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ranyai12

  • 2 years ago

Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(−t) x(0)=0 x'(0)=4

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  1. ranyai12
    • 2 years ago
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    @richyw

  2. richyw
    • 2 years ago
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    are you allowed to use a table?

  3. ranyai12
    • 2 years ago
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    this is my work and im not sure what i did wrong

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  4. ranyai12
    • 2 years ago
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    and yes i am

  5. richyw
    • 2 years ago
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    i'm already confused lol. where did you get the 18 from?

  6. ranyai12
    • 2 years ago
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    ooops thats supposed to be an 8

  7. richyw
    • 2 years ago
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    yeah if you put an 8 it looks pretty straightforward as you clearly know what to do. I never finished it though as I am cramming for calculus haha!

  8. ranyai12
    • 2 years ago
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    id you get x=(4s+6)/(s=1)(s^2+4a+8)

  9. ranyai12
    • 2 years ago
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    that part always confuses me so if you can please check that id be fine

  10. ranyai12
    • 2 years ago
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    (s-1)*

  11. richyw
    • 2 years ago
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    yes I got there that's where I quit. I'm thinking about it again now

  12. ranyai12
    • 2 years ago
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    ok thanks

  13. richyw
    • 2 years ago
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    hmm after my partial fractions I'm getting one that is hard to fix into a nice way to inverse laplace

  14. ranyai12
    • 2 years ago
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    what did you end up getting?

  15. richyw
    • 2 years ago
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    I can't figure out the answer. I really do apologize but I have a big final coming up and rushed through it. I don't mean to leave you hanging!

  16. ranyai12
    • 2 years ago
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    its ok thanks i just wanted what x= to anyway not the whole thing but I totally understand!!! Finals suck!

  17. ranyai12
    • 2 years ago
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    @jagan

  18. phi
    • 2 years ago
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    x'' + 4x' +8x= 2*exp(-t), x(0)=0, x'(0)= 4 \[ (s^2+4s+8)L[x] -(s+4)*0 - (1)*4 = \frac{2}{s+1} \] \[ (s^2+4s+8)L[x] = \frac{2}{s+1} +4 \] \[ L[x]= \frac{4s+6}{(s+1)(s^2+4s+8)} \] partial fraction expansion \[ L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s-7}{s^2+4s+8} \] complete the square on the denominator of the 2nd term to match the laplace tables rewrite the numerator \[ L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s+2-9}{(s+2)^2+2^2} \] re-write the last term as 2 fractions that match the Laplace tables. \[ L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s+2}{(s+2)^2+2^2}+\frac{9}{5}\frac{2}{(s+2)^2+2^2} \] \[ x= \frac{2}{5}e^{-t}- \frac{2}{5}e^{-2t}cos(2t)+\frac{9}{5}e^{-2t}sin(2t) \] see http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

  19. phi
    • 2 years ago
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    I just gave the highlights, as this takes quite a bit of work. Ask if you have any questions.

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