1 Solutions41-6-4-20246-4-3-2-101234tGraph ofy(t) =t+ct3for variousc2. Every solution satisfiesy(0) = 0. There is no contradiction to Theorem5 since, in standard form, the equation isy0=2ty1 =F(t; y)andF(t; y)is not continuous fort= 0.20.1. IfF(t; y) =y2thenFy(t; y) = 2y. Both are continuous on any rectan-gle that contains(t0; y0). Hence Theorem 5 applies and implies thereis a unique solution on an interval that containst0.2.y(t) = 0is a solution defined for allt;y(t) =11tis a solution definedon(1;1).21.No. Bothy1(t)andy2(t)would be solutions to the initial value problemy0=F(t; y),y(0) = 0. IfF(t; y)andFy(t; y)are both continuous near(0;0), then the initial value problem would have a unique solution byTheorem 5.22.No, Bothy1(t)andy2(t)would be solutions to the initial value problem
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Boundary value problem, Left-handedness, Lipschitz continuity