3.10.1.1 Example 1

\[ x^{2}y^{\prime \prime }+xy^{\prime }-y=x^{4}\]

Then \(p_{2}=x^{2},p_{1}=x,p_{0}=-1,f\left ( x\right ) =x^{4}\). Condition (3) becomes

\begin{align*} p_{2}^{\prime \prime }-p_{1}^{\prime }+p_{0} & =2-1-1\\ & =0 \end{align*}

Hence it is second order exact. Therefore the adjoint ode (2) is

\begin{align*} \left ( p_{2}y^{\prime }+\left ( p_{1}-p_{2}^{\prime }\right ) y\right ) ^{\prime } & =f\left ( x\right ) \\ \left ( x^{2}y^{\prime }+\left ( x-2x\right ) y\right ) ^{\prime } & =x^{4}\\ x^{2}y^{\prime }+\left ( x-2x\right ) y & =\int x^{4}dx+c\\ x^{2}y^{\prime }-xy & =\frac {x^{5}}{5}+c \end{align*}

Integrating gives

\[ x^{2}y^{\prime }+\left ( x-2x\right ) y=\int x^{4}dx+c \]

This is called the first integral of the original ode. Hence

\begin{align*} x^{2}y^{\prime }+\left ( x-2x\right ) y & =\int x^{4}dx+c_{1}\\ x^{2}y^{\prime }-xy & =\frac {x^{5}}{5}+c_{1}\end{align*}

This is linear ode. Solving this ode gives

\[ y=\frac {x^{4}}{15}-\frac {c_{1}}{2x}+c_{2}x \]

Note that this is also a Euler ode.