Can you explain how to linearise

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

Thank you

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- May 11th 2010, 03:21 AM #1

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- May 11th 2010, 04:34 AM #2

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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 11th 2010, 05:23 AM #3

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