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(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = e^{2x}$.Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{e^{2x}

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$e^{\pi/3} - e^\pi$

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(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = e^{2x}$.Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{e^{2x}}{\sin x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \cot x$ and $Q(x) = \frac{e^{2x}}{\sin x}$.The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x$.Multiplying the equation by the $I$.$F$.,we get $\frac{d}{dx}(y \sin x) = e^{2x}$.Integrating both sides with respect to $x$,we get $y \sin x = \int e^{2x} dx = \frac{e^{2x}}{2} + C$.Thus,$y = \frac{e^{2x}}{2 \sin x} + \frac{C}{\sin x}$.Using the condition $y\left(\frac{\pi}{2}\right) = 0$,we have $0 = \frac{e^\pi}{2(1)} + \frac{C}{1}$,which gives $C = -\frac{e^\pi}{2}$.Substituting $C$ back,$y = \frac{e^{2x} - e^\pi}{2 \sin x}$.Now,for $x = \frac{\pi}{6}$,$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.Therefore,$y\left(\frac{\pi}{6}\right) = \frac{e^{\pi/3} - e^\pi}{2(1/2)} = e^{\pi/3} - e^\pi$.

If the solution $y(x)$ of the differential equation $\sin x \frac{dy}{dx} + y \cos x = e^{2x}, x \in (0, \pi)$ satisfies $y\left(\frac{\pi}{2}\right) = 0$,then $y\left(\frac{\pi}{6}\right) = $

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