The system described by $x'=2x^2-8$ is linearized about the equilibrium point -2. What is the resulting linearized equation?
Answer is $x'=-8x-16$.
How? I have no idea how it went from the first equation to the 2nd. Thanks.
$\endgroup$ 23 Answers
$\begingroup$If you build the Taylor-McLaurin series at $x=-2$, you obtain $$2x^2-8=0 - 8 (x+2)+ 4 (x+2)^2/2 + O\left((x+2)^3\right)$$ Just retain the first two terms, zero and $-8(x + 2)$, which is a linear approximation of the original expression.
You get the same if you write $$2x^2-8=2\Big((x+2)-2\Big)^2-8=2\Big((x+2)^2-4(x+2)+4\Big)-8=-8(x+2)+2(x+2)^2$$
$\endgroup$ $\begingroup$Let $u=x+2$ so that $u=0$ when $x=-2$ $$x^2=(u-2)^2$$ $$x^2=4-4u+u^2$$ $x^2$ is linerized as $x^2=4-4u=4-4(x+2)$ which is brought back into the ODE :
$$x'=2x^2-8=2(4-4(x+2))-8=8-8(x+2)-8$$ $$x'=-8x-16$$
$\endgroup$ $\begingroup$$f(x) = 2 x^2-8$. We have $f(-2) = 0$ as expected. The linearized system is $y' = f'(-2) y$, which is $y'=-8y$.
If you want to express this as a system based around $-2$ rather than zero, let $z=y-2$, or $y=z+2$, which will give the equation: $z'=-8 z -16$.
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