Question

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

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

Answer 1

Answer: (A)

$x{\log }_{e}x\frac{dy}{dx}+y=x^2{\log }_{e}x,(x>1)$
$\Rightarrow \frac{dy}{dx}+\frac{y}{x\ln x}=x$
Linear differential equation
$\text{ I.F. }=e^{\int \frac{1}{x\ln x}dx}=|\ln x|$
$\therefore$ Solution of differential equation
$y|\ln x|=\int x|\ln x|dx$
$=|\ln x|\frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx$
$\Rightarrow y|\ln x|=|\ln x|\left(\frac{x^2}{2}\right)-\frac{x^2}{4}+c$
For constant
$y(2)=2\Rightarrow c=1$
So, $y(x)=\frac{x^2}{2}-\frac{x^2}{4|\ln x|}+\frac{1}{|\ln x|}$
Hence, $y(e)=\frac{e^2}{2}-\frac{e^2}{4}+1=1+\frac{e^2}{4}$