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

2

The correct answer is option (1) : 2

We have,

$(x\, log\, x) \frac{dy}{dx}+y =2x \, log \, x$

$⇒\frac{dy}{dx}+\frac{1}{x\, log \, x}y= 2 $..............(i)

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q.$

where $P=\frac{1}{xlogx}$ and $Q=2.$

I.F. $=e^{∫Pdx }=e^{∫\frac{1}{xlogx}dx}=e^{log(logx)}=log x$

Multiplying both sides of (i) by I.F = log x, we obtain

$log \, x \frac{dy}{dx}+\frac{1}{x}y = 2log x $

Integrating both sides with respect to x, we get

$y log x = ∫2log x dx + C$

$⇒y\, log x = 2\, x (log x - 1) + C$ ................(ii)

It is given that $x≥1$. Putting x = 1 in (ii), we obtain

$y ×0=2(0-1) + C⇒C=2$

Putting C+ 2 in (ii), we get

$y log x = 2x (log x-1) + 2$

Putting x= e , we get y(e) = 2.