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