JEE Main 2019 — Mathematics Calculus

Given $\frac{dy}{dx}+y\mathrm{tan}x=2x+{x}^{2}\mathrm{tan}x$dxdy​+ytanx=2x+x2tanx

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$dxdy​+Py=Q, where P=\mathrm{tan}x&Q=2x+{x}^{2}\mathrm{tan}x

Now, the integrating factor $\displaystyle I.F.={e}^{\int Pdx}={e}^{\int \mathrm{tan}x dx}$I.F.=e∫Pdx=e∫tanxdx

$$={e}^{\mathrm{lnsec}x}=\mathrm{sec}x$$=elnsecx=secx

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

$$y\cdot \mathrm{sec}x=\int (2x+{x}^{2}\mathrm{tan}x)\mathrm{sec}xdx+c$$y⋅secx=∫(2x+x2tanx)secxdx+c

$$\Rightarrow ysecx=\int 2xsec xdx+\int {x}^{2}(\mathrm{sec}x\cdot \mathrm{tan}x)dx+c$$⇒ysecx=∫2xsecxdx+∫x2(secx⋅tanx)dx+c

Using integration by parts in the second integral, we get

$$\Rightarrow ysecx=\int 2x\mathrm{sec} xdx+{x}^{2}\int (\mathrm{sec}x\cdot \mathrm{tan}x)dx-\int (\frac{d}{dx}({x}^{2})\int (\mathrm{sec}x\cdot \mathrm{tan}x))dx+c$$⇒ysecx=∫2xsecxdx+x2∫(secx⋅tanx)dx−∫(dxd​(x2)∫(secx⋅tanx))dx+c

$$\Rightarrow ysecx=\int 2x\mathrm{sec} xdx+{x}^{2}\int secxdx-\int 2xsecxdx+c$$⇒ysecx=∫2xsecxdx+x2∫secxdx−∫2xsecxdx+c

$$\Rightarrow ysecx={x}^{2}secx+c$$⇒ysecx=x2secx+c

$$\Rightarrow y={x}^{2}+c\cdot \mathrm{cos}x$$⇒y=x2+c⋅cosx

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

$$\Rightarrow 0+c=1\Rightarrow c=1$$⇒0+c=1⇒c=1

So, $y={x}^{2}+cosx$y=x2+cosx

Hence, $y(\frac{\pi }{4})=\frac{{\pi }^{2}}{16}+\frac{1}{\sqrt{2}}$y(4π​)=16π2​+2​1​ and $y(-\frac{\pi }{4})=\frac{{\pi }^{2}}{16}+\frac{1}{\sqrt{2}}$y(−4π​)=16π2​+2​1​

Differentiating $y(x)$y(x) with respect to $x$x

$$\Rightarrow {y}^{'}(x)=2x-sinx$$⇒y′(x)=2x−sinx

Hence, ${y}^{'}(\frac{\pi }{4})=\frac{\pi }{2}-\frac{1}{\sqrt{2}}$y′(4π​)=2π​−2​1​ and ${y}^{'}(-\frac{\pi }{4})=-\frac{\pi }{2}+\frac{1}{\sqrt{2}}$y′(−4π​)=−2π​+2​1​

$$\Rightarrow {y}^{'}(\frac{\pi }{4})-{y}^{'}(-\frac{\pi }{4})=\pi -\sqrt{2}.$$⇒y′(4π​)−y′(−4π​)=π−2​.