=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