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(B) The given differential equation is $\frac{dy}{dx} + 3(	an^2 x + 1)y = \sec^2 x$.Since $1 + 	an^2 x = \sec^2 x$,the equation simplifies to $\frac

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$\frac{4}{3}$

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(B) The given differential equation is $\frac{dy}{dx} + 3(	an^2 x + 1)y = \sec^2 x$.Since $1 + 	an^2 x = \sec^2 x$,the equation simplifies to $\frac{dy}{dx} + 3(\sec^2 x)y = \sec^2 x$.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 $\sec^2 x dx = \frac{du}{3}$.$y \cdot e^{3	an x} = \int e^u \frac{du}{3} + C = \frac{1}{3}e^{3	an x} + C$.Given $y(0) = \frac{1}{3} + e^3$,at $x=0$,$	an(0)=0$,so $y(0) \cdot e^0 = \frac{1}{3}e^0 + C \Rightarrow \frac{1}{3} + e^3 = \frac{1}{3} + C \Rightarrow C = e^3$.Thus,$y \cdot e^{3	an x} = \frac{1}{3}e^{3	an x} + e^3$.At $x = \frac{\pi}{4}$,$	an\left(\frac{\pi}{4}ight) = 1$.$y\left(\frac{\pi}{4}ight) \cdot e^3 = \frac{1}{3}e^3 + e^3 = \frac{4}{3}e^3$.Therefore,$y\left(\frac{\pi}{4}ight) = \frac{4}{3}$.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$,with the initial condition $y(0) = \frac{1}{3} + e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to:

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