=e$
$\Rightarrow ey=\frac{e^2}{2}\ln e-\frac{e^2}{4}$
$\Rightarrow y=\frac{e}{4}$