## ranyai12 3 years ago Help Please!! Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(-t) ; x(0)=0 and x'(0)=4

1. ranyai12

@TuringTest

2. ranyai12

this is what I got so far

3. ranyai12

after this I am stuck :(

4. ranyai12

and this is the table I used: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

5. TuringTest

use partial fractions on the $\frac2{s(s+1)(s+4)}$part and I think it should be pretty straightforward to do inverse Laplace

6. ranyai12

but what I do about the 4/s^2+4s

7. TuringTest

factor out the s and that is what I wrote

8. TuringTest

oh I see; same thing though factor out s and do partial fractions

9. ranyai12

but then you get 4/(s(s+4)-->4/s[1/s+4)--> 4e^(-4t)

10. ranyai12

is that right?

11. TuringTest

$\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4$$A=1~~~,B=-1$

12. ranyai12

oh so you do partial fractions with both of them

13. TuringTest

that's what I'd do, yeah

14. ranyai12

and for the other one it's A(s^2+4s) B(s+1) right

15. TuringTest

$\mathcal L^{-1}\{\frac1s\}=1$$-\mathcal L^{-1}\{\frac1{s+4}\}=-e^{-4t}$

16. TuringTest

for the other one it would be...

17. TuringTest

$\frac2{s(s+1)(s+4)}=\frac As+\frac B{s+1}+\frac C{s+4}$

18. TuringTest

$A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2$

19. ranyai12

then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2

20. ranyai12

right

21. TuringTest

not necessary I don't think plug in s=-1,-4,and 0 and I think you can get the values quickly

22. ranyai12

so you dont need ti distribute it?

23. TuringTest

not necessarily how did I get the other answers? I'll show you...

24. ranyai12

i didnt get the answers yet i thiught you had to distribute it then ge tthe answers but now i see that its unnecessary

25. TuringTest

$\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4$now plug in$s=0:A(0+4)+B(0)=4 \implies 4A=4\implies A=1$

26. ranyai12

B=-2/3 C=1/6

27. ranyai12

and A=1/2

28. ranyai12

thats what I got

29. TuringTest

this is the second term:$\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4$now plug in$s=-4:A(-4+4)+B(-4)=4\implies -4B=4\implies B=-1$

30. ranyai12

I didnt get those values though :(

31. TuringTest

do you see how I got mine?

32. ranyai12

i mean for the values of the other one

33. TuringTest

that I haven't checked yet, one sec...

34. TuringTest

$A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2$$s=-1:-3B=2\implies B=-\frac23$$s=-4:12C=2\implies C=\frac16$$s=0:4A=2\implies A=\frac12$yes I guess you are right :)

35. ranyai12

yay!! so would the final answer be (1/2)-(2/3)(e^-t)+(1/6)e^(-4t)-e^(-4t)

36. TuringTest

looks good to me :)

37. ranyai12

it's wrong :(

38. TuringTest

I think you forgot the one from the other term do you have infinite tries?

39. ranyai12

no I ownly have 4 more tries

40. ranyai12

but what did I forget

41. ranyai12

only*

42. TuringTest

ok let me try it again from the top; I went of what you had I should double-check everything

43. ranyai12

ok thanks

44. TuringTest

ok I think I found it...

45. TuringTest

$x''(0)+4x'+8x=2e^{-t}$$s^2X(s)-sx(0)-x'(0)+4sX(s)-4x(0)+8X(s)=\frac2{s+1}$$(s^2+4s+8)X(s)=\frac2{s+1}+4$

46. TuringTest

you dropped the 8...

47. TuringTest

unfortunately this is going to make partial fractions a real pain...

48. TuringTest

write is as a complete square$((s+2)^2+4)X(s)=\frac2{s+1}+4$you should wind up with some hyperbolic I think

49. TuringTest

hyperbolic trig functions*

50. ranyai12

can you please show me because I honestly have no clue

51. TuringTest

$X(s)={2\over{((s+2)^2+4)(s+1)}}+\frac4{((s+2)^2+4)}$$={2\over{((s+2)^2+4)(s+1)}}+2\frac2{((s+2)^2+2^2)}$the second term is form 19 on the chart in your link

52. TuringTest

$\mathcal L^{-1}\{\frac2{((s+2)^2+2^2)}\}=e^{-2t}\sin(2t)$

53. TuringTest

the other part... you're gonna have to use partial fractions fractions and it looks like it's gonna be a pain

54. TuringTest

${2\over{((s+2)^2+4)(s+1)}}={As+B\over(s+2)^2+4}+{C\over s+1}$$(As+B)(s+1)+C[(s+2)^2+4]=2$$s=-1:5C=2\implies C=\frac25$$s=0:B+\frac{16}5=2\implies B=-\frac65$

55. TuringTest

I think you'll have to get A on your own ;) I'm a bit busy

56. TuringTest

I think$(A+C)s^2=0\implies A=-C=-\frac25$but you shouod double-check me

57. TuringTest

I think that is correct