How can I solve this Cauchy equation $(x - 4)^{2}y'' - 5(x - 4)y' + 9y = 4 - x$? I am trying to solve the following Cauchy-Euler equation
$$(x - 4)^{2}y'' - 5(x - 4)y' + 9y = 4 - x$$
My first step is substituting $x - 4$ by $t$ and the equation becomes
$$t^{2}y'' - 5ty' + 9y = -t$$
before finding the value of $m$ I need to use the chain rule so the equation becomes $\mathrm{d}x/\mathrm{d}t$.  So I need help in this step. I know the chain rule but i don't know how to apply it and use it in this type of equations, so any suggestions?
 A: With the change of variables $y=(x-4)v$, then $  \begin{cases}y'=(x-4)v'+v,\\y''=(x-4)v''+2v'\end{cases}$ where the derivative are taken respect to variable $x$. Then, rewriting the ODE we have
$$(x-4)^{3}v''-3(x-4)^{2}v'+4(x-4)v=-(x-4).$$
Setting $v=(x-4)^{2}u$, then $\begin{cases}v'=(x-4)^{2}u'+2(x-4)u,\\v''=(x-4)^{2}u''+4(x-4)u'+2u\end{cases}$, where the derivative are taken respect to variable $x$. Then, rewriting the ODE we have
$$(x-4)^{5}u''+(x-4)^{4}u'=-(x-4)$$
Reduction of order with the substitution $u'=z$ then $u''=z'$ and rewriting the ODE again we have
$$z'+\frac{1}{x-4}z=\frac{-1}{(x-4)^{4}},\quad x\not=4.$$
Now the ODE is linear of first order and we can use the standard machinery and we have the general implicit solution
$$(x-4)z=\frac{1/2}{(x-4)^2}+C,\quad z=z(x)$$ and where $C$ is an arbitrary constant. Finally, substitution back in order to find the solution $y=y(x)$, that is,
$$y(x)=A(x-4)^3+B(x-4)^3 \log |x-4|-\frac{x}{4}+1,$$
with $A$ and $B$ arbitrary constants.
A: I need to use the chain rule so the equation becomes $\dfrac {dx}{dt}$
$$y'=\dfrac {dy}{dx}=\dfrac {dy}{dt}\dfrac {dt}{dx}=\dfrac {dy}{dt}\dfrac {d(x-4)}{dx}=\dfrac {dy}{dt}$$
Since $t=x-4$.
$$y''=\dfrac {dy'}{dx}=\dfrac {dy'}{dt}\dfrac {dt}{dx}=\dfrac {dy'}{dt}\dfrac {d(x-4)}{dx}=\dfrac {dy'}{dt}=\dfrac {d}{dt}\dfrac {dy}{dt}=\dfrac {d^2y}{dt^2}$$
The DE becomes:
$$t^{2}\dfrac {d^2y}{dt^2} - 5t\dfrac {dy}{dt} + 9y = -t$$
Find $m$ now.
A: Substituting $t=e^u$ to transform the ODE into a simpler one with constant coefficients.
By the chain rule,
$$t=e^u \implies \frac{dt}{du} = e^u \implies \frac{du}{dt} = \frac1t \\
\frac{dy}{dt} = \frac{dy}{du}\cdot\frac{du}{dt} = \frac1t \frac{dy}{du} \\
\frac{d^2y}{dt^2} = -\frac1{t^2}\frac{dy}{du} + \frac1t \frac{d\frac{dy}{du}}{dt} = -\frac1{t^2}\frac{dy}{du} + \frac1t \frac{d\frac{dy}{du}}{du}\frac{du}{dt} = -\frac1{t^2}\frac{dy}{du} + \frac1{t^2} \frac{d^2y}{du^2}$$
Plug everything in and you have
$$y'' - 6y' + 9y = -e^u$$
