JEE Main 2023 — Mathematics Calculus

Given,

$$(x\mathrm{cos}x)dy+(xy\mathrm{sin}x+y\mathrm{cos}x-1)dx=0$$(xcosx)dy+(xysinx+ycosx−1)dx=0

$$\Rightarrow (x\mathrm{cos}x)\frac{dy}{dx}+(yx\mathrm{sin}x+y\mathrm{cos}x)=1$$⇒(xcosx)dxdy​+(yxsinx+ycosx)=1

$$\Rightarrow \frac{dy}{dx}+y(\frac{x\mathrm{sin}x+\mathrm{cos}x}{x\mathrm{cos}x})=\frac{1}{x\mathrm{cos}x}$$⇒dxdy​+y(xcosxxsinx+cosx​)=xcosx1​

Which is a linear differential equation of form $\frac{dy}{dx}+P(x)y=Q(x)$dxdy​+P(x)y=Q(x)

Now we know that integrating factor will be $\displaystyle {e}^{\int Pdx}$e∫Pdx

$$\Rightarrow IF={e}^{\int (\mathrm{tan}x+\frac{1}{x})dx}={e}^{\mathrm{ln}|\mathrm{sec}x|+\mathrm{ln}x}=x\cdot \mathrm{sec}x$$⇒IF=e∫(tanx+x1​)dx=eln∣secx∣+lnx=x⋅secx

Hence, solution of linear differential equation is given by,

$$\Rightarrow y\cdot x\mathrm{sec}x=\int \frac{x\mathrm{sec}x}{x\mathrm{cos}x}dx$$⇒y⋅xsecx=∫xcosxxsecx​dx

$$\Rightarrow xy\mathrm{sec}x=\mathrm{tan}x+C$$⇒xysecx=tanx+C

Now using $\frac{\pi }{3}y(\frac{\pi }{3})=\sqrt{3}$3π​y(3π​)=3​ we get,

$$\Rightarrow \frac{\pi }{3}y(\frac{\pi }{3})\mathrm{sec}\frac{\pi }{3}=\mathrm{tan}\frac{\pi }{3}+C$$⇒3π​y(3π​)sec3π​=tan3π​+C

$$\Rightarrow C=\sqrt{3}$$⇒C=3​

Hence, $xy\mathrm{sec}x=\mathrm{tan}x+\sqrt{3}$xysecx=tanx+3​

$$\Rightarrow xy(x)=2\mathrm{sin}(x+\frac{\pi }{3})$$⇒xy(x)=2sin(x+3π​)

Now differentiating both side we get,

$$\Rightarrow x{y}^{'}(x)+y(x)=2\mathrm{cos}(x+\frac{\pi }{3})$$⇒xy′(x)+y(x)=2cos(x+3π​)

Now again differentiating we get,

$$\Rightarrow x{y}^{"}(x)+2{y}^{'}(x)=-2\mathrm{sin}(x+\frac{\pi }{3})$$⇒xy"(x)+2y′(x)=−2sin(x+3π​)

Thus $\frac{\pi }{6}{y}^{"}(\frac{\pi }{6})+2{y}^{'}(\frac{\pi }{6})=-2$6π​y"(6π​)+2y′(6π​)=−2

Hence $|\frac{\pi }{6}{y}^{"}(\frac{\pi }{6})+2{y}^{'}(\frac{\pi }{6})|=2$∣6π​y"(6π​)+2y′(6π​)∣=2