1. The problem gives the differential equation $$x y' = y + 2x^3 \sin^2\left(\frac{y}{x}\right).$$ 2. We need to find the solution or explore this equation step by step. 3. First, let's rewrite the equation to isolate $y'$: $$y' = \frac{y}{x} + 2x^2 \sin^2\left(\frac{y}{x}\right).$$ 4. Introduce the substitution $v = \frac{y}{x}$, hence $y = vx$. 5. Then, $y' = v + x v'$ by the product rule. 6. Substitute into the original equation: $$v + x v' = v + 2x^2 \sin^2(v).$$ 7. Cancel $v$ on both sides: $$x v' = 2x^2 \sin^2(v).$$ 8. Divide both sides by $x$ (assuming $x \neq 0$): $$v' = 2x \sin^2(v).$$ 9. Express the differential equation as: $$\frac{dv}{dx} = 2x \sin^2(v).$$ 10. Rewrite as separable equation: $$\frac{dv}{\sin^2(v)} = 2x \, dx.$$ 11. Recall that $\frac{1}{\sin^2(v)} = \csc^2(v)$. 12. Integrate both sides: $$\int \csc^2(v) \, dv = \int 2x \, dx.$$ 13. The integral on the left is: $$\int \csc^2(v) \, dv = -\cot(v) + C.$$ 14. The integral on the right is: $$\int 2x \, dx = x^{2} + C.$$ 15. Equate the two integrals (absorbing constants into one $C$): $$-\cot(v) = x^2 + C.$$ 16. Substitute back $v = \frac{y}{x}$: $$-\cot\left(\frac{y}{x}\right) = x^2 + C.$$ 17. This implicit solution relates $x$ and $y$. Final answer: $$-\cot\left(\frac{y}{x}\right) = x^2 + C.$$