anonymous
  • anonymous
Solution of y'+6y+3x=8e^t for y(0)=-1, x(0)=0 using Laplace transofrms
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Jhannybean
  • Jhannybean
@Empty
anonymous
  • anonymous
I am stuck when im at this stage: \[Y(s)(s+6)+3X(s)=\frac{ 8-s(s-1) }{ s-1 }\]
anonymous
  • 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

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IrishBoy123
  • IrishBoy123
you've got 2 different variables in there? y and x so y = x', yes?
anonymous
  • anonymous
how does y=x'?
anonymous
  • anonymous
they are just both functions of time
IrishBoy123
  • 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?
anonymous
  • anonymous
Ive just never come across two independent variables in a Laplace transform before...
ganeshie8
  • ganeshie8
there is no unique solution as we don't know any information about the function \(x(t)\)
ganeshie8
  • ganeshie8
if you have another equation, then we can solve the system by elimination and get an unique solution \((x(t),y(t))\)
anonymous
  • anonymous
actually, let me screen shot the probem, perhaps part a and b might be related. Give us a tick
ganeshie8
  • 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
anonymous
  • anonymous
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ganeshie8
  • ganeshie8
are you doing control theory ?
anonymous
  • anonymous
yep process control
IrishBoy123
  • IrishBoy123
you solve those as a pair laplace then normal algebra
ganeshie8
  • ganeshie8
pretty sure the equations in "a" and "b" form the system
anonymous
  • anonymous
ah, i thought they would be two seperate problems, but since the first didn't make sense, i was querying things.
ganeshie8
  • ganeshie8
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ganeshie8
  • ganeshie8
i might be wrong though as idk abc of process control
anonymous
  • anonymous
right, so when i've formed one equation, i get 4y'+12x=20e^t
anonymous
  • anonymous
is that the way to go?
Empty
  • Empty
It seems awkward to put the ICs with the first equation like that
ganeshie8
  • 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
Empty
  • Empty
Still if this is a "system" then where's y'?
anonymous
  • anonymous
yea true
Empty
  • Empty
Something doesn't seem right here haha
anonymous
  • anonymous
i'll give that a miss then haha
anonymous
  • anonymous
i'll follow the links process, and see if i get an answer.
ganeshie8
  • ganeshie8
no wait, it cannot be a system because you cannot have x' = something and x' = something else
anonymous
  • anonymous
perhaps the initial conditions are misspelt?
anonymous
  • 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
anonymous
  • anonymous
i think its pretty straight forward, my answer was y(t)=cosh(t)
anonymous
  • anonymous
my approach was simply taking the derivative of the integral
anonymous
  • anonymous
\[y''-y=0\]
anonymous
  • 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
ganeshie8
  • ganeshie8
simply plugin y=cosh(t) and check ?
Empty
  • 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 \[y = A e^t + Be^{-t}\]
anonymous
  • anonymous
but we are using laplace transforms here
Empty
  • Empty
Oh right I forgot about your initial conditions
anonymous
  • anonymous
haha no problem
ganeshie8
  • ganeshie8
y(0)=1 y'(0) = 0 y''-y=0
anonymous
  • anonymous
yeah, i'm assuming you have to say that y'(0)=0...
ganeshie8
  • ganeshie8
that follows trivially from the given equation : \[\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt } = y'(t)\] plugin \(t=0\)
anonymous
  • anonymous
yeah fair call
anonymous
  • anonymous
your simply integrating from 0 to 0
ganeshie8
  • ganeshie8
right thats easy.. im still clueless about ur main question on the equation with two variables.. . gtg wil check in the night
Empty
  • 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
anonymous
  • anonymous
yeah i'm baffled too...
IrishBoy123
  • IrishBoy123
if it helps \(x = \frac{5}{4}e^t - \frac{9}{4}e^{-3t}\) \(y = -\frac{1}{4}e^t + \frac{9}{4}e^{-3t}\)
anonymous
  • anonymous
hmm, how did you get to that?
anonymous
  • anonymous
perhaps i'll give it ago now, ive just finished writing the other minor problem down neatly
IrishBoy123
  • IrishBoy123
you'll kick yourself when you see this Example 4 at bottom of page
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anonymous
  • anonymous
which example am i meant to be looking at?
IrishBoy123
  • 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
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anonymous
  • anonymous
yeah right, thats very interesting. its just the IV's that don't check hey
anonymous
  • anonymous
@ganeshie8 i've tried this problem myself and it seems that there is a snap with the initial conditions, what do you think?
anonymous
  • anonymous
snag*
anonymous
  • 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?
IrishBoy123
  • IrishBoy123
hi @chris00 i'm baffled. would love to know how you eventually come out on this
ikram002p
  • ikram002p
maybe since its defined by default in some books to apply Laplace on Certain conditions , u might check =)
anonymous
  • anonymous
i'll let you guys know later on in the week when i have a tute on this. cheers guys

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