Linear transformation of a dynamical system

Question

How can I show that $x'=hx(1-x)$ can be transformed to $y'=r-y^2$ using a linear transformation (i.e. $y=mx+b$)? I tackled the problem by substituting $x'$ with $y'/m$ and after algebraic manipulation, I cant get it to simplify down to a constant term minus a square term.

Answer 1

First write $y$ as a linear transformation of $x$ and take the derivative. Then solve for $y$ in terms of $x$. From the linear transformation. Finally equate like constants with $r$, the coefficient of $y$ with $0$, and the coefficient of $y^2$ with $-1$, and solve for all your constants.

$$ \begin{align*} y&= mx + b\\ \Rightarrow y' &= mx'\\ &= mhx(1-x)\\ &= mh\frac{y-b}{m}\left(1 - \frac{y-b}{m}\right)\\ &=h(y-b)\left(\frac{m - (y-b)}{m}\right) \\ &= hy -hb -\frac{h(y-b)^2}{m}\\ &= hy(1 + \frac{2b}{m}) - hb( 1 + \frac{b}{m}) - \frac hmy^2 \end{align*} $$

Equating like terms gives you:

$$ \begin{align*} m &= h\\ b &= -\frac h2 \\ r &= \frac {h^2}{4} \end{align*} $$