JEE Main 2017 — Mathematics Calculus

We have,

$$(2+\mathrm{sin}x)\frac{dy}{dx}=-(y+1)\mathrm{cos}x$$(2+sinx)dxdy​=−(y+1)cosx

$$\Rightarrow \int \frac{dy}{(y+1)}=-\int \frac{\mathrm{cos}x}{\mathrm{sin}x+2}dx$$⇒∫(y+1)dy​=−∫sinx+2cosx​dx

Let, $\mathrm{sin}x=t$sinx=t$\Rightarrow \mathrm{cos}xdx=dt$⇒cosxdx=dt

$$\Rightarrow \mathrm{log}|y+1|=-\int \frac{dt}{t+2}+C$$⇒log∣y+1∣=−∫t+2dt​+C

where, $C$C is the constant of integration.

$$\Rightarrow \mathrm{log}|y+1|+\mathrm{log}|\mathrm{sin}x+2|-C=0$$⇒log∣y+1∣+log∣sinx+2∣−C=0

$$\Rightarrow \mathrm{log}(y+1)(\mathrm{sin}x+2)=C$$⇒log(y+1)(sinx+2)=C

Given, $y(0)=1$y(0)=1

$\Rightarrow C=\mathrm{log}4$⇒C=log4.

$$\Rightarrow \mathrm{log}(y+1)(\mathrm{sin}x+2)=\mathrm{log}4.....(i)$$⇒log(y+1)(sinx+2)=log4.....(i)

Now, put $x=\frac{\pi }{2}$x=2π​ in the equation $(i)$(i), we get, $\mathrm{log}(y+1)+\mathrm{log}(3)-\mathrm{log}(4)=0$log(y+1)+log(3)−log(4)=0

$$\Rightarrow \mathrm{log}(y+1)=\mathrm{log}(\frac{4}{3})$$⇒log(y+1)=log(34​)

$$\Rightarrow y+1=\frac{4}{3}$$⇒y+1=34​

$\Rightarrow y=\frac{4}{3}-1=\frac{1}{3}$⇒y=34​−1=31​.