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Math Help - ode

  1. #1
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    ode

    Can you explain how to linearise

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

    Thank you
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  2. #2
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    Quote Originally Posted by jerad View Post
    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, (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.
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  3. #3
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    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?
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