Differential EquationsHardExistence, Uniqueness & Bifurcations ShareExistence, Uniqueness & Bifurcations · Hard problemThe IVP y′=2ty2y' = 2ty^2y′=2ty2, y(1)=1y(1) = 1y(1)=1 has a solution on a maximal open interval (α,β)(\alpha, \beta)(α,β). Find β\betaβ.Check Show answer Walk through it step by step Practice more like this Review this concept