anonymous
  • anonymous
Differential equations question !
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[3\frac{d^3y(t)}{ dt }+\frac{ dy(t) }{ dt }+2y(t)=u(t)\]
anonymous
  • anonymous
where u(t) has this form : |dw:1436021576568:dw|
anonymous
  • anonymous
and I must calculate :\[\int\limits_{0}^{5}y(t) dt\]

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anonymous
  • anonymous
@oldrin.bataku never used that before but my question is: Does my diff eq has two solutions: y1(t) for t [0,2] y2(t) for t>2 ??
anonymous
  • anonymous
\[\int\limits_{0}^{5}y(t)dt=\int\limits_{0}^{2}y1(t)dt+\int\limits_{2}^{5}y2(t)dt\]
anonymous
  • anonymous
am I right ?
anonymous
  • anonymous
there are distinct solutions, but you need ICs to 'glue' them together
anonymous
  • anonymous
what's ICs ? Am I right with this distinct solutions ?
anonymous
  • anonymous
Initial conditions do you mean right ?
anonymous
  • anonymous
Can someone please guide me to all the steps needed(ICs,...) to help me understand this problem !
beginnersmind
  • beginnersmind
"and I must calculate : ∫05y(t)dt" Maybe you don't need to solve the differential equation in general then
beginnersmind
  • beginnersmind
If you express y(t) and integrate both sides from 0 to 5 you get a much simpler problem.
anonymous
  • anonymous
can you be more specific please
anonymous
  • anonymous
I mean how can I express something I don't know ? y(t) is the solution of that equation
anonymous
  • anonymous
I forgot that the ICs are considered null.
beginnersmind
  • beginnersmind
It's just an idea but I think it works |dw:1436022843809:dw|
beginnersmind
  • beginnersmind
Assuming all derivates at 0 are zero, I get: |dw:1436023098647:dw|
beginnersmind
  • beginnersmind
Where the integral with u(t) is known.
beginnersmind
  • beginnersmind
Ok, not sure how to make progress from here. Maybe it doesn't work after all :(
anonymous
  • anonymous
@ganeshie8 any idea ?
ganeshie8
  • ganeshie8
is the order really 3 ?
anonymous
  • anonymous
yup
anonymous
  • anonymous
actually I don't need to solve this eq but rather I need to know if my presumption is right that the eq got two distinct solutins depending on u(t); am I right ?
ganeshie8
  • ganeshie8
particular solution changes depending on u(t) but the homogeneous solution remains same
anonymous
  • anonymous
well than I will reformulate the problem
anonymous
  • anonymous
this eq needs to be solved using numerical analysis method RUNGE-Kutta in MatLab so I must transform this equation in a first degree equations
ganeshie8
  • ganeshie8
try @dan815
anonymous
  • anonymous
@dan815 are you familiar with matlab or numerical analysis ?
dan815
  • dan815
first order runge kutta = euler method right
anonymous
  • anonymous
My question is if in my script of matlab do i need 2 functions: x1=Ax+5*b and x2=Ax+cos2t*b
dan815
  • dan815
and you have no initial values?
dan815
  • dan815
uh whats that for?
anonymous
  • anonymous
Intial conditions are considered null
dan815
  • dan815
okay so first write a bunch of first order ODes
anonymous
  • anonymous
That's for transforming the equation in a first order ODE
dan815
  • dan815
|dw:1436025643499:dw|
dan815
  • dan815
|dw:1436025758838:dw|
dan815
  • dan815
|dw:1436025814119:dw|
dan815
  • dan815
solve these in order
dan815
  • dan815
|dw:1436025883462:dw|
anonymous
  • anonymous
Yeah right I think we should keep my notation as we are talking about vectors here: |dw:1436025828612:dw|
dan815
  • dan815
k thats looks fine
anonymous
  • anonymous
the matrix A that I telled you behind looks like this 0 1 0 0 0 1 -2/3 -1/3 0
dan815
  • dan815
why are you putting it in a matrix
anonymous
  • anonymous
x1=Ax+5*b and x2=Ax+cos2t*b |dw:1436026153233:dw|
anonymous
  • anonymous
MATRIXlaboratory a.k.a MatLab
dan815
  • dan815
hmm let me think okay
dan815
  • dan815
if you want to solve it all in one matrix then, ud need it in the form like
anonymous
  • anonymous
script: function x1=FCS(t,y) x1=A*x+b*5; end function x2=FCS(t,y) x2=A*x+b*cos(2t); end
dan815
  • dan815
|dw:1436026464161:dw|
dan815
  • dan815
|dw:1436026581332:dw|
dan815
  • dan815
im wondering if this is really possible in this case hmm
anonymous
  • anonymous
dan I know that I'm just trying to understand my teacher method
dan815
  • dan815
what is his method?
anonymous
  • anonymous
|dw:1436026645952:dw|
anonymous
  • anonymous
than she makes a function : function xp=FCS(t,y) xp=A*x+b*u(t); end
dan815
  • dan815
so she got a new variable for x3
dan815
  • dan815
so that she doesnt have to worry about the extra vector
anonymous
  • anonymous
then she does another script for the y function
dan815
  • dan815
|dw:1436026852588:dw|
dan815
  • dan815
now u have a new matrix
dan815
  • dan815
|dw:1436026937474:dw|