Transformation on the independent variable \(x\) method 1 ode internal name "second_order_change_of_variable_on_x_method_1"

Given ode\begin {equation} y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \tag {A} \end {equation} Let \(\tau =g\left ( x\right ) \) where \(\tau \) is the new independent variable. Applying this to (A) results in (details not shown)\begin {equation} y^{\prime \prime }\left ( \tau \right ) +p_{1}\left ( \tau \right ) y^{\prime }\left ( \tau \right ) +q_{1}\left ( \tau \right ) y\left ( \tau \right ) =r_{1}\left ( \tau \right ) \tag {1} \end {equation} Where \begin {align} p_{1}\left ( \tau \right ) & =\frac {\tau ^{\prime \prime }\left ( x\right ) +p\left ( x\right ) \tau ^{\prime }\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\tag {2}\\ q_{1}\left ( \tau \right ) & =\frac {q\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\tag {3}\\ r_{1}\left ( \tau \right ) & =\frac {r\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}} \tag {4} \end {align}

The idea of the transformation is to determine if ode (1) can be solved instead of (A).

Let \(q_{1}=c^{2}\) where \(c\) is a constant then from (2)

\begin {align} \frac {q\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}} & =c^{2}\nonumber \\ \tau ^{\prime } & =\frac {1}{c}\sqrt {q\left ( x\right ) }\tag {5}\\ \tau ^{\prime \prime } & =\frac {1}{2c}\frac {q^{\prime }\left ( x\right ) }{\sqrt {q\left ( x\right ) }} \tag {5A} \end {align}

Substituting (5,5A) in (2) finds \(p_{1}\left ( \tau \right ) \). If \(p_{1}\left ( \tau \right ) \) is a constant (does not depend on \(x\)) then (1) can be solved for \(y\left ( \tau \right ) \) and (A) is therefore solved for \(y\left ( x\right ) \).