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Hint:For the first part write asy′−ay= 0 and multiply both sides bye−at.Then explainwhy you getddt(e−aty(t))= 0, y(t0) = 0.Now you finish the argument. To show uniqueness in the second part, supposey′= (a+ib)y, y(0) = 0and verify this requiresy(t) = 0. To do this, notey′= (a−ib)y,y(0) = 0and thatddt|y(t)|2=y′(t)y(t) +y′(t)y(t)=(a+ib)y(t)y(t) + (a−ib)y(t)y(t)=2a|y(t)|2,|y|2(t0) = 0Thus from the first part|y(t)|2= 0e−2at= 0.Finally observe by a simple computation that(16.8) is solved by (16.9). For the last part, write the equation asy′−ay=fand multiply both sides bye−atand then integrate fromt0totusing the initial condition.24.↑Now considerAann×nmatrix. By Schur’s theorem there exists unitaryQsuch thatQ−1AQ=TwhereTis upper triangular. Now consider the first order initial value problemx′=Ax,x(t0) =x0.Show there exists a unique solution to this first order system.Hint:Lety=Q−1xand sothe system becomesy′=Ty,y(t0) =Q−1x0(16.10)Now lettingy= (y1,· · ·, yn)T,the bottom equation becomesy′n=tnnyn, yn(t0) =(Q−1x0)n.Then use the solution you get in this to get the solution to the initial value problem whichoccurs one level up, namelyy′n−1=t(n−1)(n−1)yn−1+t(n−1)nyn, yn−1(t0) =(Q−1x0)n−1Continue doing this to obtain a unique solution to (16.10).Saylor URL: The Saylor Foundation
16.4.EXERCISES34925.↑Now suppose Φ (t) is ann×nmatrix of the formΦ (t) =(x1(t)· · ·xn(t))(16.11)wherex′k(t) =Axk(t).Explain whyΦ′(t) =AΦ (t)if and only if Φ (t) is given in the form of (16.11). Also explain why ifc∈Fn,y(t)≡Φ (t)csolves the equationy′(t) =Ay(t).26.↑In the above problem, consider the question whether all solutions tox′=Ax(16.12)are obtained in the form Φ (t)cfor some choice ofc∈Fn.In other words, is the generalsolution to this equation Φ (t)cforc∈Fn? Prove the following theorem using linear algebra.Theorem 16.4.1SupposeΦ (t)is ann×nmatrix which satisfiesΦ′(t) =AΦ (t).Then the general solution to (16.12) isΦ (t)cif and only ifΦ (t)−1exists for somet.Fur-thermore, ifΦ′(t) =AΦ (t),then eitherΦ (t)−1exists for alltorΦ (t)−1never exists for anyt.(det (Φ (t)) is called the Wronskian and this theorem is sometimes called the Wronskian alter-native.)Hint:Suppose first the general solution is of the form Φ (t)cwherecis an arbitrary constantvector inFn. You need to verify Φ (t)−1exists for somet. In fact, show Φ (t)−1exists for everyt. Suppose then that Φ (t0)−1does not exist. Explain why there existsc∈Fnsuch that thereis no solutionxtoc= Φ (t0)xBy the existence part of Problem 24 there exists a solution tox′=Ax,x(t0) =cbut this cannot be in the form Φ (t)c. Thus for everyt, Φ (t)−1exists. Next suppose for somet0,Φ (t0)−1exists. Letz′=Azand choosecsuch thatz(t0) = Φ (t0)cThen bothz(t),Φ (t)csolvex′=Ax,x(t0) =z(t0)Apply uniqueness to concludez= Φ (t)c. Finally, consider that Φ (t)cfor