JEE Main 2024 — Mathematics Calculus

Given: $y\frac{dx}{dy}=x({\mathrm{log}}_{e}x-{\mathrm{log}}_{e}y+1),x>0,y>0$ydydx​=x(loge​x−loge​y+1),x>0,y>0

$$\Rightarrow \frac{dx}{dy}=(\frac{x}{y})[{\mathrm{log}}_{e}(\frac{x}{y})+1]$$⇒dydx​=(yx​)[loge​(yx​)+1]

Putting, $x=vy$x=vy

$$\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dx}$$⇒dydx​=v+ydxdv​

$$\Rightarrow v+y\frac{dv}{dy}=(\frac{vy}{y})[{\mathrm{log}}_{e}(\frac{vy}{y})+1]$$⇒v+ydydv​=(yvy​)[loge​(yvy​)+1]

$$\Rightarrow v+y\frac{dv}{dy}=v[{\mathrm{log}}_{e}(v)+1]$$⇒v+ydydv​=v[loge​(v)+1]

$$\Rightarrow v+y\frac{dv}{dy}=v{\mathrm{log}}_{e}(v)+v$$⇒v+ydydv​=vloge​(v)+v

$$\Rightarrow y\frac{dv}{dy}=v{\mathrm{log}}_{e}(v)$$⇒ydydv​=vloge​(v)

$$\Rightarrow \frac{dv}{v{\mathrm{log}}_{e}v}=\frac{dy}{y}$$⇒vloge​vdv​=ydy​

$$\Rightarrow \int \frac{dv}{v{\mathrm{log}}_{e}v}=\int \frac{dy}{y}$$⇒∫vloge​vdv​=∫ydy​

Putting, ${\mathrm{log}}_{e}v=t$loge​v=t

$$\Rightarrow \frac{dv}{v}=dt$$⇒vdv​=dt

$$\Rightarrow \int \frac{dt}{t}=\int \frac{dy}{y}$$⇒∫tdt​=∫ydy​

$$\Rightarrow \mathrm{log}t=\mathrm{log}y+c$$⇒logt=logy+c

$$\Rightarrow \mathrm{log}({\mathrm{log}}_{e}\frac{x}{y})=\mathrm{log}y+c$$⇒log(loge​yx​)=logy+c

Using point $(e,1)$(e,1)

$$\Rightarrow \mathrm{log}({\mathrm{log}}_{e}e)=\mathrm{log}(1)+c$$⇒log(loge​e)=log(1)+c

$$\Rightarrow c=0$$⇒c=0

$$\Rightarrow \mathrm{log}({\mathrm{log}}_{e}\frac{x}{y})=\mathrm{log}y$$⇒log(loge​yx​)=logy

$$\Rightarrow |{\mathrm{log}}_{e}\frac{x}{y}|=y$$⇒∣loge​yx​∣=y