Question

Let $y=y(x)$ be the solution of the differential equation $x \log _e x \frac{d y}{d x}+y=x^2 \log _e x,(x >1)$. 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{d y}{d x}+y=x^2 \log _e x,(x>1)$
$ \Rightarrow \frac{d y}{d x}+\frac{y}{x \ln x}=x$
Linear differential equation
$\text { I.F. }=e^{\int \frac{1}{x \ln x} d x}=|\ln x|$
$\therefore$ Solution of differential equation
$y|\ln x| =\int x|\ln x| d x$
$ =|\ln x| \frac{x^2}{2}-\int \frac{1}{x} \cdot \frac{x^2}{2} d x$
$\Rightarrow y|\ln x| =|\ln x|\left(\frac{x^2}{2}\right)-\frac{x^2}{4}+c$
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, $y(e)=\frac{ e ^2}{2}-\frac{ e ^2}{4}+1=1+\frac{ e ^2}{4}$