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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)=

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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.
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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Other answers:

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

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