I am struggling to understand the solution to the following problem:
Transform the given initial value problem into an equivalent problem with the initial point at the origin: $\frac{dy}{dt} = 4 - y^3, y(-1) = 2$
I don't understand the use of the chain rule in writing $\frac{d \omega}{ds}$ in terms of $\frac{dy}{dt}$ in the accepted solution to this problem here. In eq. (2) in the accepted solution, how are $\frac{d \omega}{dy}$ and $\frac{dy}{dt}$ related. In eq. (1), $\omega$ is the substitution for the dependent variable $y$ and $t$ is not in the equation. So why are you doing $\frac{d \omega}{dt}$? Similar question with eq. (3). I don't get how all these derivatives are related.
I learned the chain rule in terms of composite functions, such as $(f \circ g)\prime(x) = f\prime(g(x))g\prime(x)$, but I don't understand how the equations in this problem are related in a similar way. Basically, what equation(s) exactly, and in what order, is the chain rule being applied to. Thanks!
The relationship between the functions $w$ and $y$ is $$ w(s) = y(s-1)-2 . $$ This is just a matter of shifting the coordinate system to put the origin where you want it; note that $y(-1)=2$ implies that $w(0) = y(0-1)-2 = 2-2 = 0$.
Then the chain rule gives $$ w'(s) = y'(s-1) \cdot 1 - 0 = y'(s-1) , $$ which is what the notation “$dw/ds=dy/dt$” (with $t=s-1$) means.