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(D) Given the differential equation: $(2x - 10y^3) dy + y dx = 0$.Rearranging the terms,we get: $y dx = (10y^3 - 2x) dy$.Dividing by $y dy$,we obtain

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$y^5 - y^2 - 1 = 0$

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(D) Given the differential equation: $(2x - 10y^3) dy + y dx = 0$.Rearranging the terms,we get: $y dx = (10y^3 - 2x) dy$.Dividing by $y dy$,we obtain the linear differential equation in $x$:$\frac{dx}{dy} + \frac{2}{y}x = 10y^2$.Here,the integrating factor $(I.F.)$ is given by:$I.F. = e^{\int \frac{2}{y} dy} = e^{2 \ln|y|} = y^2$.The general solution is given by $x \cdot (I.F.) = \int (10y^2) \cdot (I.F.) dy + C$.$x y^2 = \int 10y^4 dy + C$.$x y^2 = 2y^5 + C$.Since the curve passes through $(0, 1)$,we substitute $x=0$ and $y=1$:$0 \cdot (1)^2 = 2(1)^5 + C \Rightarrow C = -2$.Thus,the equation of the curve is $x y^2 = 2y^5 - 2$.Since the curve passes through $(2, \beta)$,we substitute $x=2$ and $y=\beta$:$2 \beta^2 = 2 \beta^5 - 2$.Dividing by $2$,we get $\beta^2 = \beta^5 - 1$,which simplifies to $\beta^5 - \beta^2 - 1 = 0$.Therefore,$\beta$ is a root of the equation $y^5 - y^2 - 1 = 0$.

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

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