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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am stuck when im at this stage: \[Y(s)(s+6)+3X(s)=\frac{ 8s(s1) }{ s1 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i can't seem to use the initial condition of x(0)=0... i'm not sure if theres a unique solution to this question

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you've got 2 different variables in there? y and x so y = x', yes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they are just both functions of time

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1what 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
 one year ago
Best ResponseYou've already chosen the best response.0Ive just never come across two independent variables in a Laplace transform before...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3there is no unique solution as we don't know any information about the function \(x(t)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if you have another equation, then we can solve the system by elimination and get an unique solution \((x(t),y(t))\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually, let me screen shot the probem, perhaps part a and b might be related. Give us a tick

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3basically it is like solving an equation with many unknowns, there always exist infinitely many solutions when you have more variables but less equations

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3are you doing control theory ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you solve those as a pair laplace then normal algebra

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3pretty sure the equations in "a" and "b" form the system

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah, i thought they would be two seperate problems, but since the first didn't make sense, i was querying things.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440405030249:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i might be wrong though as idk abc of process control

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right, so when i've formed one equation, i get 4y'+12x=20e^t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that the way to go?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2It seems awkward to put the ICs with the first equation like that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3but 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
 one year ago
Best ResponseYou've already chosen the best response.2Still if this is a "system" then where's y'?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Something doesn't seem right here haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'll give that a miss then haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'll follow the links process, and see if i get an answer.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3no wait, it cannot be a system because you cannot have x' = something and x' = something else

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps the initial conditions are misspelt?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in 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
 one year ago
Best ResponseYou've already chosen the best response.0i think its pretty straight forward, my answer was y(t)=cosh(t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my approach was simply taking the derivative of the integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if that approach is right, then im pretty confident. and well the derivative of cosh(x) is sinh(x) as well...hence that confirms

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3simply plugin y=cosh(t) and check ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0but we are using laplace transforms here

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Oh right I forgot about your initial conditions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3y(0)=1 y'(0) = 0 y''y=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, i'm assuming you have to say that y'(0)=0...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that follows trivially from the given equation : \[\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt } = y'(t)\] plugin \(t=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your simply integrating from 0 to 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3right thats easy.. im still clueless about ur main question on the equation with two variables.. . gtg wil check in the night

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah i'm baffled too...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.0hmm, how did you get to that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps i'll give it ago now, ive just finished writing the other minor problem down neatly

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you'll kick yourself when you see this Example 4 at bottom of page

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which example am i meant to be looking at?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1example 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah right, thats very interesting. its just the IV's that don't check hey

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 i've tried this problem myself and it seems that there is a snap with the initial conditions, what do you think?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1hi @chris00 i'm baffled. would love to know how you eventually come out on this

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0maybe since its defined by default in some books to apply Laplace on Certain conditions , u might check =)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'll let you guys know later on in the week when i have a tute on this. cheers guys
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