Question

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)=

Answer 1

@Spacelimbus

Answer 2

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.

Answer 3

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

Answer 4

you mean separating t from x?

Answer 5

yea but i think im wrong

Answer 6

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

Answer 7

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

Answer 8

ok

Answer 9

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

*Repost*

Answer 10

ok what do we do after that?

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

no i havent which is why im so confused

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

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))

Answer 20

but it tried doing that before and i ended up wrong

Answer 21

just like the way I did above?

Answer 22

For the first term
\[\Large \mathcal{L}\lbrace t x''\rbrace = \frac{1}{s^2}( s^2Y(x(t))-\underbrace{sx(0)}_0-\underbrace{x'(0))}_c \]

Answer 23

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

Answer 24

but we dont have x'

Answer 25

Do you know the solution ?

Answer 26

no its an online hw assignment

Answer 27

got any idea mate @Herp_Derp ?

Answer 28

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

Answer 29

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.

Answer 30

here it is

Answer 31

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

Answer 32

@Spacelimbus Your formula is wrong.\[\large \mathcal{L}\{ t\, x''(t)\}=-s^2\tilde{x}'(s)-2s\tilde{x}(s)+x(0)\]Further,\[\large \mathcal{L}\{(3t-4)x'(t)\}=3\mathcal{L}\{tx'(t)\}-4\mathcal{L}\{x'(t)\}=-3s\tilde{x}'(s)-(3+4s)\tilde{x}(s)+x(0)\]

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

end of what

Answer 37

nvr mind

Answer 38

ok but do we o with that formula

Answer 39

@myininaya @MathSofiya