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Differential EquationsHardExistence, Uniqueness & Bifurcations

Existence, Uniqueness & Bifurcations · Hard problem

The equation t y′=2yt\,y' = 2y fails the hypotheses of the existence and uniqueness theorem at t=0t = 0. A function y(t)y(t) is differentiable on all of R\mathbb{R}, satisfies t y′=2yt\,y' = 2y for every real tt, and has y(1)=3y(1) = 3 and y(−1)=5y(-1) = 5. Find y(2)+y(−2)y(2) + y(-2).