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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

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  1. ranyai12
    • 3 years ago
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    @TuringTest

  2. ranyai12
    • 3 years ago
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    this is what I got so far

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  3. ranyai12
    • 3 years ago
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    after this I am stuck :(

  4. ranyai12
    • 3 years ago
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    and this is the table I used: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

  5. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    but what I do about the 4/s^2+4s

  7. TuringTest
    • 3 years ago
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    factor out the s and that is what I wrote

  8. TuringTest
    • 3 years ago
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    oh I see; same thing though factor out s and do partial fractions

  9. ranyai12
    • 3 years ago
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    but then you get 4/(s(s+4)-->4/s[1/s+4)--> 4e^(-4t)

  10. ranyai12
    • 3 years ago
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    is that right?

  11. TuringTest
    • 3 years ago
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    \[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]\[A=1~~~,B=-1\]

  12. ranyai12
    • 3 years ago
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    oh so you do partial fractions with both of them

  13. TuringTest
    • 3 years ago
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    that's what I'd do, yeah

  14. ranyai12
    • 3 years ago
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    and for the other one it's A(s^2+4s) B(s+1) right

  15. TuringTest
    • 3 years ago
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    \[\mathcal L^{-1}\{\frac1s\}=1\]\[-\mathcal L^{-1}\{\frac1{s+4}\}=-e^{-4t}\]

  16. TuringTest
    • 3 years ago
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    for the other one it would be...

  17. TuringTest
    • 3 years ago
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    \[\frac2{s(s+1)(s+4)}=\frac As+\frac B{s+1}+\frac C{s+4}\]

  18. TuringTest
    • 3 years ago
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    \[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]

  19. ranyai12
    • 3 years ago
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    then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2

  20. ranyai12
    • 3 years ago
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    right

  21. TuringTest
    • 3 years ago
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    not necessary I don't think plug in s=-1,-4,and 0 and I think you can get the values quickly

  22. ranyai12
    • 3 years ago
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    so you dont need ti distribute it?

  23. TuringTest
    • 3 years ago
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    not necessarily how did I get the other answers? I'll show you...

  24. ranyai12
    • 3 years ago
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    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
    • 3 years ago
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    \[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=0:A(0+4)+B(0)=4 \implies 4A=4\implies A=1\]

  26. ranyai12
    • 3 years ago
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    B=-2/3 C=1/6

  27. ranyai12
    • 3 years ago
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    and A=1/2

  28. ranyai12
    • 3 years ago
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    thats what I got

  29. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    I didnt get those values though :(

  31. TuringTest
    • 3 years ago
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    do you see how I got mine?

  32. ranyai12
    • 3 years ago
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    i mean for the values of the other one

  33. TuringTest
    • 3 years ago
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    that I haven't checked yet, one sec...

  34. TuringTest
    • 3 years ago
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    \[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]\[s=-1:-3B=2\implies B=-\frac23\]\[s=-4:12C=2\implies C=\frac16\]\[s=0:4A=2\implies A=\frac12\]yes I guess you are right :)

  35. ranyai12
    • 3 years ago
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    yay!! so would the final answer be (1/2)-(2/3)(e^-t)+(1/6)e^(-4t)-e^(-4t)

  36. TuringTest
    • 3 years ago
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    looks good to me :)

  37. ranyai12
    • 3 years ago
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    it's wrong :(

  38. TuringTest
    • 3 years ago
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    I think you forgot the one from the other term do you have infinite tries?

  39. ranyai12
    • 3 years ago
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    no I ownly have 4 more tries

  40. ranyai12
    • 3 years ago
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    but what did I forget

  41. ranyai12
    • 3 years ago
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    only*

  42. TuringTest
    • 3 years ago
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    ok let me try it again from the top; I went of what you had I should double-check everything

  43. ranyai12
    • 3 years ago
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    ok thanks

  44. TuringTest
    • 3 years ago
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    ok I think I found it...

  45. TuringTest
    • 3 years ago
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    \[x''(0)+4x'+8x=2e^{-t}\]\[s^2X(s)-sx(0)-x'(0)+4sX(s)-4x(0)+8X(s)=\frac2{s+1}\]\[(s^2+4s+8)X(s)=\frac2{s+1}+4\]

  46. TuringTest
    • 3 years ago
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    you dropped the 8...

  47. TuringTest
    • 3 years ago
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    unfortunately this is going to make partial fractions a real pain...

  48. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    hyperbolic trig functions*

  50. ranyai12
    • 3 years ago
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    can you please show me because I honestly have no clue

  51. TuringTest
    • 3 years ago
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    \[X(s)={2\over{((s+2)^2+4)(s+1)}}+\frac4{((s+2)^2+4)}\]\[={2\over{((s+2)^2+4)(s+1)}}+2\frac2{((s+2)^2+2^2)}\]the second term is form 19 on the chart in your link

  52. TuringTest
    • 3 years ago
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    \[\mathcal L^{-1}\{\frac2{((s+2)^2+2^2)}\}=e^{-2t}\sin(2t)\]

  53. TuringTest
    • 3 years ago
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    the other part... you're gonna have to use partial fractions fractions and it looks like it's gonna be a pain

  54. TuringTest
    • 3 years ago
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    \[{2\over{((s+2)^2+4)(s+1)}}={As+B\over(s+2)^2+4}+{C\over s+1}\]\[(As+B)(s+1)+C[(s+2)^2+4]=2\]\[s=-1:5C=2\implies C=\frac25\]\[s=0:B+\frac{16}5=2\implies B=-\frac65\]

  55. TuringTest
    • 3 years ago
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    I think you'll have to get A on your own ;) I'm a bit busy

  56. TuringTest
    • 3 years ago
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    I think\[(A+C)s^2=0\implies A=-C=-\frac25\]but you shouod double-check me

  57. TuringTest
    • 3 years ago
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    I think that is correct

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