Question

If $x \log x \frac{d y}{d x}+y=\log x^{2}$ and $y(e)=0$, then
$y\left(e^{2}\right)=$

• A. 0
• B. 1
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

We have,
$x \log x \frac{d y}{d x}+y=\log x^{2}$
$\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2 \log x}{x \log x}$
$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x}$
$I F=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}$
$IF =\log x$
Solution of different equation is
$ y \log x =\int \frac{2 \log x}{x} d x+c $
$ y \log x =(\log x)^{2}+c$
put $ x =e, y=0$ we get
$ 0 =1+c \Rightarrow c=-1 $
$ \therefore y \log x =(\log x)^{2}-1 $
put $x =e^{2} $
$ y \log e^{2} =\left(\log e^{2}\right)^{2}-1 $
$2 y =4-1 $
$ \Rightarrow y =\frac{3}{2} $
$ \therefore y\left(e^{2}\right) =\frac{3}{2} $