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ranyai12 Group Title

Use Laplace transforms to find a nontrivial solution to tx''+(3t−4)x'+3x=0, x(0)=0 Require that x(1)=e^(−3) to find any arbitrary constant in your solution. x(t)=

  • one year ago
  • one year ago

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  1. ranyai12 Group Title
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    @Spacelimbus

    • one year ago
  2. Spacelimbus Group Title
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    I will have to work through this myself, but I am sure you have to use the fact that the LaPlace Transform is a linear operator, and then apply what this identity: \[\Large \mathcal{L} \lbrace x'(t) \rbrace =s Y(x(t))-x(0)\] which suits your initial conditions.

    • one year ago
  3. ranyai12 Group Title
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    ok but I thought you had to separate each part and do the laplace of that but im not a hundred percent sure because when I tried doing a different one I got it wrong

    • one year ago
  4. Spacelimbus Group Title
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    you mean separating t from x?

    • one year ago
  5. ranyai12 Group Title
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    yea but i think im wrong

    • one year ago
  6. Spacelimbus Group Title
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    I would try it the regular way to be honest and just chain the functions. Let me see.

    • one year ago
  7. Spacelimbus Group Title
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    they are both in the same domain, time, so it would work from what I can tell.

    • one year ago
  8. ranyai12 Group Title
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    ok

    • one year ago
  9. Spacelimbus Group Title
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    \[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))-sx(0)-x'(0)) \] *Repost*

    • one year ago
  10. ranyai12 Group Title
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    ok what do we do after that?

    • one year ago
  11. Spacelimbus Group Title
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    I am a bit lost here because of the initial conditions, we need them to be at t=0, so a change of variable could sole the problem we have with x'(0) maybe.

    • one year ago
  12. Spacelimbus Group Title
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    otherwise I can't see how to find a solution, x(0)=0, but we require x'(0)

    • one year ago
  13. Spacelimbus Group Title
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    Otherwise it will be impossible to change from the s domain into the t domain.

    • one year ago
  14. Spacelimbus Group Title
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    have you tackled a problem like this before? where they require you to change the setup?

    • one year ago
  15. ranyai12 Group Title
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    no i havent which is why im so confused

    • one year ago
  16. Spacelimbus Group Title
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    is it maybe x'(1)=e^(-3) ?

    • one year ago
  17. ranyai12 Group Title
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    no its x(0)=0 and x(1)=e^(-3)

    • one year ago
  18. Spacelimbus Group Title
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    otherwise the only thing I can recommend you to do is carry out the steps normally to the end, maybe they want to keep it constant.

    • one year ago
  19. Spacelimbus Group Title
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    Did you see what I have done above? You just apply the LaPlace transform and then at the end you solve for Y(x(t))

    • one year ago
  20. ranyai12 Group Title
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    but it tried doing that before and i ended up wrong

    • one year ago
  21. Spacelimbus Group Title
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    just like the way I did above?

    • one year ago
  22. Spacelimbus Group Title
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    For the first term \[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))-\underbrace{sx(0)}_0-\underbrace{x'(0))}_c \]

    • one year ago
  23. Spacelimbus Group Title
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    doing all of that to the three terms and then solving for big Y

    • one year ago
  24. ranyai12 Group Title
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    but we dont have x'

    • one year ago
  25. Spacelimbus Group Title
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    Do you know the solution ?

    • one year ago
  26. ranyai12 Group Title
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    no its an online hw assignment

    • one year ago
  27. Spacelimbus Group Title
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    got any idea mate @Herp_Derp ?

    • one year ago
  28. ranyai12 Group Title
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    I have the work for a different problem from my friend. would that help?

    • one year ago
  29. Spacelimbus Group Title
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    I don't know yet, but you can post it when you like, I haven't figured out yet how this problem could be tackled with the missing initial condition for the derivative. My assumption was that it could be derived at the end. But it doesn't seem to work out for me.

    • one year ago
  30. ranyai12 Group Title
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    here it is

    • one year ago
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  31. ranyai12 Group Title
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    oh btw that whole ine is u(t-2)(t-)e^(-2)(e^(-2t))

    • one year ago
  32. Herp_Derp Group Title
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    @Spacelimbus Your formula is wrong.\[\large \mathcal{L}\{ t\, x''(t)\}=-s^2\tilde{x}'(s)-2s\tilde{x}(s)+x(0)\]Further,\[\large \mathcal{L}\{(3t-4)x'(t)\}=3\mathcal{L}\{tx'(t)\}-4\mathcal{L}\{x'(t)\}=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+x(0)\]

    • one year ago
  33. Herp_Derp Group Title
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    Where\[\tilde{x}(s)=\mathcal{L}\{x(t)\}_{(s)}\]Is the transform of x.

    • one year ago
  34. ranyai12 Group Title
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    ok but this one ended up being correct. My friend solved for it and it ended up right but idk what she did exactly and she left back home for the rest of the summer and regarding to what you both said im totally lost im sorry

    • one year ago
  35. Herp_Derp Group Title
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    The end may be cut off on your browser:\[\ldots=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+4x(0)\]

    • one year ago
  36. ranyai12 Group Title
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    end of what

    • one year ago
  37. ranyai12 Group Title
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    nvr mind

    • one year ago
  38. ranyai12 Group Title
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    ok but do we o with that formula

    • one year ago
  39. ranyai12 Group Title
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    @myininaya @MathSofiya

    • one year ago
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