JEE Main 2022 — Mathematics Calculus

Given, $({\mathrm{sin}}^{2}2x)\frac{dy}{dx}+(8{\mathrm{sin}}^{2}2x+2\mathrm{sin}4x)y=$(sin22x)dxdy​+(8sin22x+2sin4x)y=

$$2{e}^{-4x}(2\mathrm{sin}2x+\mathrm{cos}2x)$$2e−4x(2sin2x+cos2x)

Now rewriting the given differential equation we get,

$$\Rightarrow \frac{dy}{dx}+(8+4\mathrm{cot}2x)y=\frac{2{e}^{-4x}}{{\mathrm{sin}}^{2}2x}(2\mathrm{sin}x+\mathrm{cos}2x)$$⇒dxdy​+(8+4cot2x)y=sin22x2e−4x​(2sinx+cos2x)

Which is a linear differential equation.

Now calculating $\displaystyle IF={e}^{\int (8+4\mathrm{cot}2x)\mathrm{dx}}={e}^{8x+2\mathrm{ln}(\mathrm{sin}2x)}$IF=e∫(8+4cot2x)dx=e8x+2ln(sin2x)

$$={e}^{8x}\cdot {\mathrm{sin}}^{2}2x$$=e8x⋅sin22x

Now solution of differential equation is given by,

$$y\times IF=\int \frac{2{e}^{-4x}(2\mathrm{sin}2x+\mathrm{cos}2x)}{{\mathrm{sin}}^{2}2x}\times IFdx$$y×IF=∫sin22x2e−4x(2sin2x+cos2x)​×IFdx

$$\Rightarrow y({e}^{8x}\cdot {\mathrm{sin}}^{2}2x)=\int 2{e}^{4x}(2\mathrm{sin}2x+\mathrm{cos}2x)dx$$⇒y(e8x⋅sin22x)=∫2e4x(2sin2x+cos2x)dx

$$\Rightarrow y({e}^{8x}\cdot {\mathrm{sin}}^{2}2x)={e}^{4x}\cdot \mathrm{sin}2x+C$$⇒y(e8x⋅sin22x)=e4x⋅sin2x+C

Now given $y(\frac{\pi }{4})={e}^{-\pi }\Rightarrow C=0$y(4π​)=e−π⇒C=0

So, the equation of curve is given by $y=\frac{{e}^{-4x}}{\mathrm{sin}2x}$y=sin2xe−4x​

So, the value of $y(\frac{\pi }{6})=\frac{{e}^{-4\frac{\pi }{6}}}{\mathrm{sin}(2\cdot \frac{\pi }{6})}=\frac{2}{\sqrt{3}}{e}^{-\frac{2\pi }{3}}$y(6π​)=sin(2⋅6π​)e−46π​​=3​2​e−32π​