JEE Main 2019 — Mathematics Calculus

The given $D.E.$D.E. can be written as$\frac{dy}{dx}-y\mathrm{tan}x=6x\mathrm{sec}x...(i),$dxdy​−ytanx=6xsecx...(i), which is a liner differential equation of the form $\frac{dy}{dx}+Py=Q,$dxdy​+Py=Q, where P&Q are the function of $x$x or constants.

Here, P=-\mathrm{tan}x&Q=6xsecx

Integrating factor $\displaystyle ={e}^{\int Pdx}={e}^{-\int \mathrm{tan}x dx}$=e∫Pdx=e−∫tanxdx

$$={e}^{-(-\mathrm{ln}(\mathrm{cos}x))}={e}^{\mathrm{ln}(\mathrm{cos}x)}=\mathrm{cos}x$$=e−(−ln(cosx))=eln(cosx)=cosx

And, the solution of the linear differential equation is,

$$y(I.F.)=\int Q(I.F.)dx+c$$y(I.F.)=∫Q(I.F.)dx+c

$$\Rightarrow y\cdot \mathrm{cos}x=\int (6x\mathrm{sec}x)\cdot \mathrm{cos}xdx$$⇒y⋅cosx=∫(6xsecx)⋅cosxdx

$$\Rightarrow y\cdot \mathrm{cos}x=\int 6xdx$$⇒y⋅cosx=∫6xdx

$$\Rightarrow y\cdot \mathrm{cos}x=3{x}^{2}+c...(ii)$$⇒y⋅cosx=3x2+c...(ii)

Given $y(\frac{\pi }{3})=0,$y(3π​)=0, hence

$$0=3{(\frac{\pi }{3})}^{2}+c$$0=3(3π​)2+c

$$\Rightarrow c=-\frac{{\pi }^{2}}{3}$$⇒c=−3π2​

Thus, the curve is, $y\cdot \mathrm{cos}x=3{x}^{2}-\frac{{\pi }^{2}}{3}$y⋅cosx=3x2−3π2​

Now, put $x=\frac{\pi }{6},$x=6π​, to get

$$y(\frac{\pi }{6})\cdot \mathrm{cos}(\frac{\pi }{6})=3{(\frac{\pi }{6})}^{2}-\frac{{\pi }^{2}}{3}$$y(6π​)⋅cos(6π​)=3(6π​)2−3π2​

$$\Rightarrow y(\frac{\pi }{6})\cdot \frac{\sqrt{3}}{2}=3\cdot \frac{{\pi }^{2}}{36}-\frac{{\pi }^{2}}{3}$$⇒y(6π​)⋅23​​=3⋅36π2​−3π2​

$$\Rightarrow y(\frac{\pi }{6})\cdot \frac{\sqrt{3}}{2}=\frac{{\pi }^{2}}{12}-\frac{{\pi }^{2}}{3}$$⇒y(6π​)⋅23​​=12π2​−3π2​

$$\Rightarrow y(\frac{\pi }{6})\cdot \frac{\sqrt{3}}{2}=-\frac{{\pi }^{2}}{4}$$⇒y(6π​)⋅23​​=−4π2​

$$\Rightarrow y(\frac{\pi }{6})=-\frac{{\pi }^{2}}{2\sqrt{3}}.$$⇒y(6π​)=−23​π2​.