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

- anonymous

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

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- Jhannybean

@Empty

- anonymous

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

- 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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## More answers

- IrishBoy123

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

- anonymous

how does y=x'?

- anonymous

they are just both functions of time

- 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

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

- ganeshie8

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

- ganeshie8

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

- anonymous

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

- 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

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- ganeshie8

are you doing control theory ?

- anonymous

yep process control

- IrishBoy123

you solve those as a pair
laplace then normal algebra

- ganeshie8

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

- anonymous

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

- ganeshie8

|dw:1440405030249:dw|

- ganeshie8

i might be wrong though as idk abc of process control

- anonymous

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

- anonymous

is that the way to go?

- Empty

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

- 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

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

- anonymous

yea true

- Empty

Something doesn't seem right here haha

- anonymous

i'll give that a miss then haha

- anonymous

i'll follow the links process, and see if i get an answer.

- ganeshie8

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

- anonymous

perhaps the initial conditions are misspelt?

- 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

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

- anonymous

my approach was simply taking the derivative of the integral

- anonymous

\[y''-y=0\]

- 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

simply plugin y=cosh(t) and check ?

- 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

but we are using laplace transforms here

- Empty

Oh right I forgot about your initial conditions

- anonymous

haha no problem

- ganeshie8

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

- anonymous

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

- 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

yeah fair call

- anonymous

your simply integrating from 0 to 0

- ganeshie8

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

- 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

yeah i'm baffled too...

- 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

hmm, how did you get to that?

- anonymous

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

- IrishBoy123

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

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- anonymous

which example am i meant to be looking at?

- 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

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

- 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

snag*

- 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

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

- ikram002p

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

- 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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