linearise:

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

Question

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

anonymous · Answer

i am unsure how to linearise this when only 1 variable is constant. could someone enlighten me how to linearise this with multiple variables?

anonymous · Answer

sorry that should be db/dt

anonymous · Answer

not dy/dt

freckles · Answer

hey what does linearise mean exactly?

freckles · Answer

does that mean to solve the differential equation or something else?

anonymous · Answer

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.

anonymous · Answer

|dw:1438924011903:dw|

anonymous · Answer

so we try to fit that curve with linear tangents

anonymous · Answer

where the point we know as a basis is the steady state condition.

anonymous · Answer

http://facstaff.cbu.edu/rprice/lectures/lineariz.html

anonymous · Answer

its along these lines..

anonymous · Answer

this is mearly the fundementals for laplace transforms

freckles · Answer

@zzr0ck3r do you know how to do this ?

zzr0ck3r · Answer

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...

freckles · Answer

I was trying to find something easy to follow online but I can't find anything.

anonymous · Answer

haha thanks for helping anyway

anonymous · Answer

we are doing this for process control but i think ive nutted it out

freckles · Answer

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.

anonymous · Answer

to linearise ba^2 we need to partial differentiate each term at steady state

freckles · Answer

$$z=ba^2 \ z_b=a^2 \ z_a=2ab ?$$

anonymous · Answer

so we get|dw:1438925115302:dw|