3.9.6.0.4 Example 4 \(y^{\prime \prime }-4xy^{\prime }+\left ( 4x^{2}-2\right ) y=0\)

\begin{equation} y^{\prime \prime }-4xy^{\prime }+\left ( 4x^{2}-2\right ) y=0 \tag {1}\end{equation}

In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=-4x,q=\left ( 4x^{2}-2\right ) ,r=0\). Hence (6A) is

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

Now we check if Liouville ode invariant \(q_{1}\) is constant or a constant divided by \(x^{2}\).

\begin{align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =\left ( 4x^{2}-2\right ) -\frac {1}{2}\left ( -4x\right ) ^{\prime }-\frac {1}{4}\left ( -4x\right ) ^{2}\\ & =\left ( 4x^{2}-2\right ) +2-\frac {1}{4}\left ( 16x^{2}\right ) \\ & =4x^{2}-2+2-4x^{2}\\ & =0 \end{align*}

Since \(q_{1}\) is constant, then we can use the change of the variable \(y=v\left ( x\right ) z\left ( x\right ) \) which is

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

Substituting the above into the original ODE (1) gives

\begin{align*} y^{\prime \prime }-4xy^{\prime }+\left ( 4x^{2}-2\right ) y & =0\\ \left ( ve^{x^{2}}\right ) ^{\prime \prime }-4x\left ( ve^{x^{2}}\right ) ^{\prime }+\left ( 4x^{2}-2\right ) ve^{x^{2}} & =0 \end{align*}

Carrying out the simplification gives

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

Which is constant coefficient ode. This is easily solved giving the solution

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

Hence

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