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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)=

  • 2 years ago
  • 2 years ago

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  1. ranyai12 Group Title
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    @Spacelimbus

    • 2 years 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.

    • 2 years 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

    • 2 years ago
  4. Spacelimbus Group Title
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    you mean separating t from x?

    • 2 years ago
  5. ranyai12 Group Title
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    yea but i think im wrong

    • 2 years 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.

    • 2 years 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.

    • 2 years ago
  8. ranyai12 Group Title
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    ok

    • 2 years 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*

    • 2 years ago
  10. ranyai12 Group Title
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    ok what do we do after that?

    • 2 years 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.

    • 2 years 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)

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

    • 2 years 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?

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

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

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

    • 2 years 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.

    • 2 years 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))

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

    • 2 years ago
  21. Spacelimbus Group Title
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    just like the way I did above?

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

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

    • 2 years ago
  24. ranyai12 Group Title
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    but we dont have x'

    • 2 years ago
  25. Spacelimbus Group Title
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    Do you know the solution ?

    • 2 years ago
  26. ranyai12 Group Title
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    no its an online hw assignment

    • 2 years ago
  27. Spacelimbus Group Title
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    got any idea mate @Herp_Derp ?

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

    • 2 years 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.

    • 2 years ago
  30. ranyai12 Group Title
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    here it is

    • 2 years 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))

    • 2 years 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,\[\large \mathcal{L}\{(3t-4)x'(t)\}=3\mathcal{L}\{tx'(t)\}-4\mathcal{L}\{x'(t)\}=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+x(0)\]

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

    • 2 years 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

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

    • 2 years ago
  36. ranyai12 Group Title
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    end of what

    • 2 years ago
  37. ranyai12 Group Title
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    nvr mind

    • 2 years ago
  38. ranyai12 Group Title
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    ok but do we o with that formula

    • 2 years ago
  39. ranyai12 Group Title
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    @myininaya @MathSofiya

    • 2 years ago
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