ode

May 2009
15
0
Can you explain how to linearise

x''+2ax'-x-x^3=Fcos(bt) about x=+1 and -1

Thank you
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Can you explain how to linearise

x''+2ax'-x-x^3=Fcos(bt) about x=+1 and -1

Thank you
To "linearize" a function about x= a, replace it by its tangent line approximation there.

For example, \(\displaystyle (x^3)'= 3x^2\) so, in particular, at x= 1, its derivative is 3. The tangent approximation at x= 1 is y= 3(x- 1)+ 1= 3x- 2.

The linearization of x"+ 2ax'- x- x^3= F cos(bt) about x= 1 is x"+ 2ax'- x- 3x- 2= x"+ 2ax'- 4x- 2= F cos(bt) or x"+ 2ax'- 4x= 2+ F cos(bt).

Do the same at x= -1. At x= -1, the tangent approximation is y= 3(x+1)-1= 3x+ 2.
 
May 2009
15
0
sorry I don't understand tangent approximation. can u explain and the how are we getting the linearised equation.
I think it is x''+2ax'-4x=Fcos(bt)
am I right?