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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)=
 one year ago
 one year 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)=
 one year ago
 one year ago

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SpacelimbusBest 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.
 one year ago

ranyai12Best 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
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
you mean separating t from x?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
yea but i think im wrong
 one year ago

SpacelimbusBest 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.
 one year ago

SpacelimbusBest 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.
 one year ago

SpacelimbusBest 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*
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok what do we do after that?
 one year ago

SpacelimbusBest 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.
 one year ago

SpacelimbusBest 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)
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Otherwise it will be impossible to change from the s domain into the t domain.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
have you tackled a problem like this before? where they require you to change the setup?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
no i havent which is why im so confused
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
is it maybe x'(1)=e^(3) ?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
no its x(0)=0 and x(1)=e^(3)
 one year ago

SpacelimbusBest 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.
 one year ago

SpacelimbusBest 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))
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
but it tried doing that before and i ended up wrong
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
just like the way I did above?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
For the first term \[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))\underbrace{sx(0)}_0\underbrace{x'(0))}_c \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
doing all of that to the three terms and then solving for big Y
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Do you know the solution ?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
no its an online hw assignment
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
got any idea mate @Herp_Derp ?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
I have the work for a different problem from my friend. would that help?
 one year ago

SpacelimbusBest 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.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
oh btw that whole ine is u(t2)(t)e^(2)(e^(2t))
 one year ago

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

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

ranyai12Best 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
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
ok but do we o with that formula
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
@myininaya @MathSofiya
 one year ago
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