ranyai12 Group Title 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

1. ranyai12 Group Title

@Spacelimbus

2. Spacelimbus Group Title

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.

3. ranyai12 Group Title

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

4. Spacelimbus Group Title

you mean separating t from x?

5. ranyai12 Group Title

yea but i think im wrong

6. Spacelimbus Group Title

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

7. Spacelimbus Group Title

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

8. ranyai12 Group Title

ok

9. Spacelimbus Group Title

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

10. ranyai12 Group Title

ok what do we do after that?

11. Spacelimbus Group Title

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.

12. Spacelimbus Group Title

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

13. Spacelimbus Group Title

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

14. Spacelimbus Group Title

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

15. ranyai12 Group Title

no i havent which is why im so confused

16. Spacelimbus Group Title

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

17. ranyai12 Group Title

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

18. Spacelimbus Group Title

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.

19. Spacelimbus Group Title

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))

20. ranyai12 Group Title

but it tried doing that before and i ended up wrong

21. Spacelimbus Group Title

just like the way I did above?

22. Spacelimbus Group Title

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

23. Spacelimbus Group Title

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

24. ranyai12 Group Title

but we dont have x'

25. Spacelimbus Group Title

Do you know the solution ?

26. ranyai12 Group Title

no its an online hw assignment

27. Spacelimbus Group Title

got any idea mate @Herp_Derp ?

28. ranyai12 Group Title

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

29. Spacelimbus Group Title

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.

30. ranyai12 Group Title

here it is

31. ranyai12 Group Title

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

32. Herp_Derp Group Title

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

33. Herp_Derp Group Title

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

34. ranyai12 Group Title

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

35. Herp_Derp Group Title

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

36. ranyai12 Group Title

end of what

37. ranyai12 Group Title

nvr mind

38. ranyai12 Group Title

ok but do we o with that formula

39. ranyai12 Group Title

@myininaya @MathSofiya