If $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}, x \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$, and

$y\left(\frac{\pi}{4}\right)=\frac{4}{3}$, then $y\left(-\frac{\pi}{4}\right)$ equals:


  1. (1) $\frac{1}{3}+e^{6}$

  2. (2) $\frac{1}{3}$

  3. (3) $-\frac{4}{3}$

  4. (4) $\frac{1}{3}+\mathrm{e}^{3}$

Correct Option: 1


Given, $\frac{d y}{d x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}$

$\frac{d y}{d x}=\sec ^{2} x(1-3 y)$

$\Rightarrow \int \frac{d y}{(1-3 y)}=\int \sec ^{2} x d x$

$\Rightarrow-\frac{1}{3} \ln |1-3 y|=\tan x+C$.......(1)

$\because \quad y\left(\frac{\pi}{4}\right)=\frac{4}{3}$ (Given)

$\Rightarrow-\frac{1}{3} \ln |1-4|=\tan \frac{\pi}{4}+C$

$\Rightarrow-\frac{1}{3} \ln 3=C+1 \Rightarrow C=-1-\frac{1}{3} \ln 3$

$\therefore$ in eq. (i), we get

$\frac{-1}{3} \ln |1-3 y|=\tan x-1-\frac{1}{3} \ln 3$

Put, $x=-\frac{\pi}{4}$

$\Rightarrow-\frac{1}{3} \ln |1-3 y|=\tan \left(-\frac{\pi}{4}\right)-1-\frac{1}{3} \ln 3$

$=-1-1-\frac{1}{3} \ln 3$

$\Rightarrow \ln |1-3 y|=6+\ln 3$

$\Rightarrow \ln \left|\frac{1}{3}-y\right|=6 \Rightarrow\left|\frac{1}{3}-y\right|=e^{6} \Rightarrow y=\frac{1}{3} \pm e^{6}$

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