Question

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

• A. e
• B. 0
• C. 2
• D. 2e

Answer 1

Answer: (C)

Given differential equation is
$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=2$
This is a linear differential equation.
$\therefore $ IF $=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x$
Now, the solution of given differential equation is given by
$y \cdot \log x=\int \log x \cdot 2 d x $
$\Rightarrow y \cdot \log x=2 \int \log x d x $
$\Rightarrow y \cdot \log x=2[x \log x-x]+c $
At $ x=1 \Rightarrow c=2 $
$\Rightarrow y \cdot \log x=2[x \log x-x]+2$
At $ x=e, y=2(e-e)+2$
$\Rightarrow y=2$