JEE Main 2021 — Mathematics Calculus

Given, ${cosec}^{2}xdy+2dx=(1+y\mathrm{cos}2x){cosec}^{2}xdx$cosec2xdy+2dx=(1+ycos2x)cosec2xdx

$$\Rightarrow \frac{dy}{{\mathrm{sin}}^{2}x}+2dx=(\frac{1+y\mathrm{cos}2x}{{\mathrm{sin}}^{2}x})dx$$⇒sin2xdy​+2dx=(sin2x1+ycos2x​)dx

$$\Rightarrow \frac{dy}{dx}+2{\mathrm{sin}}^{2}x=1+y\mathrm{cos}2x$$⇒dxdy​+2sin2x=1+ycos2x

$$\Rightarrow \frac{dy}{dx}+2{\mathrm{sin}}^{2}x-1=y\mathrm{cos}2x$$⇒dxdy​+2sin2x−1=ycos2x

Using, $\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x,$cos2x=1−2sin2x, we get

$$\frac{dy}{dx}+(-\mathrm{cos}2x)=y\mathrm{cos}2x$$dxdy​+(−cos2x)=ycos2x

$$\Rightarrow \frac{dy}{dx}+(-\mathrm{cos}2x)y=\mathrm{cos}2x$$⇒dxdy​+(−cos2x)y=cos2x

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$dxdy​+Py=Q, where P&Q are the functions of $x$x or constants.

Thus, P=-\mathrm{cos}2x&Q=\mathrm{cos}2x

Now, we have integrating factor I.F.$\displaystyle ={e}^{\int Pdx}={e}^{\int -cos2xdx}={e}^{-(\frac{\mathrm{sin}2x}{2})}$=e∫Pdx=e∫−cos2xdx=e−(2sin2x​)

And, the solution of the given differential equation is

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

$$\Rightarrow y\cdot {e}^{-(\frac{\mathrm{sin}2x}{2})}=\int (\mathrm{cos}2x)\cdot {e}^{-(\frac{\mathrm{sin}2x}{2})}dx+c$$⇒y⋅e−(2sin2x​)=∫(cos2x)⋅e−(2sin2x​)dx+c

Put $\frac{\mathrm{sin}2x}{2}=t,\Rightarrow \frac{2\mathrm{cos}2x}{2}dx=dt$2sin2x​=t,⇒22cos2x​dx=dt

$$\Rightarrow y\cdot {e}^{-(\frac{\mathrm{sin}2x}{2})}=\int {e}^{-t}dt+c$$⇒y⋅e−(2sin2x​)=∫e−tdt+c

$$\Rightarrow y\cdot {e}^{-(\frac{\mathrm{sin}2x}{2})}=-{e}^{-t}dt+c$$⇒y⋅e−(2sin2x​)=−e−tdt+c

$$\Rightarrow y\cdot {e}^{-(\frac{\mathrm{sin}2x}{2})}=-{e}^{-(\frac{\mathrm{sin}2x}{2})}+c$$⇒y⋅e−(2sin2x​)=−e−(2sin2x​)+c

Given, $y(\frac{\pi }{4})=0$y(4π​)=0

$$\Rightarrow 0\cdot {e}^{-(\frac{\mathrm{sin}(\frac{\pi }{2})}{2})}=-{e}^{-(\frac{\mathrm{sin}(\frac{\pi }{2})}{2})}+c$$⇒0⋅e−(2sin(2π​)​)=−e−(2sin(2π​)​)+c

$$\Rightarrow c=-{e}^{-(\frac{1}{2})}$$⇒c=−e−(21​)

$$\Rightarrow y\cdot {e}^{-(\frac{\mathrm{sin}2x}{2})}=-{e}^{-(\frac{\mathrm{sin}2x}{2})}+{e}^{-\frac{1}{2}}$$⇒y⋅e−(2sin2x​)=−e−(2sin2x​)+e−21​

Now, at $x=0$x=0

$$\Rightarrow y(0)\cdot {e}^{-(\frac{\mathrm{sin}0}{2})}=-{e}^{-(\frac{\mathrm{sin}0}{2})}+{e}^{-\frac{1}{2}}$$⇒y(0)⋅e−(2sin0​)=−e−(2sin0​)+e−21​

$$\Rightarrow y(0)\cdot {e}^{0}=-{e}^{0}+{e}^{-\frac{1}{2}}$$⇒y(0)⋅e0=−e0+e−21​

$$\Rightarrow y(0)=-1+{e}^{-\frac{1}{2}}$$⇒y(0)=−1+e−21​

$$\Rightarrow y(0)+1={e}^{-\frac{1}{2}}$$⇒y(0)+1=e−21​

$$\Rightarrow {(y(0)+1)}^{2}={e}^{-1}.$$⇒(y(0)+1)2=e−1.