Example 2 \[ y^{\prime \prime }+\frac {1}{x}y^{\prime }+\frac {1}{x}y=0 \] Here \(p=1,q=\frac {1}{x},r=\frac {1}{x},f\left ( x\right ) =0\). The condition of exactness is\begin {align*} p^{\prime \prime }-q^{\prime }+r & =0\\ 0-\left ( -\frac {1}{x^{2}}\right ) +\frac {1}{x} & =0 \end {align*}

Is not satisfied. Hence the ode is not exact. The adjoint ode (4) to find the integrating factor becomes\begin {align*} \mu ^{\prime \prime }p+\mu ^{\prime }\left ( p^{\prime }-q\right ) +\mu \left ( p^{\prime \prime }-q^{\prime }+r\right ) & =0\\ \mu ^{\prime \prime }+\mu ^{\prime }\left ( -\frac {1}{x}\right ) +\mu \left ( -\frac {1}{x^{2}}+\frac {1}{x}\right ) & =0\\ \mu ^{\prime \prime }-\frac {1}{x}\mu ^{\prime }-\mu \left ( \frac {1-x}{x^{2}}\right ) & =0\\ x^{2}\mu ^{\prime \prime }-x\mu ^{\prime }-\left ( 1-x\right ) \mu & =0 \end {align*}

Which has solutions \(\mu \) as bessel functions. We see that trying to find an integrating factor using this method is not practical, as it leads to an ode just as hard to solve as the original one. We could just have solved \(y^{\prime \prime }+\frac {1}{x}y^{\prime }+\frac {1}{x}y=0\) directly, since this is Bessel ode. Unless there is a short cut to solving the ODE to find the integrating factor, this method is not practical. See section below for simpler method

The main difficulty when second order is not exact, is in finding the integrating factor \(\mu \left ( x\right ) \) which itself requires solving another second order ode. The whole point of an ODE being exact is that it is a complete differential which means the order is reduced by one to make it easier to solve. This means solving a second order ode becomes solving a first order ode when the ode is exact.