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Mathematics LibreTexts

11.53: A.8.4- Section 8.4 Answers

( \newcommand{\kernel}{\mathrm{null}\,}\)

1. 1+u(t4)(t1);1s+e4s(1s2+3s)

2. t+u(t1)(1t);1ess2

3. 2t1u(t2)(t1);(2s21s)e2s(1s2+1s)

4. 1+u(t1)(t+1);1s+es(1s2+2s)

5. t1+u(t2)(5t);1s21se2s(1s23s)

6. t2(1u(t1));2s3es(2s3+2s2+1s)

7. u(t2)(t2+3t);e2s(2s3+7s2+10s)

8. t2+2+u(t1)(tt22);2s3+2ses(2s3+1s2+2s)

9. tet+u(t1)(ettet);1e(s1)(s1)2

10. et+u(t1)(e2tet);1e(s+1)s+1+e(s+2)s+2

11. t+2u(t2)(t2)u(t3)(t5);1s2+2e2ss2+e3s(2s1s2)

12. [u(t1)u(t2)]t;es(1s2+1s)e2s(1s2+2s)

13. t+u(t1)(t2t)u(t2)t2;1s2+es(2s3+1s2)e2s(2s3+4s2+4s)

14. t+u(t1)(22t)+u(t2)(4+t);1s22ess2+e2s(1s2+6s)

15. sint+u(tπ/2)sint+u(tπ)(cost2sint);1+eπ2sseπs(s2)s2+1

16. 22u(t1)t+u(t3)(5t2);2ses(2s2+2s)+e3s(5s2+13s)

17. 3+u(t2)(3t1)+u(t4)(t2);3s+e2s(3s2+5s)+e4s(1s2+2s)

18. (t+1)2+u(t1)(2t+3);2s3+2s2+1s+es(2s2+5s)

19. u(t2)e2(t2)={0,0t<2,e2(t2),t2

20. u(t1)(1e(t1))={0,0t<1,1e(t1),t1

21. u(t1)(t1)22+u(t2)(t2)={0,0t<1,(t1)22,1t<2,t232,t2

22. 2+t+u(t1)(4t)+u(t3)(t2)={2+t,0t<1,6,1t<3,t+4,t3

23. 5t+u(t3)(7t15)+32u(t6)(t6)2={5t,0t<3,6t10,3t<6,4412t+32t2,t6

24. u(tπ)e2(tπ)(2cost5sint)={0,0t<π,e2(tπ)(2cost5sint)tπ

25. 1cost+u(tπ/2)(3sint+cost)={1cost,0t<π2,1+3sint,tπ2

26. u(t2)(4e(t2)4e2(t2)+2e(t2)={0,0t<2,4e(t2)4e2(t2)+2e(t2),t2

27. 1+t+u(t1)(2t+1)+u(t3)(3t5)={t+1,0t<1,3t+2,1t<3,6t3,t3

28. 1t2+u(t2)(t22+2t+1)+u(t4)(t4)={1t2,0t<2,3t22+2t+2,2t<4,3t22+3t2,t4

29. eτss

30. For each t only finitely many terms are nonzero.

33. 1+m=1u(tm);1s(1es)

34. 1+2m=1(1)mu(tm);1s;1es1+es

35. 1+m=1(2m+1)u(tm);es(1+es)s(1es)2

36. m=1(1)m(2m1)u(tm);1s(1es)(1+es)2


This page titled 11.53: A.8.4- Section 8.4 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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