981 Solutions30.(1;1)31.(3;1)32.(1;2),(2;0),(0 2),(2;1)33.The initial condition occurs att= 0which is precisely wherea2(t) =t2has a zero. Theorem 6 does not apply.34.In this casey(t0) = 0andy0(t0) = 0. The functiony(t) = 0,t2Iis asolution to the initial value problem. By the uniqueness part of Theorem6y= 0is the only solution.35.The assumptions say thaty1(t0) =y2(t0)andy01(t0) =y02(t0). Bothy1andy2therefore satisfies the same initial conditions. By the uniquenesspart of Theorem 6y1=y2.Section 4.21.dependent;2tand5tare multiples of each other.2.independent3.independent4.dependent;e2t+ 1 =e1e2tande2t3=e3e2t, they are multiples of eachother.5.independent6.dependent;lnt2= 2 lntandlnt5= 5 lnt, they are multiples of each other.7.dependent;
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