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

Answer 1

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

$\frac{dy}{dx}+\frac{y}{x}=\ell nx$
Integration factor $e^{\int \frac{1}{x}\mathrm{dx}}=x$
$xy=\int x\ell nxdx+C$

$=\int x\ell nx=\ell n\times \frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}=\ell nx\times \frac{x^2}{2}-\frac{x^2}{4}+C$
solving initial values
$xy=\frac{x^2}{2}\ell nx-\frac{x^2}{4}+C,$ for $2y\left(2\right)=2\ell n2-1$ $\Rightarrow C=0$
then
$y=\frac{x}{2}\ell nx-\frac{x}{4}$
$y\left(e\right)=\frac{e}{4}$