Differential equations question !

- anonymous

Differential equations question !

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- 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

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

- anonymous

am I right ?

- anonymous

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

- anonymous

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

- anonymous

Initial conditions do you mean right ?

- anonymous

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

- beginnersmind

"and I must calculate :
âˆ«05y(t)dt"
Maybe you don't need to solve the differential equation in general then

- beginnersmind

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

- anonymous

can you be more specific please

- anonymous

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

- anonymous

I forgot that the ICs are considered null.

- beginnersmind

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

- beginnersmind

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

- beginnersmind

Where the integral with u(t) is known.

- beginnersmind

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

- anonymous

@ganeshie8 any idea ?

- ganeshie8

is the order really 3 ?

- anonymous

yup

- 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

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

- anonymous

well than I will reformulate the problem

- 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

try @dan815

- anonymous

@dan815 are you familiar with matlab or numerical analysis ?

- dan815

first order runge kutta = euler method right

- 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

and you have no initial values?

- dan815

uh whats that for?

- anonymous

Intial conditions are considered null

- dan815

okay so first write a bunch of first order ODes

- anonymous

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

- dan815

|dw:1436025643499:dw|

- dan815

|dw:1436025758838:dw|

- dan815

|dw:1436025814119:dw|

- dan815

solve these in order

- dan815

|dw:1436025883462:dw|

- anonymous

Yeah right I think we should keep my notation as we are talking about vectors here:
|dw:1436025828612:dw|

- dan815

k thats looks fine

- anonymous

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

- dan815

why are you putting it in a matrix

- anonymous

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

- anonymous

MATRIXlaboratory a.k.a MatLab

- dan815

hmm let me think okay

- dan815

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

- 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

|dw:1436026464161:dw|

- dan815

|dw:1436026581332:dw|

- dan815

im wondering if this is really possible in this case hmm

- anonymous

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

- dan815

what is his method?

- anonymous

|dw:1436026645952:dw|

- anonymous

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

- dan815

so she got a new variable for x3

- dan815

so that she doesnt have to worry about the extra vector

- anonymous

then she does another script for the y function

- dan815

|dw:1436026852588:dw|