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

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- anonymous

@Spacelimbus

- 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

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

you mean separating t from x?

- anonymous

yea but i think im wrong

- anonymous

I would try it the regular way to be honest and just chain the functions. Let me see.

- anonymous

they are both in the same domain, time, so it would work from what I can tell.

- anonymous

ok

- anonymous

\[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))-sx(0)-x'(0)) \]
*Repost*

- anonymous

ok what do we do after that?

- 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

otherwise I can't see how to find a solution, x(0)=0, but we require x'(0)

- anonymous

Otherwise it will be impossible to change from the s domain into the t domain.

- anonymous

have you tackled a problem like this before? where they require you to change the setup?

- anonymous

no i havent which is why im so confused

- anonymous

is it maybe x'(1)=e^(-3) ?

- anonymous

no its x(0)=0 and x(1)=e^(-3)

- 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

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

but it tried doing that before and i ended up wrong

- anonymous

just like the way I did above?

- 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

doing all of that to the three terms and then solving for big Y

- anonymous

but we dont have x'

- anonymous

Do you know the solution ?

- anonymous

no its an online hw assignment

- anonymous

got any idea mate @Herp_Derp ?

- anonymous

I have the work for a different problem from my friend. would that help?

- 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

here it is

##### 1 Attachment

- anonymous

oh btw that whole ine is u(t-2)(t-)e^(-2)(e^(-2t))

- 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

Where\[\tilde{x}(s)=\mathcal{L}\{x(t)\}_{(s)}\]Is the transform of x.

- 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

The end may be cut off on your browser:\[\ldots=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+4x(0)\]

- anonymous

end of what

- anonymous

nvr mind

- anonymous

ok but do we o with that formula

- anonymous

@myininaya @MathSofiya

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