JEE-Mains 2015 (17/30)

It can be assumed that $\log(x)$ refers to the natural logarithm.

For the moment we will just look at the case where $x \gt 1.$ For this domain, we can divide through by $x\log(x)$ without worrying about division by zero. The linear DE then becomes

$\dfrac{dy}{dx} + \dfrac{1}{x\log(x)}y = 2.$

We then have an integrating factor of

$\Large e^{\int \frac{1}{x\log(x)} dx}$ $= e^{\log(\log(x))} = \log(x),$

where the integral was solved using the substitution $u = \log(x).$ Multiplying through by $\log(x)$ gives us the equation

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

$\Longrightarrow \dfrac{d}{dx}(y\log(x)) = 2\log(x)$

$\Longrightarrow y\log(x) = 2x(\log(x) - 1) + C \Longrightarrow y = 2x - \dfrac{2x - C}{\log(x)},$

for some constant $C,$ where the integration of $\log(x)$ was done by parts, (see comments for explanation).

Now for this solution to be valid as $x \rightarrow 1$ we will require that $(2x - C) \rightarrow 0$ at this value to deal with the division by zero issue. This gives us $C = 2,$ and thus the general solution

$y = 2x - \dfrac{2x - 2}{\log(x)},$ for which $y(e) = 2e - \dfrac{2e - 2}{1} = \boxed{2}.$

Comments: Note that $\lim_{x \rightarrow 1} \left( 2x - \dfrac{2x - 2}{\log(x)}\right) = \lim_{x \rightarrow 1} \left(2 - \dfrac{2}{\frac{1}{x}}\right) = 2 - 2 = 0,$

where L'Hopital's rule was used to evaluate the limit of the fractional term. Thus our choice of $C = 2$ gives us a unique function $y(x)$ defined for $x \ge 1,$ for any other choice of $C$ would have made $y$ indeterminate at $x = 1.$

For the integration by parts solution, let $u = \log(x)$ and $dv = dx.$ Then $du = \dfrac{1}{x} dx$ and $v = x,$ and so

$\displaystyle\int \log(x) dx = x\log(x) - \int x*\dfrac{1}{x} dx = x(\log(x) - 1) + C.$