Differential Equations Part 1

Given, differential equation
(2x−10y3)dy+ydx=0
⇒

dx

dy

+

2x

y

=10y2 ...(i)
This is Linear differential equation
Integrating factor IF=e2‌∫

1

y

dy=y2
Solution of differential Eq. (i),
x.y2=∫10y2.y2dy+C
⇒xy2=2y5+C ...(ii)
Solution Eq. (ii) passes through (0, 1)
⇒0.12=215+C
⇒ C = − 2
∴ Solution of Eq. (i) is
xy2=2y5−2
Now, this equation passes through (2, β).
∴2.β2=2β5−2
⇒β5−β2−1=0
⇒ β is root of the equation y5−y2−1=0