anonymous
  • anonymous
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)=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Spacelimbus
anonymous
  • anonymous
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.
anonymous
  • anonymous
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

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anonymous
  • anonymous
you mean separating t from x?
anonymous
  • anonymous
yea but i think im wrong
anonymous
  • anonymous
I would try it the regular way to be honest and just chain the functions. Let me see.
anonymous
  • anonymous
they are both in the same domain, time, so it would work from what I can tell.
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))-sx(0)-x'(0)) \] *Repost*
anonymous
  • anonymous
ok what do we do after that?
anonymous
  • anonymous
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.
anonymous
  • anonymous
otherwise I can't see how to find a solution, x(0)=0, but we require x'(0)
anonymous
  • anonymous
Otherwise it will be impossible to change from the s domain into the t domain.
anonymous
  • anonymous
have you tackled a problem like this before? where they require you to change the setup?
anonymous
  • anonymous
no i havent which is why im so confused
anonymous
  • anonymous
is it maybe x'(1)=e^(-3) ?
anonymous
  • anonymous
no its x(0)=0 and x(1)=e^(-3)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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))
anonymous
  • anonymous
but it tried doing that before and i ended up wrong
anonymous
  • anonymous
just like the way I did above?
anonymous
  • anonymous
For the first term \[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))-\underbrace{sx(0)}_0-\underbrace{x'(0))}_c \]
anonymous
  • anonymous
doing all of that to the three terms and then solving for big Y
anonymous
  • anonymous
but we dont have x'
anonymous
  • anonymous
Do you know the solution ?
anonymous
  • anonymous
no its an online hw assignment
anonymous
  • anonymous
got any idea mate @Herp_Derp ?
anonymous
  • anonymous
I have the work for a different problem from my friend. would that help?
anonymous
  • anonymous
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.
anonymous
  • anonymous
here it is
1 Attachment
anonymous
  • anonymous
oh btw that whole ine is u(t-2)(t-)e^(-2)(e^(-2t))
anonymous
  • anonymous
@Spacelimbus Your formula is wrong.\[\large \mathcal{L}\{ t\, x''(t)\}=-s^2\tilde{x}'(s)-2s\tilde{x}(s)+x(0)\]Further,\[\large \mathcal{L}\{(3t-4)x'(t)\}=3\mathcal{L}\{tx'(t)\}-4\mathcal{L}\{x'(t)\}=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+x(0)\]
anonymous
  • anonymous
Where\[\tilde{x}(s)=\mathcal{L}\{x(t)\}_{(s)}\]Is the transform of x.
anonymous
  • anonymous
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
anonymous
  • anonymous
The end may be cut off on your browser:\[\ldots=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+4x(0)\]
anonymous
  • anonymous
end of what
anonymous
  • anonymous
nvr mind
anonymous
  • anonymous
ok but do we o with that formula
anonymous
  • anonymous
@myininaya @MathSofiya

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