1. linearizing second order DE's

Recently my controls teacher gave me a problem set. In it he asks us to linearize the following equation: y* = y - y^3 +u (y* is the first derivative of y) about the point y+=1. I found the solution was to take the partial derivative of f(y) where f(y) = y - y^3 evaluated at y+ and to multiply it by the change in y. I then found its transfer function to be: 1/(s+2).

A similar problem is with the equation: y** +3y* +2y = x^3 about the point x+ = 0. I am not sure how to do the problem but my guess is that I will take two partials of x^3 evaluated at x+ and find the transfer function from there. Am I way off the mark here? Do I need to do anything about the 2y term?

2. Re: linearizing second order DE's

suppose you convert the 2nd order DE into a system of 2 first order DEs

$u_1 = y$

$u_2 = y^\prime$

$\begin{pmatrix}u_1^\prime \\ u_2^\prime\end{pmatrix} = \begin{pmatrix}u_2 \\ x^3 - 2u_1 - 3u_2 \end{pmatrix}$

you indicated you know how to linearize a first order DE, so just apply that to the two resulting DEs.

3. Re: linearizing second order DE's

Okay would I evaluate them separately at x+ and then multiply the two respective transfer functions together?

4. Re: linearizing second order DE's

So I have two equations y1* = y2 and y2* = x^3-3y*-2y. One of them seems linear already. Can I just linearize the non-linear one and then multiply the two transfer functions together?

5. Re: linearizing second order DE's

I am confused by "x+". Do you mean just "x"? Yes, one of the equations is already linear. Yes, you can simply linearize the other equation. That gives you a system of two linear equations in two unknown functions. Why would you want to multiply them together?