How to normalise equations of the form $dy/dx=B$ and $d^2y/dx^2=A$?

Question

So I am trying to normalise equations of the form, $$dy/dx=B \mbox{ and } d^{2}y/dx^{2}=A$$ If I define $y^{*}$ as; $$y^{*}=By \Rightarrow dy^{*}/dy=B $$

Is it also then true that, $$d(dy^{*})/dy = B = d^{2}y^{*}/dy = B $$

Answer 1

Your notation is making me uncomfortable. We can say that

$$ y^* = B y \implies dy^* = B dy $$ that I am alright with. You seem to be interested in how to relate $ \frac{d^2 y^* }{dx^2} $ and $ \frac{d^2 y}{dx^2} $ for which we can see

$$ \frac{d^2 y^* }{dx^2} = \frac{d^2}{dx^2} y^* = \frac{d^2}{dx^2} (B y ) = B \frac{d^2 y}{dx^2} $$

which is, I think what you are after. I would not say $ d^2 y^* / dy^2 = B $ as that is false, though if you wanted to see how to do it with the chain rule, I think they way to think of it is

$$ \frac{d^2 y^* }{ dx^2 } = \frac{ d}{dx} \left( \frac{ dy^* }{ dx} \right) = \frac{d}{dx} \left( \frac{ dy^* }{dy} \frac{dy}{dx} \right) = B \frac{d^2 y}{dx^2} $$ Where the last identity follows since $dy^*/dy$ is independent of $x$.

Though...

I also think a good way to think of these sorts of things is dimensionally. Given the equations you have, calling the dimension of $y$, $[Y]$ and the dimension of $x$, $[X]$, you have that

$$ \left[ \frac{dy}{dx} \right] = [ Y X^{-1} ] = [B] $$ and $$ \left[ \frac{ d^2y}{dx^2} \right] = [ Y X^{-2} ] = [A] $$ which would suggest that you should normalize not as you suggested but with

$$ y^* = \frac{A}{B^2} y \qquad x^* = \frac{A}{B} x $$ if you were trying to nondimensionalize your equations. This gives you new equations of the form $$ \frac{dy^*}{dx^*} = 1 \qquad \frac{d^2 y^*}{d(x^*)^2} = 1 $$ which are nice and tidy.