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

We seek a solution of the differential equation $$x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y$$ satisfying the initial condition $$y(2) = 0$$, and we wish to determine $$\tan(y(1)).$$

Dividing both sides by $$\cos^2 y$$ gives

$$x\frac{1}{\cos^2 y}\frac{dy}{dx} - \frac{\sin 2y}{\cos^2 y} = x^3(2 - x^3),$$

that is,

$$x\sec^2 y \frac{dy}{dx} - \frac{2\sin y\cos y}{\cos^2 y} = x^3(2 - x^3),$$

which simplifies to

$$x\sec^2 y \frac{dy}{dx} - 2\tan y = x^3(2 - x^3).$$

Introduce the substitution $$v = \tan y$$ so that $$\frac{dv}{dx} = \sec^2 y \frac{dy}{dx}.$$ The equation then becomes

$$xv' - 2v = x^3(2 - x^3),$$

or equivalently

$$v' - \frac{2}{x}v = x^2(2 - x^3).$$

This is a first‐order linear ODE with integrating factor $$e^{\int -2/x\,dx} = x^{-2}.$$ Multiplying through by $$x^{-2}$$ yields

$$\frac{d}{dx}\Bigl(\frac{v}{x^2}\Bigr) = \frac{x^2(2 - x^3)}{x^2} = 2 - x^3,$$

and integrating both sides gives

$$\frac{v}{x^2} = 2x - \frac{x^4}{4} + C,\quad\text{so}\quad v = 2x^3 - \frac{x^6}{4} + Cx^2.$$

Using the initial condition $$y(2)=0$$ implies $$v(2)=\tan0=0$$, hence

$$0 = 2\cdot8 - \frac{64}{4} + 4C = 16 - 16 + 4C = 4C\quad\Longrightarrow\quad C=0.$$

Thus

$$\tan y = v = 2x^3 - \frac{x^6}{4}.$$

Finally, at $$x=1$$ we obtain

$$\tan(y(1)) = 2\cdot1^3 - \frac{1^6}{4} = 2 - \frac{1}{4} = \frac{7}{4}.$$

Therefore, the required value is $$\tan(y(1)) = \frac{7}{4}.$$