If the solution curve of the differential equation 2 x - 10 y 3 d y + y d x = 0 , passes through the points…

Given that: $\left(2x-10y^3\right)dy+ydx=0$

$\Rightarrow \frac{dx}{dy}=10y^2-\frac{2x}{y}$

$\Rightarrow \frac{dx}{dy}+\frac{2x}{y}=10y^2$

$\Rightarrow \text{ I.F. }=e^{2\int \frac{1}{y}dx}=y^2$

$\Rightarrow x\cdot y^2=\int 10y^{4}dy+C$ 

$\Rightarrow x\cdot y^2=2y^5+c$

$\mathrm{Put} x=0 \& y=1\Rightarrow c=-2$

$\Rightarrow x\cdot y^2=2y^5-2$

Passing Through $(2,\beta )$
$\Rightarrow 2\cdot {\beta }^2=2{\beta }^5-2$

$\Rightarrow {\beta }^5-{\beta }^2-1=0$
root of the equations
$y^5-y^2-1=0$