3.9.6.0.3 Example 3 \(x^{2}y^{\prime \prime }-x\left ( x+2\right ) y^{\prime }+\left ( x+2\right ) y=2x^{3}\)

\begin{align} x^{2}y^{\prime \prime }-x\left ( x+2\right ) y^{\prime }+\left ( x+2\right ) y & =2x^{3}\nonumber \\ y^{\prime \prime }-\frac {x+2}{x}y^{\prime }+\frac {x+2}{x^{2}}y & =2x \tag {1}\end{align}

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

\begin{align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{\int \frac {x+2}{2x}dx}\\ & =xe^{\frac {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}\\ & =\frac {\left ( x+2\right ) }{x^{2}}-\frac {1}{2}\left ( xe^{\frac {x}{2}}\right ) ^{\prime }-\frac {1}{4}\left ( -\frac {x+2}{x}\right ) ^{2}\\ & =-\frac {1}{4}\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 ( xe^{\frac {x}{2}}\right ) \end{align*}

Substituting the above into the original ODE (1) gives

\begin{align*} y^{\prime \prime }-\frac {x+2}{x}y^{\prime }+\frac {x+2}{x}y & =2x\\ \left ( v\left ( xe^{\frac {x}{2}}\right ) \right ) ^{\prime \prime }-\frac {x+2}{x}\left ( v\left ( xe^{\frac {x}{2}}\right ) \right ) ^{\prime }+\frac {x+2}{x^{2}}v\left ( xe^{\frac {x}{2}}\right ) & =2x \end{align*}

Carrying out the simplification gives

\[ 4v^{\prime \prime }-v=8e^{-\frac {x}{2}}\]

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

\[ v=c_{1}\sinh \left ( \frac {x}{2}\right ) +c_{2}\cosh \left ( \frac {x}{2}\right ) -2xe^{\frac {-x}{2}}\]

Hence

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