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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am unsure how to linearise this when only 1 variable is constant. could someone enlighten me how to linearise this with multiple variables?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry that should be db/dt

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hey what does linearise mean exactly?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1does that mean to solve the differential equation or something else?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0clearly, this is a nonlinear function, so to make it linear we try to make a general equation of all the tangents where our condition is at steady state.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438924011903:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we try to fit that curve with linear tangents

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where the point we know as a basis is the steady state condition.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its along these lines..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is mearly the fundementals for laplace transforms

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@zzr0ck3r do you know how to do this ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Nah, 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...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I was trying to find something easy to follow online but I can't find anything.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha thanks for helping anyway

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we are doing this for process control but i think ive nutted it out

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to linearise ba^2 we need to partial differentiate each term at steady state

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[z=ba^2 \\ z_b=a^2 \\ z_a=2ab ?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we getdw:1438925115302:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so thats a linear approximation of ab^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you apply that with all the other non linear terms

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hmm...weird I see where you got the last two terms

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so hey would sin(ca) be \[ \approx \sin( c a_{ss})+c \cos(ca)a'\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1though that isn't linear :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i havent figured the sin term yet haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0might have to use an iterative procedure for that...hmm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is interesting. its only the sin term which is weird to linearise
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