0

if I'm not wrong, if I have a differential equation like this:

$Dl=a_0k_1 + k_1a_1x + k_1a_2x^2 - k_1l$

and I want to linearize it, I can linearize only the quadratic part (ergo, $k_1a_2x^2$ ). Right?

But then, when I carry out the linearization around $x_0=0$ of that part, I will end up with:

$Dl=a_0k_1 + k_1a_1x + 2k_1a_2x - k_1l$

Am I right? What Am I doing wrong?

Thank you very much!

Huaweu149
  • 11
  • The general formula is $f(x_{0} + h) \approx f(x_0) + hf'(x_{0})$ where $h\approx0$.

    You want an approximation of $f(x) = x^2$ around $x_{0}$. Setting $x = x_{0} + h$ we get $f(x) \approx {x_{0}}^2 + h2x_{0} = 2xx_{0} - {x_{0}}^2$. For $x_{0} = 0$ we simply get $f(x) \approx 0$.

    The linearization of a quadratic around $0$ is $0$.

     – user3257842 Apr 21 '23 at 06:25
  • @user3257842 Thank you very much, linearizing a quadratic around $0$ in indeed $0$, even though it's clear only now that you showed me. Now I have a follow-up question: can you help me in finding another equilibrium point $x_0$ different from $0$? – Huaweu149 Apr 22 '23 at 05:56

0 Answers0