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(B) The given differential equation is $(y+1) 	an ^{2} x \,dx+	an x \,dy+y \,dx=0$.Rearranging the terms,we get $	an x \,dy + (y 	an^2 x + y + 	a

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$\frac{\pi}{4}-1$

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(B) The given differential equation is $(y+1) 	an ^{2} x \,dx+	an x \,dy+y \,dx=0$.Rearranging the terms,we get $	an x \,dy + (y 	an^2 x + y + 	an^2 x) \,dx = 0$.Dividing by $	an x \,dx$,we get $\frac{dy}{dx} + \frac{y(	an^2 x + 1) + 	an^2 x}{	an x} = 0$.Since $1+	an^2 x = \sec^2 x$,we have $\frac{dy}{dx} + y \frac{\sec^2 x}{	an x} = -	an x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sec^2 x}{	an x}$ and $Q(x) = -	an x$.The integrating factor $IF = e^{\int \frac{\sec^2 x}{	an x} \,dx} = e^{\ln(	an x)} = 	an x$.The general solution is $y \cdot IF = \int Q(x) \cdot IF \,dx + C$.$y 	an x = \int -	an^2 x \,dx + C = \int (1 - \sec^2 x) \,dx + C = x - 	an x + C$.Thus,$y = \frac{x}{	an x} - 1 + \frac{C}{	an x}$.Given $\lim_{x 	o 0+} x y(x) = 1$,we have $\lim_{x 	o 0+} x \left( \frac{x}{	an x} - 1 + \frac{C}{	an x} ight) = 1$.Since $\lim_{x 	o 0} \frac{x}{	an x} = 1$,the limit becomes $1(1) - 0 + C(1) = 1$,which implies $1 + C = 1$,so $C = 0$.Thus,$y(x) = \frac{x}{	an x} - 1$.Evaluating at $x = \frac{\pi}{4}$,$y\left(\frac{\pi}{4}ight) = \frac{\pi/4}{1} - 1 = \frac{\pi}{4} - 1$.

Let $y=y(x)$ be a solution curve of the differential equation $(y+1) \tan ^{2} x \,dx+\tan x \,dy+y \,dx=0$ for $x \in \left(0, \frac{\pi}{2}\right)$. If $\lim _{x \rightarrow 0+} x y(x)=1$,then the value of $y\left(\frac{\pi}{4}\right)$ is:

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