JEE Main 2021 — Mathematics Calculus

Given:

$$(y+1){\mathrm{tan}}^{2}xdx+\mathrm{tan}xdy+ydx=0$$(y+1)tan2xdx+tanxdy+ydx=0

$$\Rightarrow [(y+1){\mathrm{tan}}^{2}x+y]dx+\mathrm{tan}xdy=0$$⇒[(y+1)tan2x+y]dx+tanxdy=0

$$\Rightarrow \mathrm{tan}x(\frac{dy}{dx})+(y+1){\mathrm{tan}}^{2}x+y=0$$⇒tanx(dxdy​)+(y+1)tan2x+y=0

$$\Rightarrow \frac{dy}{dx}+(1+y)\mathrm{tan}x=-y\mathrm{cot}x$$⇒dxdy​+(1+y)tanx=−ycotx

$$\Rightarrow \frac{dy}{dx}+y(\mathrm{tan}x+\mathrm{cot}x)=-\mathrm{tan}x$$⇒dxdy​+y(tanx+cotx)=−tanx

This is a linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$dxdy​+P(x)y=Q(x).

$$\text{ I.F }={e}^{\int (\mathrm{tan}x+\mathrm{cot}x)dx}$$ I.F =e∫(tanx+cotx)dx

$$={e}^{\int (\frac{{\mathrm{tan}}^{2}x+1}{\mathrm{tan}x})dx}$$=e∫(tanxtan2x+1​)dx

$$={e}^{\int (\frac{{\mathrm{sec}}^{2}x}{\mathrm{tan}x})dx}$$=e∫(tanxsec2x​)dx

$$={e}^{{\mathrm{log}}_{e}|\mathrm{tan}x|}$$=eloge​∣tanx∣

$$=\mathrm{tan}x(\because x\in (0,\frac{\pi }{2}))$$=tanx(∵x∈(0,2π​))

Solution is

$$y\mathrm{tan}x=\int -{\mathrm{tan}}^{2}xdx+c$$ytanx=∫−tan2xdx+c

$$\Rightarrow y\mathrm{tan}x=\int (1-{\mathrm{sec}}^{2}x)dx+c$$⇒ytanx=∫(1−sec2x)dx+c

$$\Rightarrow y\mathrm{tan}x=x-\mathrm{tan}x+c$$⇒ytanx=x−tanx+c

Now,

$$\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}xy=1$$x→0+lim​xy=1

$$\Rightarrow \underset{x\rightarrow {0}^{+}}{\mathrm{lim}}(\frac{x}{\mathrm{tan}x})(x-\mathrm{tan}x+c)=1$$⇒x→0+lim​(tanxx​)(x−tanx+c)=1

$$\Rightarrow 1(0-0+c)=1\Rightarrow c=1$$⇒1(0−0+c)=1⇒c=1

Then the function is

$$y\mathrm{tan}x=x-\mathrm{tan}x+1$$ytanx=x−tanx+1

Put $x=\frac{\pi }{4}$x=4π​,

$$y(\frac{\pi }{4})\mathrm{tan}(\frac{\pi }{4})=(\frac{\pi }{4})-\mathrm{tan}(\frac{\pi }{4})+1$$y(4π​)tan(4π​)=(4π​)−tan(4π​)+1

$$\Rightarrow y(\frac{\pi }{4})=\frac{\pi }{4}$$⇒y(4π​)=4π​