3.9.8.1 Example 1. \(xy^{\prime \prime }+2y^{\prime }-xy=0\)

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 {2}{x}\\ q & =-1\\ 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 \(q=-1\), therefore from (3)

\[ \tau ^{\prime }=\frac {1}{c}\sqrt {-1}\]

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 {0+\left ( 2x^{-1}\right ) \frac {1}{c}\sqrt {-1}}{\frac {-1}{c^{2}}}\\ & =-2x^{-1}\sqrt {-1}c \end{align*}

Which is not a constant. So this transformation failed.

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 2x^{-1}dx}dx\\ & =\int x^{-2}dx\\ & =\frac {-1}{x}\end{align*}

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

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

Which is not constant and nor a constant divided by \(\tau ^{2}\). So this transformation did not work.

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

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

Substituting this in the given ode results in new ode where the dependent variable is \(v\) and not \(y\) which can be found to be

\[ v^{\prime \prime }\left ( x\right ) +\left ( p+\frac {2}{z}z^{\prime }\left ( x\right ) \right ) v^{\prime }\left ( x\right ) +\frac {1}{z}\left ( z^{\prime \prime }\left ( x\right ) +pz^{\prime }\left ( x\right ) +qz\left ( x\right ) \right ) v\left ( x\right ) =\frac {r}{z}\]

Let \(p+\frac {2}{z}z^{\prime }\left ( x\right ) =0\). Solving gives \(z=e^{-\int \frac {p}{2}dx}\). With this choice the above ode becomes

\[ v^{\prime \prime }+\frac {1}{z}\left ( z^{\prime \prime }+pz^{\prime }+qz\right ) v=\frac {r}{z}\]

Applying \(z=e^{-\int \frac {p}{2}dx}\) to the above reduces it to

\begin{equation} v^{\prime \prime }+q_{1}v=r_{1} \tag {6}\end{equation}

Where

\begin{align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ r_{1} & =re^{\frac {1}{2}\int pdx}\end{align*}

If \(q_{1}\) turns out to be constant or a constant divided by \(x^{2}\) with this choice of \(z\), then \(v\) is solved for from (6) and the solution to the original ode is obtained. Applying this method on the given ode gives

\begin{align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{-\int x^{-1}dx}\\ & =e^{-\ln x}\\ & =x^{-1}\end{align*}

Hence

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

Since \(q_{1}\) is constant, then this transformation works. Eq (6) now becomes

\[ v^{\prime \prime }-v=0 \]

The solution is

\[ v=c_{1}e^{-x}+c_{2}e^{x}\]

Therefore, since \(z=x^{-1}\) then

\begin{align*} y & =v\left ( x\right ) z\left ( x\right ) \\ & =\frac {1}{x}\left ( c_{1}e^{-x}+c_{2}e^{x}\right ) \end{align*}

This example shows that change of variable on the independent variable did not work, but change of variable on the dependent variable (general case) worked.

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

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

For some \(n\), This transformation changes the ode to an ode with a missing \(y\), which can be easily solved as two first order ode’s. Substituting this in (A) results in the following ode where the dependent variable is \(v\) and not \(y\)

\begin{equation} 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}\end{equation}

If it happens that

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

For some \(n\), then (7) becomes

\begin{equation} x^{n}v^{\prime \prime }+\left ( 2x^{n-1}n+x^{n}p\right ) v^{\prime }=r \tag {7B}\end{equation}

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

\[ u^{\prime }+\frac {\left ( 2x^{n-1}n+x^{n}p\right ) }{x^{n}}u=r \]

Applying (7A) on this example ode gives

\begin{align*} \left ( n\left ( n-1\right ) x^{n-2}+n\left ( \frac {2}{x}\right ) x^{n-1}+\left ( -1\right ) x^{n}\right ) & =0\\ n\left ( n-1\right ) x^{n-2}+2nx^{n-2}-x^{n} & =0\\ \left ( n+n^{2}\right ) x^{n-2}-x^{n} & =0 \end{align*}

It is clear that there exists no integer or rational number \(n\) which makes the LHS above zero. Hence this special transformation did not work.

This is an example where only the change of variable on the dependent variable (general case) worked.