JEE Main 2015 — Mathematics Calculus

Given, $(x\mathrm{log}x)\frac{dy}{dx}+y=2x\mathrm{log}x$(xlogx)dxdy​+y=2xlogx

$$\Rightarrow \frac{dy}{dx}+(\frac{1}{x\mathrm{log}x})y=2$$⇒dxdy​+(xlogx1​)y=2

Integrating factor $\displaystyle I.F={e}^{\int Pdx}$I.F=e∫Pdx$\displaystyle ={e}^{\int \frac{1}{x\mathrm{log}x}dx}={e}^{\int \frac{1/x}{\mathrm{log}x}dx}={e}^{\mathrm{log}(\mathrm{log}x)}$=e∫xlogx1​dx=e∫logx1/x​dx=elog(logx)$=\mathrm{log}x$=logx

The solution of the equation becomes

$\displaystyle y\cdot {e}^{\int Pdx}=\int Q\cdot {e}^{\int Pdx}dx+c$y⋅e∫Pdx=∫Q⋅e∫Pdxdx+c, where, $c$c is the constant of integration.

$$\Rightarrow y\mathrm{log x}=\int 2\mathrm{log x}dx+c$$⇒ylogx=∫2logxdx+c

$$\Rightarrow y\mathrm{log x}=2(\mathrm{x log x}-x)+c$$⇒ylogx=2(xlogx−x)+c

Let, $P(1,{y}_{1})$P(1,y1​) be any point on the curve.

$\Rightarrow {y}_{1}\cdot (0)=2(0-1)+c$⇒y1​⋅(0)=2(0−1)+c$\Rightarrow c=2$⇒c=2

$$\Rightarrow y\mathrm{log}x=2(x\mathrm{log}x-x)+2$$⇒ylogx=2(xlogx−x)+2

Now, putting $x=e$x=e, we get, $y=2$y=2.