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(D) Given the differential equation: $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{	an x}{2y}$.Multiply by $2y$: $2y \frac{dy}{dx} + y^2 \sec x = 	an

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$y^2 = 1 - \frac{x}{\sec x + 	an x}$

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(D) Given the differential equation: $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{	an x}{2y}$.Multiply by $2y$: $2y \frac{dy}{dx} + y^2 \sec x = 	an x$.Let $y^2 = t$,then $2y \frac{dy}{dx} = \frac{dt}{dx}$.The equation becomes: $\frac{dt}{dx} + t \sec x = 	an x$.This is a linear differential equation of the form $\frac{dt}{dx} + Pt = Q$,where $P = \sec x$ and $Q = 	an x$.Integrating factor $IF = e^{\int \sec x dx} = e^{\ln(\sec x + 	an x)} = \sec x + 	an x$.The solution is $t(IF) = \int Q(IF) dx + C$.$t(\sec x + 	an x) = \int 	an x(\sec x + 	an x) dx + C$.$t(\sec x + 	an x) = \int (\sec x 	an x + 	an^2 x) dx + C$.$t(\sec x + 	an x) = \int (\sec x 	an x + \sec^2 x - 1) dx + C$.$t(\sec x + 	an x) = \sec x + 	an x - x + C$.Given $y(0) = 1$,then $t(0) = 1^2 = 1$. At $x=0$,$\sec 0 = 1, 	an 0 = 0$.$1(1 + 0) = 1 + 0 - 0 + C \Rightarrow C = 0$.Thus,$t(\sec x + 	an x) = \sec x + 	an x - x$.$t = 1 - \frac{x}{\sec x + 	an x}$.Since $t = y^2$,we have $y^2 = 1 - \frac{x}{\sec x + 	an x}$.

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{\tan x}{2y}$,where $0 \le x < \frac{\pi}{2}$,and $y(0) = 1$,is given by

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