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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: $p(x)=\frac{1}{2}q^{'}(x)$

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What have you tried? I would suggest letting $x=f(z)$ then writing all of the derivatives with respect to $x$ in terms of derivatives with respect to $z$. –  Antonio Vargas Jan 25 '13 at 21:09
    
@AntonioVargas I've tried something (see edit), but I'm still not sure if I'm following the correct path. –  tim_a Jan 26 '13 at 9:08
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