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\[3\frac{d^3y(t)}{ dt }+\frac{ dy(t) }{ dt }+2y(t)=u(t)\]

where u(t) has this form :
|dw:1436021576568:dw|

and I must calculate :\[\int\limits_{0}^{5}y(t) dt\]

\[\int\limits_{0}^{5}y(t)dt=\int\limits_{0}^{2}y1(t)dt+\int\limits_{2}^{5}y2(t)dt\]

am I right ?

there are distinct solutions, but you need ICs to 'glue' them together

what's ICs ?
Am I right with this distinct solutions ?

Initial conditions do you mean right ?

Can someone please guide me to all the steps needed(ICs,...) to help me understand this problem !

If you express y(t) and integrate both sides from 0 to 5 you get a much simpler problem.

can you be more specific please

I mean how can I express something I don't know ?
y(t) is the solution of that equation

I forgot that the ICs are considered null.

It's just an idea but I think it works
|dw:1436022843809:dw|

Assuming all derivates at 0 are zero, I get: |dw:1436023098647:dw|

Where the integral with u(t) is known.

Ok, not sure how to make progress from here. Maybe it doesn't work after all :(

@ganeshie8 any idea ?

is the order really 3 ?

yup

particular solution changes depending on u(t)
but the homogeneous solution remains same

well than I will reformulate the problem

first order runge kutta = euler method right

My question is if in my script of matlab do i need 2 functions:
x1=Ax+5*b
and
x2=Ax+cos2t*b

and you have no initial values?

uh whats that for?

Intial conditions are considered null

okay so first write a bunch of first order ODes

That's for transforming the equation in a first order ODE

|dw:1436025643499:dw|

|dw:1436025758838:dw|

|dw:1436025814119:dw|

solve these in order

|dw:1436025883462:dw|

k thats looks fine

the matrix A that I telled you behind looks like this
0 1 0
0 0 1
-2/3 -1/3 0

why are you putting it in a matrix

x1=Ax+5*b
and
x2=Ax+cos2t*b
|dw:1436026153233:dw|

MATRIXlaboratory a.k.a MatLab

hmm let me think okay

if you want to solve it all in one matrix then, ud need it in the form like

script:
function x1=FCS(t,y)
x1=A*x+b*5;
end
function x2=FCS(t,y)
x2=A*x+b*cos(2t);
end

|dw:1436026464161:dw|

|dw:1436026581332:dw|

im wondering if this is really possible in this case hmm

dan I know that I'm just trying to understand my teacher method

what is his method?

|dw:1436026645952:dw|

than she makes a function :
function xp=FCS(t,y)
xp=A*x+b*u(t);
end

so she got a new variable for x3

so that she doesnt have to worry about the extra vector

then she does another script for the y function

|dw:1436026852588:dw|

now u have a new matrix

|dw:1436026937474:dw|