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

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

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I am stuck when im at this stage: \[Y(s)(s+6)+3X(s)=\frac{ 8-s(s-1) }{ s-1 }\]
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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you've got 2 different variables in there? y and x so y = x', yes?
how does y=x'?
they are just both functions of time
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?
Ive just never come across two independent variables in a Laplace transform before...
there is no unique solution as we don't know any information about the function \(x(t)\)
if you have another equation, then we can solve the system by elimination and get an unique solution \((x(t),y(t))\)
actually, let me screen shot the probem, perhaps part a and b might be related. Give us a tick
basically it is like solving an equation with many unknowns, there always exist infinitely many solutions when you have more variables but less equations
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are you doing control theory ?
yep process control
you solve those as a pair laplace then normal algebra
pretty sure the equations in "a" and "b" form the system
ah, i thought they would be two seperate problems, but since the first didn't make sense, i was querying things.
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i might be wrong though as idk abc of process control
right, so when i've formed one equation, i get 4y'+12x=20e^t
is that the way to go?
It seems awkward to put the ICs with the first equation like that
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
Still if this is a "system" then where's y'?
yea true
Something doesn't seem right here haha
i'll give that a miss then haha
i'll follow the links process, and see if i get an answer.
no wait, it cannot be a system because you cannot have x' = something and x' = something else
perhaps the initial conditions are misspelt?
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
i think its pretty straight forward, my answer was y(t)=cosh(t)
my approach was simply taking the derivative of the integral
\[y''-y=0\]
if that approach is right, then im pretty confident. and well the derivative of cosh(x) is sinh(x) as well...hence that confirms
simply plugin y=cosh(t) and check ?
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 \[y = A e^t + Be^{-t}\]
but we are using laplace transforms here
Oh right I forgot about your initial conditions
haha no problem
y(0)=1 y'(0) = 0 y''-y=0
yeah, i'm assuming you have to say that y'(0)=0...
that follows trivially from the given equation : \[\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt } = y'(t)\] plugin \(t=0\)
yeah fair call
your simply integrating from 0 to 0
right thats easy.. im still clueless about ur main question on the equation with two variables.. . gtg wil check in the night
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
yeah i'm baffled too...
if it helps \(x = \frac{5}{4}e^t - \frac{9}{4}e^{-3t}\) \(y = -\frac{1}{4}e^t + \frac{9}{4}e^{-3t}\)
hmm, how did you get to that?
perhaps i'll give it ago now, ive just finished writing the other minor problem down neatly
you'll kick yourself when you see this Example 4 at bottom of page
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which example am i meant to be looking at?
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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yeah right, thats very interesting. its just the IV's that don't check hey
@ganeshie8 i've tried this problem myself and it seems that there is a snap with the initial conditions, what do you think?
snag*
@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?
hi @chris00 i'm baffled. would love to know how you eventually come out on this
maybe since its defined by default in some books to apply Laplace on Certain conditions , u might check =)
i'll let you guys know later on in the week when i have a tute on this. cheers guys

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