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(B) Given the differential equation: $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y$.Dividing by $x\cos^{2}y$, we get: $\sec^{2}y\frac{dy}{dx}-\frac

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$\frac{7}{4}$

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(B) Given the differential equation: $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y$.Dividing by $x\cos^{2}y$, we get: $\sec^{2}y\frac{dy}{dx}-\frac{\sin(2y)}{x\cos^{2}y}=x^{2}(2-x^{3})$.Since $\sin(2y)=2\sin y\cos y$, the equation becomes: $\sec^{2}y\frac{dy}{dx}-\frac{2	an y}{x}=x^{2}(2-x^{3})$.Let $	an y=t$, then $\sec^{2}y\frac{dy}{dx}=\frac{dt}{dx}$.The equation becomes a linear differential equation: $\frac{dt}{dx}-\frac{2}{x}t=x^{2}(2-x^{3})$.The integrating factor is $I.F. = e^{\int-\frac{2}{x}dx} = e^{-2\ln x} = \frac{1}{x^{2}}$.Multiplying by the $I.F.$, we get: $\frac{d}{dx}(\frac{t}{x^{2}}) = 2-x^{3}$.Integrating both sides: $\frac{t}{x^{2}} = \int(2-x^{3})dx = 2x-\frac{x^{4}}{4}+C$.Substituting $t=	an y$: $\frac{	an y}{x^{2}} = 2x-\frac{x^{4}}{4}+C$.Given $y(2)=0$, so $	an(0)=0$: $\frac{0}{4} = 2(2)-\frac{16}{4}+C \Rightarrow 0 = 4-4+C \Rightarrow C=0$.Thus, $	an y = 2x^{3}-\frac{x^{6}}{4}$.For $x=1$, $	an(y(1)) = 2(1)^{3}-\frac{1^{6}}{4} = 2-\frac{1}{4} = \frac{7}{4}$.

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y,$ for $x\ne0.$ If $y(2)=0,$ then $\tan(y(1))$ is equal to

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