Chain rule for differentiation with a constant, e.g. $y(x(t)), x = \alpha t$

Question

I'm solving the following equation: $$xy'' +y' +k^2xy = 0$$ By setting $x = \alpha t$ and choosing $\alpha$ such that the equation becomes a Bessel equation, shown below with the general solution: $$xy'' +y' +xy = 0$$ $$y = AJ_o(x)+BY_o(x)$$ So if I plug in $x = \alpha t$: $$\alpha ty'' +y'+k^2\alpha ty = 0$$ and since $y = y(x(t))$, we have to do chain differentiation and $$y'' = \frac{d^2y}{dx^2}(\frac{dx}{dt})^2$$ $$y' = \frac{dy}{dx}\frac{dx}{dt}$$ Where $\frac{dx}{dt} = \alpha$, leading to: $$\alpha^3 ty'' +\alpha y'+k^2\alpha ty = 0$$ But I think this is where I'm messing up and I'm supposed to divide by $\alpha$ to get $$\frac{1}{\alpha} ty'' +\frac{1}{\alpha} y'+k^2\alpha ty = 0$$ And then I can multiply by $\alpha$ and set $\alpha = \frac{1}{k}$ and it's back in good form: $$ty'' + y'+ty = 0$$ So how am I messing up the chain rule? Thank you for any help!

Answer 1

If you are confused, don't use the same symbol $y'$ for two different notations :

$y'=\frac{dy}{dx}\quad$ and $\quad y'=\frac{dy}{dt}\quad$ .

Instead of the ambiguous $\quad xy'' +y' +k^2xy = 0\quad $ a less ambiguous notation is : $$x\frac{d^2y}{dx^2} +\frac{dy}{dx} +k^2xy = 0$$

Also, to avoid confusion, use a different symbol for the functions $$y(x)=Y(t)$$ $x = \alpha t \quad\to\quad y(\alpha t)=Y(t) \quad\text{or}\quad y(x)=Y\left(\frac{x}{\alpha} \right) $

$\frac{dy}{dx}=\frac{dY}{dt}\frac{dt}{dx}=\frac{1}{\alpha}\frac{dY}{dt}$

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\frac{d}{dt}\left(\frac{1}{\alpha}\frac{dY}{dt}\right)\frac{dt}{dx}=\frac{1}{\alpha^2}\frac{d^2Y}{dt^2}$

$$x\frac{d^2y}{dx^2} +\frac{dy}{dx} +k^2xy = (\alpha t)\left( \frac{1}{\alpha^2}\frac{d^2Y}{dt^2}\right) +\left( \frac{1}{\alpha}\frac{dY}{dt}\right) +k^2(\alpha t)Y = 0$$

$$ t\frac{d^2Y}{dt^2} + \frac{dY}{dt} +k^2\alpha^2 tY = 0$$ This allows to find a convenient value of $\alpha$ so that $k^2\alpha^2=1$

$$\alpha=\frac{1}{k} \quad\to\quad t\frac{d^2Y}{dt^2} + \frac{dY}{dt}+ tY = 0$$ The general solution of this Bessel ODE is : $$Y(t)=AJ_0(t)+BY_0(t)$$ Coming back to the variable $\quad x=\alpha t=\frac{1}{k}t \quad\to\quad t=kx \quad \text{and}\quad Y(t)=Y(kx)=y(x)$ $$y(x)=AJ_0(kx)+BY_0(kx)$$ NOTE : Do not confuse

$\quad y(x)=AJ_0(kx)+BY_0(kx)\quad$ which is solution of $\quad x\frac{d^2y}{dx^2} +\frac{dy}{dx} +k^2xy = 0$

and

$\quad y(x)=AJ_0(x)+BY_0(x)\quad$ which is solution of $\quad x\frac{d^2y}{dx^2} +\frac{dy}{dx} +xy = 0$

Of course, when one becomes very familiar with the notation $y'$ and with the functions $y(x)$ and $y(t)$ and dummy variables, in case of change of variable, he can use the "ambiguous" notations without mistake.