If the solution curve of the differential equation $$(2x - 10y^3)dy + ydx = 0$$, passes through the points $$(0, 1)$$ and $$(2, \beta)$$, then $$\beta$$ is a root of the equation?

We begin with the differential equation

$$\bigl(2x-10y^{3}\bigr)\,dy + y\,dx = 0.$$

For convenience we regard $$x$$ as a function of $$y$$ (that is, we shall write $$x=x(y)$$ and differentiate with respect to $$y$$). Writing the differentials explicitly we have

$$y\,dx + \bigl(2x-10y^{3}\bigr)\,dy = 0.$$

Dividing by $$dy$$ gives a first-order linear ordinary differential equation in $$x$$:

$$y\,\dfrac{dx}{dy} + 2x - 10y^{3} = 0.$$

Re-arranging,

$$y\,\dfrac{dx}{dy} = 10y^{3} - 2x,$$

and hence

$$\dfrac{dx}{dy} + \dfrac{2}{y}\,x = 10y^{2}.$$

This is of the standard linear form

$$\dfrac{dx}{dy} + P(y)\,x = Q(y),\qquad\text{with }P(y)=\dfrac{2}{y},\;Q(y)=10y^{2}.$$

For such an equation the integrating factor is obtained from

$$\mu(y)=e^{\int P(y)\,dy}.$$

Computing the integral,

$$\int P(y)\,dy = \int\dfrac{2}{y}\,dy = 2\ln y,$$

so

$$\mu(y)=e^{2\ln y}=y^{2}.$$

Multiplying the differential equation by this integrating factor we get

$$y^{2}\dfrac{dx}{dy}+y^{2}\left(\dfrac{2}{y}\right)x = y^{2}\cdot10y^{2}.$$

Simplifying each term,

$$y^{2}\dfrac{dx}{dy}+2yx = 10y^{4}.$$

Notice that the left-hand side is exactly the derivative of the product $$y^{2}x$$ with respect to $$y$$, because

$$\dfrac{d}{dy}\bigl(y^{2}x\bigr)=y^{2}\dfrac{dx}{dy}+2yx.$$

Hence we can write

$$\dfrac{d}{dy}\bigl(y^{2}x\bigr)=10y^{4}.$$

Integrating both sides with respect to $$y$$,

$$\int\dfrac{d}{dy}\bigl(y^{2}x\bigr)\,dy=\int10y^{4}\,dy,$$

which yields

$$y^{2}x = 10\cdot\dfrac{y^{5}}{5} + C = 2y^{5}+C,$$

where $$C$$ is the constant of integration. Solving for $$x$$ gives

$$x = \dfrac{2y^{5}+C}{y^{2}} = 2y^{3} + \dfrac{C}{y^{2}}.$$

To determine $$C$$ we apply the first point through which the curve passes, namely $$(0,1)$$. Substituting $$x=0$$ and $$y=1$$ we have

$$0 = 2\cdot1^{3} + \dfrac{C}{1^{2}} \quad\Longrightarrow\quad C = -2.$$

Thus the particular solution is

$$x = 2y^{3} - \dfrac{2}{y^{2}}.$$

Now we use the second point $$(2,\beta)$$. Substituting $$x=2$$ and $$y=\beta$$ in the above relation yields

$$2 = 2\beta^{3} - \dfrac{2}{\beta^{2}}.$$

Multiplying both sides by $$\beta^{2}$$ to clear the denominator,

$$2\beta^{2} = 2\beta^{5} - 2.$$

Bringing all terms to one side,

$$2\beta^{5} - 2\beta^{2} - 2 = 0.$$

Factoring out the common factor $$2$$,

$$2\bigl(\beta^{5} - \beta^{2} - 1\bigr) = 0.$$

Since $$2\neq0$$, we must have

$$\beta^{5} - \beta^{2} - 1 = 0.$$

This is exactly the polynomial equation listed as Option B.

Hence, the correct answer is Option B.