If frac{dy}{dx} + frac{3}{cos^2 x} y = frac{1 | vedclass

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(A) The given differential equation is $\frac{dy}{dx} + (3 \sec^2 x) y = \sec^2 x$.This is a linear differential equation of the form $\frac{dy}{dx} +

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$\frac{1}{3} + e^6$

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(A) The given differential equation is $\frac{dy}{dx} + (3 \sec^2 x) y = \sec^2 x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$,where $P(x) = 3 \sec^2 x$ and $Q(x) = \sec^2 x$.The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int 3 \sec^2 x dx} = e^{3 	an x}$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \cdot e^{3 	an x} = \int \sec^2 x \cdot e^{3 	an x} dx + C$.Let $u = 3 	an x$,then $du = 3 \sec^2 x dx$,so $\sec^2 x dx = \frac{du}{3}$.$y \cdot e^{3 	an x} = \int e^u \cdot \frac{du}{3} + C = \frac{1}{3} e^{3 	an x} + C$.Dividing by $e^{3 	an x}$,we get $y = \frac{1}{3} + C e^{-3 	an x}$.Given $y\left( \frac{\pi}{4} ight) = \frac{4}{3}$,we have $\frac{4}{3} = \frac{1}{3} + C e^{-3 	an(\pi/4)} = \frac{1}{3} + C e^{-3}$.$1 = C e^{-3} \Rightarrow C = e^3$.Thus,$y(x) = \frac{1}{3} + e^3 \cdot e^{-3 	an x} = \frac{1}{3} + e^{3 - 3 	an x}$.For $x = -\frac{\pi}{4}$,$y\left( -\frac{\pi}{4} ight) = \frac{1}{3} + e^{3 - 3 	an(-\pi/4)} = \frac{1}{3} + e^{3 - 3(-1)} = \frac{1}{3} + e^6$.

If $\frac{dy}{dx} + \frac{3}{\cos^2 x} y = \frac{1}{\cos^2 x}$,$x \in \left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ and $y\left( \frac{\pi}{4} \right) = \frac{4}{3}$,then $y\left( -\frac{\pi}{4} \right)$ equals

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