Example 1 \[ x^{2}y^{\prime \prime }+xy^{\prime }-y=x^{4}\] Then \(p=x^{2},q=x,r=-1,f\left ( x\right ) =x^{4}\). Condition (2) becomes\begin {align*} p^{\prime \prime }-q^{\prime }+r & =0\\ 2-1-1 & =0\\ 0 & =0 \end {align*}

Hence it is second order exact. Therefore the adjoint ode (3) is\begin {align*} \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*}

The first integral is\begin {align*} 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*}

This is linear ode. Solving this ode gives \[ y=\frac {x^{4}}{15}-\frac {c}{2x}+c_{2}x \] Note that this is also a Euler ode.