3.9.8.5 Example 5. \(x^{2}y^{\prime \prime }-xy^{\prime }+\left ( -x^{2}-\frac {1}{4}\right ) y=0\)

Writing the ode in normal form gives

\begin{align*} y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y & =r\qquad \\ p & =\frac {-1}{x}\qquad x\neq 0\\ q & =-\frac {x^{2}+\frac {1}{4}}{x^{2}}\\ r & =0 \end{align*}

Trying change of variable on the independent variable as above.  Let \(\tau =g\left ( x\right ) \) where \(\tau \) will be the new independent variable. Applying this transformation results in

\begin{equation} y^{\prime \prime }+p_{1}y^{\prime }+q_{1}y=r_{1} \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 & =\frac {1}{c}\int \sqrt {q}dx \tag {5}\end{align}

If with this \(\tau \), then \(p_{1}\) turns out to be constant, then it means (1) is second order constant coefficient ode which is easily solved. Applying this on the given ode then (5)

\begin{align*} \tau & =\frac {1}{c}\int \sqrt {-\frac {x^{2}+\frac {1}{4}}{x^{2}}}dx\\ & =\frac {1}{2c}\sqrt {-4x^{2}-1}+\arctan \left ( \frac {1}{\sqrt {-4x^{2}-1}}\right ) \end{align*}

Hence (2) now becomes

\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 ( 8x^{2}+4\right ) c}{\left ( -4x^{2}-1\right ) ^{\frac {3}{2}}}\end{align*}

Which is not constant. Therefore this transformation did not work.

Approach 2  Let \(p_{1}=0\). If with this choice now \(q_{1}\) becomes 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 {1}{x}dx}dx\\ & =\int e^{\ln x}dx\\ & =\int xdx\\ & =\frac {x^{2}}{2}\end{align*}

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

\begin{align*} q_{1} & =\frac {q}{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\\ & =\frac {-\frac {x^{2}+\frac {1}{4}}{x^{2}}}{x^{2}}\\ & =-\frac {1}{x^{4}}\left ( x^{2}+\frac {1}{4}\right ) \end{align*}

Which is not constant. Trying change of variable on the dependent variable (first method). This method assumes

\[ y=v\left ( x\right ) z\left ( x\right ) \]

The Liouville ode invariant is

\begin{align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =-\frac {x^{2}+\frac {1}{4}}{x^{2}}-\frac {1}{2}\frac {d}{dx}\left ( \frac {-1}{x}\right ) -\frac {1}{4}\left ( \frac {-1}{x}\right ) ^{2}\\ & =-\frac {1}{x^{2}}\left ( x^{2}+1\right ) \end{align*}

Which is not constant. Hence this method does not work. One way to solve this is as a Bessel ODE. I have many examples how to do this on my main page.

3.9.8.5.0.1 Example 6. \(\left ( x^{2}-1\right ) y^{\prime \prime }-2xy^{\prime }+2y=0\) Writing the ode in normal form gives

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

Trying change of variable on the independent variable as above.  Let \(\tau =g\left ( x\right ) \) where \(\tau \) will be the new independent variable. Applying this transformation results in

\begin{equation} y^{\prime \prime }+p_{1}y^{\prime }+q_{1}y=r_{1} \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 & =\frac {1}{c}\int \sqrt {q}dx \tag {5}\end{align}

If with this \(\tau \), then \(p_{1}\) turns out to be constant, then it means (1) is second order constant coefficient ode. Applying this on the given ode (5) becomes

\begin{align*} \tau & =\frac {1}{c}\int \sqrt {\frac {2}{x^{2}-1}}dx\\ & =\frac {1}{c}\sqrt {2}\ln \left ( x+\sqrt {x^{2}}-1\right ) \end{align*}

Hence (2) reduces to

\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 {3\sqrt {2}cx}{\sqrt {\frac {1}{x^{2}-1}}\left ( 2x^{2}-2\right ) }\end{align*}

Which is not constant. This transformation did not work.

Approach 2  Let \(p_{1}=0\). If with this choice now \(q_{1}\) becomes constant or a constant divided by \(\tau ^{2}\) then (2) can be easily 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 {2x}{x^{2}-1}dx}dx\\ & =\int \left ( x^{2}-1\right ) dx \end{align*}

Hence \(q_{1}\) becomes

\begin{align*} q_{1} & =\frac {q}{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\\ & =\frac {\frac {2}{x^{2}-1}}{\left ( x^{2}-1\right ) ^{2}}\\ & =\frac {2}{\left ( x^{2}-1\right ) ^{3}}\end{align*}

Which is not constant. This transformation did not work.

Trying change of variable on the dependent variable (first method). This method assumes that

\[ y=v\left ( x\right ) z\left ( x\right ) \]

The Liouville ode invariant is

\begin{align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\left ( \frac {2}{x^{2}-1}\right ) -\frac {1}{2}\frac {d}{dx}\left ( \frac {-2x}{x^{2}-1}\right ) -\frac {1}{4}\left ( \frac {-2x}{x^{2}-1}\right ) ^{2}\\ & =-\frac {3}{\left ( x^{2}-1\right ) ^{2}}\end{align*}

Which is not constant and not constant divided by \(x^{2}\). Hence this transformation also did not work.

Trying the Lagrange adjoint ode method. From above the adjoint ode is

\[ z^{\prime \prime }-\frac {d\left ( zp\right ) }{dx}+zq=0 \]

For some unknown function \(z\left ( x\right ) \). Hence it becomes

\begin{align*} z^{\prime \prime }-\frac {d}{dx}\left ( z\left ( \frac {-2x}{x^{2}-1}\right ) \right ) +z\left ( \frac {2}{x^{2}-1}\right ) & =0\\ z^{\prime \prime }-\left ( -\frac {2z^{\prime }x}{x^{2}-1}+\frac {4zx^{2}}{\left ( x^{2}-1\right ) ^{2}}-\frac {2z}{x^{2}-1}\right ) +z\left ( \frac {2}{x^{2}-1}\right ) & =0\\ z^{\prime \prime }+\frac {2x}{x^{2}-1}z^{\prime }-\frac {4x^{2}+4\left ( x^{2}-1\right ) }{\left ( x^{2}-1\right ) ^{2}}z & =0 \end{align*}

Clearly this is just as hard to solve as the original ode So this method does it work.

Trying integrating factor method. For this to work the condition is that \(\frac {1}{2}\left ( p^{\prime }+\frac {1}{2}p^{2}\right ) =q\). Applying this on the current ode gives

\begin{align*} \frac {1}{2}\left ( p^{\prime }+\frac {1}{2}p^{2}\right ) & =q\\ \frac {1}{2}\left ( \frac {d}{dx}\left ( \frac {-2x}{x^{2}-1}\right ) +\frac {1}{2}\left ( \frac {-2x}{x^{2}-1}\right ) ^{2}\right ) & =\frac {2}{x^{2}-1}\\ \frac {\left ( 2x^{2}+1\right ) }{\left ( x^{2}-1\right ) ^{2}} & =\frac {2}{x^{2}-1}\\ \frac {2x^{2}+1}{x^{2}-1} & =2 \end{align*}

Which is not true. Hence there is no integrating factor.

Trying transformation on the dependent variable (second method). This method assumes

\[ y=v\left ( x\right ) x^{n}\]

This works only if (7A) given in the introduction is satisfied.

\begin{equation} \left ( n\left ( n-1\right ) x^{n-2}+npx^{n-1}+qx^{n}\right ) =0 \tag {7A}\end{equation}

Applying this on the current ode example gives

\[ \left ( n\left ( n-1\right ) x^{n-2}+n\left ( \frac {-2x}{x^{2}-1}\right ) x^{n-1}+\left ( \frac {2}{x^{2}-1}\right ) x^{n}\right ) =0 \]

Trying \(n=1\) the above becomes

\[ \left ( \left ( \frac {-2x}{x^{2}-1}\right ) +\left ( \frac {2}{x^{2}-1}\right ) x\right ) =0 \]

Hence this transformation works for \(n=1\). Therefore \(y=v\left ( x\right ) x\). eq (7) in the introduction now reduces to

\begin{align} x^{n}v^{\prime \prime }+\left ( 2x^{n-1}n+x^{n}p\right ) v^{\prime }+\left ( n\left ( n-1\right ) x^{n-2}+npx^{n-1}+qx^{n}\right ) v & =r\tag {7}\\ v^{\prime \prime }+\frac {\left ( xp+2\right ) }{x}v^{\prime } & =0\nonumber \end{align}

Which now can be solved using substitution \(u=v^{\prime }\).

\[ u^{\prime }+\frac {\left ( xp+2\right ) }{x}u=r \]

Which is linear first order ode. Once \(u\) is found, then \(v\) is found by integration. Hence \(y\) is now found. Hence

\[ u^{\prime }-\frac {2}{x^{3}-x}u=0 \]

Which has the solution \(u=c_{1}\frac {x^{2}}{x^{2}-1}\). Hence \(v^{\prime }=c_{1}\frac {x^{2}}{x^{2}-1}\). Integrating gives \(v=c_{1}\left ( x+\frac {1}{x}\right ) +c_{2}\). Therefore \(y=xv=c_{1}\left ( x^{2}+1\right ) +c_{2}x\)

This was an example where only the transformation on the dependent second method \(y=v\left ( x\right ) x^{n}\) worked.