Linearization of an equation around fixed points

darkspym7
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Homework Statement


Find the linearization of the equation y' = y(-1+4y-3y^2) about each of the fixed points


The Attempt at a Solution


I think this is correct for finding fixed points:
Set y' = 0 = y(-1+4y+3y^2), so the fixed points are y = 0, 1/3, 1

What exactly does it mean by linearization of the equation around each of the fixed points?
 
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linearization

the linearization of a function f about a, (linearization at x=a) is
L(x)= f(a)+f'(a)(x-a) Its pretty much like a taylor series approximation.
 
example

find the linearization of f(x)=x^2 about x=3

L(x)=f(3)+f'(3)(x-3)
L(x)=9+6(x-3)=6x-9
 
But using those fixed points, f'(a) would always be 0. Are those the correct fixed points?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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