## anonymous one year ago Solution of y'+6y+3x=8e^t for y(0)=-1, x(0)=0 using Laplace transofrms

1. Jhannybean

@Empty

2. anonymous

I am stuck when im at this stage: $Y(s)(s+6)+3X(s)=\frac{ 8-s(s-1) }{ s-1 }$

3. anonymous

i can't seem to use the initial condition of x(0)=0... i'm not sure if theres a unique solution to this question

4. IrishBoy123

you've got 2 different variables in there? y and x so y = x', yes?

5. anonymous

how does y=x'?

6. anonymous

they are just both functions of time

7. IrishBoy123

what i am saying is that this would make more sense if that were true but i do not have sight of the original question to confirm if they are independent functions, what are you solving for. don't you need another equation in x and y?

8. anonymous

Ive just never come across two independent variables in a Laplace transform before...

9. ganeshie8

there is no unique solution as we don't know any information about the function $$x(t)$$

10. ganeshie8

if you have another equation, then we can solve the system by elimination and get an unique solution $$(x(t),y(t))$$

11. anonymous

actually, let me screen shot the probem, perhaps part a and b might be related. Give us a tick

12. ganeshie8

basically it is like solving an equation with many unknowns, there always exist infinitely many solutions when you have more variables but less equations

13. anonymous

14. ganeshie8

are you doing control theory ?

15. anonymous

yep process control

16. IrishBoy123

you solve those as a pair laplace then normal algebra

17. ganeshie8

pretty sure the equations in "a" and "b" form the system

18. anonymous

ah, i thought they would be two seperate problems, but since the first didn't make sense, i was querying things.

19. ganeshie8

|dw:1440405030249:dw|

20. ganeshie8

i might be wrong though as idk abc of process control

21. anonymous

right, so when i've formed one equation, i get 4y'+12x=20e^t

22. anonymous

is that the way to go?

23. Empty

It seems awkward to put the ICs with the first equation like that

24. ganeshie8

but if those two equations form a single system, then you may use below method to solve it using laplace transform http://tutorial.math.lamar.edu/Classes/DE/SystemsLaplace.aspx

25. Empty

Still if this is a "system" then where's y'?

26. anonymous

yea true

27. Empty

Something doesn't seem right here haha

28. anonymous

i'll give that a miss then haha

29. anonymous

30. ganeshie8

no wait, it cannot be a system because you cannot have x' = something and x' = something else

31. anonymous

perhaps the initial conditions are misspelt?

32. anonymous

in the mean time, i have another question that i think i've solved but just need someone to check it. $\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt }$ with initial conditions y(0)=1

33. anonymous

i think its pretty straight forward, my answer was y(t)=cosh(t)

34. anonymous

my approach was simply taking the derivative of the integral

35. anonymous

$y''-y=0$

36. anonymous

if that approach is right, then im pretty confident. and well the derivative of cosh(x) is sinh(x) as well...hence that confirms

37. ganeshie8

simply plugin y=cosh(t) and check ?

38. Empty

Yeah I guess there are a couple different solutions you could have here though for the most general cases (second order equation needs 2 constants to be most general) $y=A \sinh t + B \cosh t$ or $y = A e^t + Be^{-t}$

39. anonymous

but we are using laplace transforms here

40. Empty

41. anonymous

haha no problem

42. ganeshie8

y(0)=1 y'(0) = 0 y''-y=0

43. anonymous

yeah, i'm assuming you have to say that y'(0)=0...

44. ganeshie8

that follows trivially from the given equation : $\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt } = y'(t)$ plugin $$t=0$$

45. anonymous

yeah fair call

46. anonymous

your simply integrating from 0 to 0

47. ganeshie8

right thats easy.. im still clueless about ur main question on the equation with two variables.. . gtg wil check in the night

48. Empty

I thought, what if I subtract the second from the first and you get: $x+y=e^t$ Then you can see this is wrong because plugging in the intitial conditions will give: $-1=1$ So weird hahaha

49. anonymous

yeah i'm baffled too...

50. IrishBoy123

if it helps $$x = \frac{5}{4}e^t - \frac{9}{4}e^{-3t}$$ $$y = -\frac{1}{4}e^t + \frac{9}{4}e^{-3t}$$

51. anonymous

hmm, how did you get to that?

52. anonymous

perhaps i'll give it ago now, ive just finished writing the other minor problem down neatly

53. IrishBoy123

you'll kick yourself when you see this Example 4 at bottom of page

54. anonymous

which example am i meant to be looking at?

55. IrishBoy123

example 4 shows the method here are my scribbles there is indeed a snag in this question. the iv for y on P3 of my scribbles is not met but the same iv is not used in the solution i would love to understand why this as the solution just feels right but has this issue is it a problem with the question or have i messed up some stupid algebra? i really hope this helps rather than hinders

56. anonymous

yeah right, thats very interesting. its just the IV's that don't check hey

57. anonymous

@ganeshie8 i've tried this problem myself and it seems that there is a snap with the initial conditions, what do you think?

58. anonymous

snag*

59. anonymous

@Irishboy123 it seems that you don't even need to use the initial condition that y(0)=0 to solve the problem... so perhaps this is misleading information?

60. IrishBoy123

hi @chris00 i'm baffled. would love to know how you eventually come out on this

61. ikram002p

maybe since its defined by default in some books to apply Laplace on Certain conditions , u might check =)

62. anonymous

i'll let you guys know later on in the week when i have a tute on this. cheers guys