JEE Main 2014 — Mathematics Calculus

The given differential equation $\frac{dy}{dx}+y\mathrm{tan}x=\mathrm{sin}2x$dxdy​+ytanx=sin2x is linear differential equation of the form

$\frac{dy}{dx}+P(x)y=Q(x)$dxdy​+P(x)y=Q(x) with $P(x)=\mathrm{tan}x$P(x)=tanx, $Q(x)=\mathrm{sin}2x$Q(x)=sin2x

Now, $\displaystyle I.F.={e}^{\int P(x)dx}={e}^{\int \mathrm{tan}xdx}={e}^{\mathrm{ln}(\mathrm{sec}x)}=\mathrm{sec}x$I.F.=e∫P(x)dx=e∫tanxdx=eln(secx)=secx

And, the solution is $\displaystyle y(I.F.)=\int Q(x)(I.F.)dx+c$y(I.F.)=∫Q(x)(I.F.)dx+c

$$\Rightarrow y\mathrm{sec}x=\int \mathrm{sec}x\mathrm{sin}2xdx+c$$⇒ysecx=∫secxsin2xdx+c

$$\Rightarrow y\mathrm{sec}x=\int \frac{1}{\mathrm{cos}x}(2\mathrm{sin}x\mathrm{cos}x)dx+c$$⇒ysecx=∫cosx1​(2sinxcosx)dx+c

$$\Rightarrow y\mathrm{sec}x=\int 2\mathrm{sin}xdx+c$$⇒ysecx=∫2sinxdx+c

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

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

$$\Rightarrow 1\cdot \mathrm{sec}0=-2\mathrm{cos}0+c$$⇒1⋅sec0=−2cos0+c

$$\Rightarrow 1=-2+c$$⇒1=−2+c

$$\Rightarrow c=3$$⇒c=3

$$\Rightarrow y\mathrm{sec}x=-2\mathrm{cos}x+3$$⇒ysecx=−2cosx+3

Now, at $x=\pi$x=π

$$y(\pi )\mathrm{sec}\pi =-2\mathrm{cos}\pi +3$$y(π)secπ=−2cosπ+3

$$\Rightarrow y(\pi )(-1)=-2(-1)+3$$⇒y(π)(−1)=−2(−1)+3

$$\Rightarrow y(\pi )=-5.$$⇒y(π)=−5.