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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
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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ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@Spacelimbus
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
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

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
you mean separating t from x?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
yea but i think im wrong
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
I would try it the regular way to be honest and just chain the functions. Let me see.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
they are both in the same domain, time, so it would work from what I can tell.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
\[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))sx(0)x'(0)) \] *Repost*
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok what do we do after that?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
otherwise I can't see how to find a solution, x(0)=0, but we require x'(0)
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Otherwise it will be impossible to change from the s domain into the t domain.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
have you tackled a problem like this before? where they require you to change the setup?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
no i havent which is why im so confused
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
is it maybe x'(1)=e^(3) ?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
no its x(0)=0 and x(1)=e^(3)
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but it tried doing that before and i ended up wrong
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
just like the way I did above?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
doing all of that to the three terms and then solving for big Y
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but we dont have x'
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Do you know the solution ?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
no its an online hw assignment
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
got any idea mate @Herp_Derp ?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I have the work for a different problem from my friend. would that help?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
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

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
here it is
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
oh btw that whole ine is u(t2)(t)e^(2)(e^(2t))
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
@Spacelimbus Your formula is wrong.\[\large \mathcal{L}\{ t\, x''(t)\}=s^2\tilde{x}'(s)2s\tilde{x}(s)+x(0)\]Further,\[\large \mathcal{L}\{(3t4)x'(t)\}=3\mathcal{L}\{tx'(t)\}4\mathcal{L}\{x'(t)\}=3s\tilde{x}'(s)(3+4s)\tilde{x}(s)+x(0)\]
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
Where\[\tilde{x}(s)=\mathcal{L}\{x(t)\}_{(s)}\]Is the transform of x.
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
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

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
The end may be cut off on your browser:\[\ldots=3s\tilde{x}'(s)(3+4s)\tilde{x}(s)+4x(0)\]
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
end of what
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok but do we o with that formula
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@myininaya @MathSofiya
 2 years ago
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