JEE Main 2019 — Mathematics Calculus

Given $\frac{dy}{dx}+(3{\mathrm{sec}}^{2}x)y={\mathrm{sec}}^{2}x$dxdy​+(3sec2x)y=sec2x

This is linear differential equation, of the type $\frac{dy}{dx}+Py=Q,$dxdy​+Py=Q, where P&Q are functions of $x$x or constants.

Here, P=3se{c}^{2}x&Q=se{c}^{2}x

Now, integrating factor $\displaystyle I.F.={e}^{\int Pdx}$I.F.=e∫Pdx

$$={e}^{\int 3{\mathrm{sec}}^{2}x dx}={e}^{3\mathrm{tan}x}$$=e∫3sec2xdx=e3tanx

Hence, the solution of the differential equation is $\displaystyle y(I.F.)=\int Q(I.F.)dx+c$y(I.F.)=∫Q(I.F.)dx+c

$$\Rightarrow y\cdot {e}^{3\mathrm{tan}x}=\int {e}^{3\mathrm{tan}x}\cdot {\mathrm{sec}}^{2}xdx$$⇒y⋅e3tanx=∫e3tanx⋅sec2xdx

Let, $\mathrm{tan}x=t,\Rightarrow se{c}^{2}xdx=dt$tanx=t,⇒sec2xdx=dt

$$\Rightarrow y\cdot {e}^{3\mathrm{tan}x}=\int {e}^{3t}dt+c$$⇒y⋅e3tanx=∫e3tdt+c

$$\Rightarrow y\cdot {e}^{3\mathrm{tan}x}=\frac{{e}^{3t}}{3}+c$$⇒y⋅e3tanx=3e3t​+c

$$\Rightarrow y\cdot {e}^{3\mathrm{tan}x}=\frac{{e}^{3\mathrm{tan}x}}{3}+c$$⇒y⋅e3tanx=3e3tanx​+c

$$\Rightarrow y=c{e}^{-3\mathrm{tan}x}+\frac{1}{3}$$⇒y=ce−3tanx+31​

Given, $y(\frac{\pi }{4})=\frac{4}{3}\Rightarrow$y(4π​)=34​⇒

$$\frac{4}{3}=c{e}^{-3}+\frac{1}{3}$$34​=ce−3+31​

$$\Rightarrow c={e}^{3}$$⇒c=e3

Thus, $y={e}^{3}\cdot {e}^{-3\mathrm{tan}x}+\frac{1}{3}$y=e3⋅e−3tanx+31​

Hence, $y(-\frac{\pi }{4})={e}^{3}\cdot {e}^{-3(\mathrm{tan}(-\frac{\pi }{4}))}+\frac{1}{3}$y(−4π​)=e3⋅e−3(tan(−4π​))+31​

$$\Rightarrow y(-\frac{\pi }{4})={e}^{3}\cdot {e}^{-3(-1)}+\frac{1}{3}$$⇒y(−4π​)=e3⋅e−3(−1)+31​

$$={e}^{3}\cdot {e}^{3}+\frac{1}{3}={e}^{6}+\frac{1}{3}.$$=e3⋅e3+31​=e6+31​.