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anonymous
 4 years ago
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)=
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you mean separating t from x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea but i think im wrong

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

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

anonymous
 4 years ago
Best 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*

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok what do we do after that?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Otherwise it will be impossible to change from the s domain into the t domain.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you tackled a problem like this before? where they require you to change the setup?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i havent which is why im so confused

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it maybe x'(1)=e^(3) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no its x(0)=0 and x(1)=e^(3)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0otherwise 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
 4 years ago
Best ResponseYou've already chosen the best response.0Did 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
 4 years ago
Best ResponseYou've already chosen the best response.0but it tried doing that before and i ended up wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just like the way I did above?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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
 4 years ago
Best ResponseYou've already chosen the best response.0doing all of that to the three terms and then solving for big Y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you know the solution ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no its an online hw assignment

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0got any idea mate @Herp_Derp ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have the work for a different problem from my friend. would that help?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh btw that whole ine is u(t2)(t)e^(2)(e^(2t))

anonymous
 4 years ago
Best 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,\[\large \mathcal{L}\{(3t4)x'(t)\}=3\mathcal{L}\{tx'(t)\}4\mathcal{L}\{x'(t)\}=3s\tilde{x}'(s)(3+4s)\tilde{x}(s)+x(0)\]

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok but do we o with that formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@myininaya @MathSofiya
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