Change of Variables in the Differential Equation $\epsilon \frac{d^2u}{dt^2}-a(t)\frac{du}{dt}+b(t)u=0$ For the ODE 
$$\epsilon \frac{d^2u}{dt^2}-a(t)\frac{du}{dt}+b(t)u=0$$
what does one have to do in order to express it in terms of $T=\frac{1-t}{e^\nu}$?
Here's my attempt: 
$$\epsilon \frac{d^2u}{d[(1-t)/\epsilon^\nu)]^2}-a(T)\frac{du}{d[(1-t)/\epsilon^\nu)]}+b(T)u=\\
\,\,\,\epsilon^{\nu^2+1} \frac{d^2u}{d[(1-t)]^2}-\epsilon^\nu a(T)\frac{du}{d[(1-t)]}+b(T)u=0$$ 
But I need to express this in the form $\epsilon^\xi u'' -\epsilon^\zeta a u' + bu=0$, and I'm quite tangled up in the variables. Would appreciate some advice.
 A: Basically, you want to write $u(t)$ as a function of the variable $T(t)$, so say
\begin{equation}
u(t) = U(T(t)),\qquad (1)
\end{equation}
with the specific choice for $T(t)$
\begin{equation}
 T(t) = e^{-\nu}(1-t). \qquad (2)
\end{equation}
For example, if you had the function $u(t) = -\nu+\log (1-t)$, you could write this as $U(T(t))$ with $U(T) = \log T$.
Now, using $(1)$, you can re-express the derivative of $u$ to $t$ as
\begin{equation}
 \frac{\text{d} u}{\text{d} t} = \frac{\text{d}}{\text{d} t} U(T(t)) = U'(T(t))\,T'(t)\qquad (3)
\end{equation}
by the chain rule. Here, $U'(T(t))$ is the derivative of $U$ to $T$, evaluated at the value $T(t)$; $T'(t)$ is the derivative of $T$ to $t$, which is in this case $(2)$ equal to
\begin{equation}
T'(t) = \frac{\text{d}}{\text{d} t} \Big(e^{-\nu}(1-t)\Big) = - e^{-\nu}.
\end{equation}
Using $(3)$, we can also derive an expression for the second derivative of $u$ to $t$, as
\begin{equation}
 \frac{\text{d}^2 u}{\text{d} t^2} = \frac{\text{d}}{\text{d} t} \left( \frac{\text{d} u}{\text{d} t}\right) = \frac{\text{d}}{\text{d} t} \Big( U'(T(t))\,T'(t)\Big) = U''(T(t))\,\left[T'(t)\right]^2 + U'(T(t))\, T''(t)
\end{equation}
by the product rule and the chain rule. Note that in our particular choice $(2)$ for $T(t)$, $T'(t)$ is constant, so $T''(t) = 0$. Therefore, we see that
\begin{align}
 \frac{\text{d} u}{\text{d} t} &= -e^{-\nu} \frac{\text{d}U}{\text{d} T},\\
 \frac{\text{d}^2 u}{\text{d} t^2} &= e^{-2 \nu} \frac{\text{d}^2 U}{\text{d} T^2}.
\end{align}
Substituting this in the original ODE yields
\begin{equation}
\epsilon\, e^{-2 \nu} \frac{\text{d}^2 U}{\text{d} T^2} + A(T)\, e^{-\nu} \frac{\text{d}U}{\text{d} T} + B(T)\, U = 0,
\end{equation}
with $a(t) = A(T(t))$ and $b(t) = B(T(t))$.
A: First notice that
$$\matrix{
   {T = T(t) \to } & {T = {{1 - t} \over {{e^\nu }}}}  \cr 
 } $$
Then define
$$\eqalign{
  & f = f(t)  \cr 
  & u = u(t) = \left( {f \circ T} \right)(t) \cr} $$
and according to chain-rule you can conclude that
$$\eqalign{
  & {{du} \over {dt}} = \left( {{{df} \over {dt}} \circ T} \right){{dT} \over {dt}} =  - {e^{ - \nu }}\left( {{{df} \over {dt}} \circ T} \right)  \cr 
  & {{{d^2}u} \over {d{t^2}}} = {d \over {dt}}\left( {{{df} \over {dt}} \circ T} \right){{dT} \over {dt}} + \left( {{{df} \over {dt}} \circ T} \right){{{d^2}T} \over {d{t^2}}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\, = \left( {{{{d^2}f} \over {d{t^2}}} \circ T} \right){\left( {{{dT} \over {dt}}} \right)^2} + \left( {{{df} \over {dt}} \circ T} \right){{{d^2}T} \over {d{t^2}}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\, = {e^{ - 2\nu }}\left( {{{{d^2}f} \over {d{t^2}}} \circ T} \right) \cr} $$
Next, substituting this into the ODE we find that
$$\eqalign{
  & \varepsilon {e^{ - 2\nu }}\left( {{{{d^2}f} \over {d{t^2}}} \circ T} \right) + {e^{ - \nu }}a.\left( {{{df} \over {dt}} \circ T} \right) + b.\left( {f \circ T} \right) = 0  \cr 
  & \left[ {\varepsilon {e^{ - 2\nu }}\left( {{{{d^2}f} \over {d{t^2}}} \circ T} \right) + {e^{ - \nu }}a.\left( {{{df} \over {dt}} \circ T} \right) + b.\left( {f \circ T} \right)} \right] \circ {T^{ - 1}} = 0 \circ T^{-1}  \cr 
  & \varepsilon {e^{ - 2\nu }}\left( {{{{d^2}f} \over {d{t^2}}} \circ T \circ {T^{ - 1}}} \right) + {e^{ - \nu }}\left( {a \circ {T^{ - 1}}} \right).\left( {{{df} \over {dt}} \circ T \circ {T^{ - 1}}} \right) + \left( {b \circ {T^{ - 1}}} \right).\left( {f \circ T \circ {T^{ - 1}}} \right) = 0 \cr} $$
Finally, considering that $T \circ {T^{ - 1}} = 1$ and $g \circ 1 =g$ where $1$ is the identity function and $g$ is any arbitrary function,  we will get
$$\varepsilon {e^{ - 2\nu }}{{{d^2}f} \over {d{t^2}}} + {e^{ - \nu }}\left( {a \circ {T^{ - 1}}} \right).{{df} \over {dt}} + \left( {b \circ {T^{ - 1}}} \right).f = 0$$
