3.9.8.5.3 Example 9  \(x^{2}y^{\prime \prime }-\left ( 2a-1\right ) xy^{\prime }+a^{2}y=0\)

The above is standard Euler ode. But below shows how to apply these transformations if one did not know this.

Trying change of variable on independent variable first.  Let \(\tau =g\left ( x\right ) \) where \(z\) will be the new independent variable. Writing the ode in normal form gives

\begin{align*} y^{\prime \prime }+py^{\prime }+qy & =r\\ p & =\frac {\left ( 1-2a\right ) }{x}\\ q & =\frac {a^{2}}{x^{2}}\\ r & =0 \end{align*}

Applying \(\tau =g\left ( x\right ) \) transformation on the above ode gives

\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}

Approach 1. Let \(q_{1}=c^{2}\) where \(c^{2}\) is some constant. This implies

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

If \(p_{1}\) is constant using this \(\tau \) then (1) is a second order constant coefficient ode which can be solved easily. This ode has \(q=\frac {a^{2}}{x^{2}}\), therefore from (5) assuming positive

\begin{align*} \tau ^{\prime } & =\frac {1}{c}\sqrt {\frac {a^{2}}{x^{2}}}\\ & =\frac {a}{cx}\end{align*}

Hence \(p_{1}\) becomes using (2)

\begin{align*} p_{1}\left ( \tau \right ) & =\frac {\tau ^{\prime \prime }\left ( x\right ) +p\tau ^{\prime }\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\\ & =\frac {\left ( 1-2a\right ) c}{x}\end{align*}

Which is not a constant. So this transformation failed.

Approach 2  Let \(p_{1}=0\). If with this choice \(q_{1}\) becomes a constant or a constant divided by \(\tau ^{2}\) then (2) can be integrated. \(p_{1}=0\) implies from (2) that

\begin{align*} \tau ^{\prime \prime }+p\tau ^{\prime } & =0\\ \tau & =\int e^{-\int pdx}dx\\ & =\int e^{-\int \frac {\left ( 1-2a\right ) }{x}dx}dx\\ & =\int x^{2a-1}dx\\ & =\frac {x^{2a}}{2a}\end{align*}

Using this then \(q_{1}\) becomes

\begin{align} q_{1} & =\frac {q}{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\nonumber \\ & =\frac {\left ( \frac {a^{2}}{x^{2}}\right ) }{\left ( x^{2a-1}\right ) ^{2}}\nonumber \\ & =\frac {a^{2}}{x^{2}x^{4a-2}}\nonumber \\ & =\frac {a^{2}}{x^{4a}} \tag {6}\end{align}

Which is not constant. But \(\tau ^{2}=\left ( \frac {x^{2a}}{2a}\right ) ^{2}=\frac {x^{4a}}{4a^{2}}\). Hence \(q_{1}=\frac {1}{4}\frac {1}{\tau ^{2}}\). Hence this transformation works. Eq (2) becomes

\begin{align} y^{\prime \prime }+p_{1}y^{\prime }+q_{1}y & =0\tag {1}\\ y^{\prime \prime }+\frac {1}{4}\frac {1}{\tau ^{2}}y & =0\nonumber \\ 4\tau ^{2}y^{\prime \prime }+y & =0\nonumber \end{align}

Which is standard Euler ode which can be solved easily. Giving

\[ y\left ( \tau \right ) =A\sqrt {\tau }+B\sqrt {\tau }\ln \left ( \tau \right ) \]

But \(\tau =\frac {x^{2a}}{2a}\). Hence the above becomes

\begin{align*} y\left ( x\right ) & =A\sqrt {\frac {x^{2a}}{2a}}+B\sqrt {\frac {x^{2a}}{2a}}\ln \left ( \frac {x^{2a}}{2a}\right ) \\ & =A\sqrt {\frac {x^{2a}}{2a}}+B\sqrt {\frac {x^{2a}}{2a}}\ln \left ( \frac {x^{2a}}{2a}\right ) \\ & =A_{1}x^{a}+B_{1}x^{a}\ln \left ( \frac {x^{2a}}{2a}\right ) \end{align*}