Differential Equation Transformation

The given differential equation is:

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

Since $x \neq 0$, we can divide by $\cos^2y$ (assuming $\cos y \neq 0$).

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

Using $\sec^2y = \frac{1}{\cos^2y}$ and $\sin 2y = 2\sin y \cos y$, we get:

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

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

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

Substitution for Linearization

Let $t = \tan y$. Then, differentiating with respect to $x$ gives:

$\frac{dt}{dx} = \sec^2y \frac{dy}{dx}$

Substituting $t$ and $\frac{dt}{dx}$ into the equation:

$x\frac{dt}{dx} - 2t = x^3(2-x^3)$

This is a linear first-order ordinary differential equation in $t$. Rearranging it:

$\frac{dt}{dx} - \frac{2}{x}t = x^2(2-x^3)$

Solving the Linear ODE

The integrating factor (IF) is:

$IF = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = e^{\ln(x^{-2})} = x^{-2}$

Multiply the linear ODE by the integrating factor:

$x^{-2}\frac{dt}{dx} - 2x^{-3}t = x^{-2} \cdot x^2(2-x^3)$

The left side is the derivative of the product $(t \cdot IF)$:

$\frac{d}{dx}(t x^{-2}) = 2 - x^3$

Integrate both sides with respect to $x$:

$\int \frac{d}{dx}(t x^{-2}) dx = \int (2 - x^3) dx$

$t x^{-2} = 2x - \frac{x^4}{4} + C$

Where $C$ is the constant of integration.

Solve for $t$:

$t = x^2 \left( 2x - \frac{x^4}{4} + C \right) = 2x^3 - \frac{x^6}{4} + C x^2$

Applying the Initial Condition

Substitute back $t = \tan y$:

$\tan y = 2x^3 - \frac{x^6}{4} + C x^2$

We are given the condition $y(2) = 0$. Substitute $x=2$ and $y=0$:

$\tan(0) = 2(2^3) - \frac{2^6}{4} + C (2^2)$

$0 = 2(8) - \frac{64}{4} + C(4)$

$0 = 16 - 16 + 4C$

$0 = 4C \implies C = 0$

Finding the Final Value

The specific solution is:

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

We need to find $\tan(y(1))$. Substitute $x=1$ into the specific solution:

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

$\tan(y(1)) = 2(1) - \frac{1}{4}$

$\tan(y(1)) = 2 - \frac{1}{4}$

$\tan(y(1)) = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}$