JEE Main 2021 — Mathematics Calculus

Given,

$$\mathrm{sec}y\frac{dy}{dx}-\mathrm{sin}(x+y)-\mathrm{sin}(x-y)=0$$secydxdy​−sin(x+y)−sin(x−y)=0

$$\Rightarrow \mathrm{sec}y\frac{dy}{dx}=2\mathrm{sin}x\mathrm{cos}y$$⇒secydxdy​=2sinxcosy

$$\Rightarrow {\mathrm{sec}}^{2}ydy=2\mathrm{sin}xdx$$⇒sec2ydy=2sinxdx

$$\Rightarrow \int {\mathrm{sec}}^{2}ydy=\int 2\mathrm{sin}xdx$$⇒∫sec2ydy=∫2sinxdx

$$\Rightarrow \mathrm{tan}y=-2\mathrm{cos}x+c$$⇒tany=−2cosx+c

Given, $y(0)=0$y(0)=0

So, $0=-2\mathrm{cos}0+c$0=−2cos0+c

$$\Rightarrow c=2$$⇒c=2

So, the particular solution is

$$\mathrm{tan}y=-2\mathrm{cos}x+2$$tany=−2cosx+2

at $x=\frac{\pi }{2},\mathrm{tan}y=-2\mathrm{cos}\frac{\pi }{2}+2$x=2π​,tany=−2cos2π​+2

$$\Rightarrow \mathrm{tan}y=2\Rightarrow {\mathrm{sec}}^{2}y=5$$⇒tany=2⇒sec2y=5

Here, differentiate both sides of $\mathrm{tan}y=-2\mathrm{cos}x+2,$tany=−2cosx+2, we get,

$${\mathrm{sec}}^{2}y\frac{dy}{dx}=2\mathrm{sin}x$$sec2ydxdy​=2sinx

At $x=\frac{\pi }{2}$x=2π​

$$5\frac{dy}{dx}=2\mathrm{sin}(\frac{\pi }{2})$$5dxdy​=2sin(2π​)

$$\Rightarrow 5\frac{dy}{dx}=2$$⇒5dxdy​=2