Derivation using Leibniz's notation with change of variable Given the following differential equation:
$x^2 \cdot \frac{d^2Y}{dx^2} + p \cdot x \cdot \frac{dY}{dx} + q \cdot Y = 0$ with $p,q \in \mathbb{R}$
I have to prove that by using the change of variable $ x = e^t$ the equation will have constant coefficients.
I was able to prove it by taking $Z(t) = Y(e^t) = Y(x)$, deriving $Z$ and then making the necessary substitutions.
Our proffesor also showed us this other way of solving it:
$\frac{dY}{dx} = \frac{dY}{dt} \cdot \frac{dt}{dx} = \frac{dY}{dt} \cdot (\frac{dx}{dt})^{-1} = \frac{dY}{dt} \cdot (e^t)^{-1} = \frac{dY}{dx} \cdot x^{-1}$ 
$\implies$ $p \cdot x \cdot \frac{dY}{dx} = p \cdot \frac{dY}{dt} \cdot x^{-1} = p \cdot \frac{dY}{dt}$
My problem is I was not able to use the same method to write $\frac{d^2Y}{dx^2}$ in function of $t$. How can I use this notation to solve the problem?
 A: I don't think your professor's procedure simplifies things. In the first place we have to establish a formula relating differentiation with respect to $x$ (denoted by a ${}'\>$) with differentiation with respect to $t$ (denoted by a $\dot{}\>$), valid for any function $$u:\quad {\mathbb R}_{>0}\to{\mathbb R},\qquad x\mapsto u(x)\ ,$$ resp., its pullback $$\tilde u:\quad {\mathbb R}\to{\mathbb R},\qquad t\mapsto \tilde u(t):=u\bigl(e^t\bigr)\ .$$
For simplicity this pullback is again denoted by $u$.
The substitution $$x:=e^t\qquad(-\infty<t<\infty)$$ together with the chain rule leads to
$$\dot u={du\over dt}={du\over dx}\cdot{dx\over dt}=u' \>e^t=x\>u'\ .\tag{1}$$
Applying this principle to $\dot u=x\>u'$ we obtain
$$\ddot u=(\dot u)^{\textstyle\cdot}=(x\>u')^{\textstyle\cdot}=x\>(x\>u')'=x\>u'+x^2\>u''\ .\tag{2}$$
From $(1)$ and $(2)$ one deduces
$$xu'=\dot u\ , \quad x^2u''=\ddot u-\dot u\ .$$
In this way the given ODE in terms of the independent variable $x$ transforms into
$$\ddot y-\dot y+p\>\dot y+q\>y=0\ ,$$
or
$$\ddot y+(p-1)\>\dot y+q\>y=0\ .$$
