linearise: dy/dt=a[(1-b)/b]+ba^2+sin(c*a) ; where c is constant.

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linearise: dy/dt=a[(1-b)/b]+ba^2+sin(c*a) ; where c is constant.

Mathematics
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i am unsure how to linearise this when only 1 variable is constant. could someone enlighten me how to linearise this with multiple variables?
sorry that should be db/dt
not dy/dt

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hey what does linearise mean exactly?
does that mean to solve the differential equation or something else?
clearly, this is a non-linear function, so to make it linear we try to make a general equation of all the tangents where our condition is at steady state.
|dw:1438924011903:dw|
so we try to fit that curve with linear tangents
where the point we know as a basis is the steady state condition.
http://facstaff.cbu.edu/rprice/lectures/lineariz.html
its along these lines..
this is mearly the fundementals for laplace transforms
@zzr0ck3r do you know how to do this ?
Nah, this is some physics/engineering b.s. :) I spent a summer doing it at Uof O but that was 4 years ago and I forgot...
I was trying to find something easy to follow online but I can't find anything.
haha thanks for helping anyway
we are doing this for process control but i think ive nutted it out
you think you nutted it out? lol I take that as you did it! :) I would like to see your solution if and when you get time.
to linearise ba^2 we need to partial differentiate each term at steady state
\[z=ba^2 \\ z_b=a^2 \\ z_a=2ab ?\]
so we get|dw:1438925115302:dw|
so thats a linear approximation of ab^2
so you apply that with all the other non linear terms
hmm...weird I see where you got the last two terms
so hey would sin(ca) be \[ \approx \sin( c a_{ss})+c \cos(ca)a'\]
though that isn't linear :p
yeah i havent figured the sin term yet haha
might have to use an iterative procedure for that...hmm
this is interesting. its only the sin term which is weird to linearise
suggestions?

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