Question

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