I have to find a necessary and a sufficient condition for the functions $p$ and $q$ so that the linear differential equation : $y''+p(x)y'+q(x)y=0$ can be converted in a linear differential equation with constant coefficients by changing the independent variable of the equation.
Can someone help with this or point me in the right direction? Thanks in advance!
[EDIT:] First of all, I'm not sure if I should post this as an answer to my own question or as an edit. My apologies.
With the tip of Antonio I've did the following, but I'm still not sure if it's completely correct. $\Phi(t)=y(x(t))$
$\Phi^{'}(t)=y^{'}(x(t)).x^{'}(t)$
$\Phi^{''}(t)=y^{''}(x(t)).(x^{'}(t))^{2}+y^{'}(x(t)).x^{''}(t)$
$y^{''}(t)+p(x)y^{'}(t)+q(x)y(t)=0$
$\frac{1}{q(x)}y^{''}(t)+\frac{p(x)}{q(x)}y^{'}(t)+y(t)=0$
$\frac{1}{q(x)}=(x^{'}(t))^{2}\Longrightarrow x^{'}(t)=\frac{1}{\sqrt{q(x)}}$
Condition 1: $\forall x:q(x)>0 $
$x^{''}(t)=\frac{x^{'}(t).q^{'}(x)}{2\sqrt{q(x)}}$
$x^{''}(t)=\frac{q^{'}(x)}{2.q(x)}$
Condition 2: $p(x)=\frac{1}{2}q^{'}(x)$
