4.3.2.0 Linear second order not exact but solved by ﬁnding mu(x) integrating factor.

As mentioned above, an exact ode is one which has a corresponding adjoint ODE. In the case when the ode was exact, we did not use an integrating factor (this is the same as saying the integrating factor was \(1\)), i.e. \(\mu \left ( x\right ) =1\).

But if the ode is not exact, then we look for integrating factor \(\mu \left ( x\right ) \) that when multiplied by the ode makes it exact and hence will have an adjoint ODE. Given

\begin{equation} py^{\prime \prime }+q\left ( x\right ) y^{\prime }+r\left ( x\right ) y=f\left ( x\right ) \tag {1}\end{equation}

Which is assumed not to be exact. Multiplying both sides by \(\mu \left ( x\right ) \) gives \(\mu \left ( py^{\prime \prime }+q\left ( x\right ) y^{\prime }+r\left ( x\right ) y\right ) =\mu f\left ( x\right ) \). Let

\begin{equation} \mu \left ( py^{\prime \prime }+q\left ( x\right ) y^{\prime }+r\left ( x\right ) y\right ) =\left ( \mu py^{\prime }+By\right ) ^{\prime } \tag {2}\end{equation}

\begin{align*} \mu \left ( py^{\prime \prime }+q\left ( x\right ) y^{\prime }+r\left ( x\right ) y\right ) & =\mu ^{\prime }py^{\prime }+\mu p^{\prime }y^{\prime }+\mu py^{\prime \prime }+B^{\prime }y+By^{\prime }\\ \mu py^{\prime \prime }+\mu qy^{\prime }+\mu ry & =\mu py^{\prime \prime }+y^{\prime }\left ( \mu ^{\prime }p+\mu p^{\prime }+B\right ) +yB^{\prime }\end{align*}

\begin{align} \mu q & =\mu ^{\prime }p+\mu p^{\prime }+B\tag {2A}\\ \mu r & =B^{\prime } \tag {2B}\end{align}

\[ \mu ^{\prime }q+\mu q^{\prime }=\mu ^{\prime \prime }p+\mu ^{\prime }p^{\prime }+\mu ^{\prime }p^{\prime }+\mu p^{\prime \prime }+B^{\prime }\]

\begin{equation} \mu ^{\prime }q+\mu q^{\prime }=\mu ^{\prime \prime }p+\mu ^{\prime }p^{\prime }+\mu ^{\prime }p^{\prime }+\mu p^{\prime \prime }+\mu r \tag {3}\end{equation}

\begin{equation} \mu ^{\prime \prime }p+\mu ^{\prime }\left ( 2p^{\prime }-q\right ) +\mu \left ( p^{\prime \prime }-q^{\prime }+r\right ) =0 \tag {4}\end{equation}

The integrating factor \(\mu \) is the solution to the above ODE (called the adjoint ode also). Note that in (4), the term \(p^{\prime \prime }-q^{\prime }+r\) will not be zero, as this is the condition for exactness, and this ode is not exact (else we will not need an integrating factor to start with).

We can obtain (4) directly from \(py^{\prime \prime }+qy^{\prime }+ry=0\). Since the relation between an ode and its adjoint ode is the following: given

\begin{align*} \left ( \left ( p\mu \right ) ^{\prime }-q\mu \right ) ^{\prime }+r\mu & =0\\ \left ( p\mu \right ) ^{\prime \prime }-\left ( q\mu \right ) ^{\prime }+r\mu & =0\\ \left ( p^{\prime }\mu +p\mu ^{\prime }\right ) ^{\prime }-\left ( q^{\prime }\mu +q\mu ^{\prime }\right ) +r\mu & =0\\ p^{\prime \prime }\mu +p^{\prime }\mu ^{\prime }+p^{\prime }\mu ^{\prime }+p\mu ^{\prime \prime }-q^{\prime }\mu -q\mu ^{\prime }+r\mu & =0\\ p\mu ^{\prime \prime }+\mu ^{\prime }\left ( 2p^{\prime }-q\right ) +\mu \left ( p^{\prime \prime }-q^{\prime }+r\right ) & =0 \end{align*}

We see this is the same as (4). In summary, an ode \(py^{\prime \prime }+qy^{\prime }+ry=0\) has adjoint ode \(\left ( p\mu \right ) ^{\prime \prime }-\left ( q\mu \right ) ^{\prime }+r\mu =0\) where the solution to the adjoint ode makes the ﬁrst ode exact. Once the integrating factor \(\mu \) is found then the ﬁrst integral is given by

\begin{align*} B & =\mu q-\mu ^{\prime }p-\mu p^{\prime }\\ & =\mu \left ( q-p^{\prime }\right ) -\mu ^{\prime }p \end{align*}

\begin{equation} py^{\prime \prime }+qy^{\prime }+ry=\left ( \mu py^{\prime }+\left ( \mu \left ( q-p^{\prime }\right ) -\mu ^{\prime }p\right ) y\right ) ^{\prime } \tag {5}\end{equation}

\[ \mu \left ( py^{\prime \prime }+qy^{\prime }+ry\right ) -y\overline {\left ( py^{\prime \prime }+qy^{\prime }+ry\right ) }=\frac {d}{dx}\left ( P\left ( y,u\right ) \right ) \]

Where the bar above the ode means its complex conjugate. The function \(P\left ( y,u\right ) \) is called the bilinear concomitant (see Murphy book, page 93). And is given by

Unfortunately, all this does not help us in solving the adjoint ode (4) in order to ﬁnd the integrating factor \(\mu \). Since it will also be a second order ode which can be as hard to solve as the original ode. So this method is not practical as far as I can see unless the adjoint ODE comes out very simple to solve, but in all the examples I looked at, this was not the case.

\(p=1,q=-4x,r=\left ( 4x^{2}-2\right ) \,.\) Let us ﬁrst check if the ode is exact or not as is. The condition for exactness is

The LHS is not zero. This means the ode is not exact. Therefore we need to try to ﬁnd an integration factor \(\mu \left ( x\right ) \) to make the ode exact. (4) becomes

\begin{align*} \mu ^{\prime \prime }p+\mu ^{\prime }\left ( 2p^{\prime }-q\right ) +\mu \left ( p^{\prime \prime }-q^{\prime }+r\right ) & =0\\ \mu ^{\prime \prime }+\mu ^{\prime }\left ( 4x\right ) +\mu \left ( 4+\left ( 4x^{2}-2\right ) \right ) & =0\\ \mu ^{\prime \prime }+4x\mu ^{\prime }+\mu \left ( 2+4x^{2}\right ) & =0 \end{align*}

We see in practice that ﬁnding the integrating factor leads to yet another second order ode which is as hard to solve as the original ode. The solution to this ode can be found to be \(e^{-x^{2}},xe^{-x^{2}}\). We only need one integrating factor. Hence let

To see this let us check the condition again now. Here \(p=e^{-x^{2}},q=-4xe^{-x^{2}},r=\left ( 4x^{2}-2\right ) e^{-x^{2}}\). Hence

\begin{align*} p^{\prime \prime }-q^{\prime }+r & =0\\ \left ( 4e^{-x^{2}}x^{2}-2e^{-x^{2}}\right ) -\left ( 8e^{-x^{2}}x^{2}-4e^{-x^{2}}\right ) +\left ( 4x^{2}-2\right ) e^{-x^{2}} & =0\\ 0 & =0 \end{align*}

\[ \left ( \mu py^{\prime }+\left ( \mu \left ( q-p^{\prime }\right ) -\mu ^{\prime }p\right ) y\right ) ^{\prime }=0 \]

\begin{align*} e^{-x^{2}}y^{\prime }+\left ( -4xe^{-x^{2}}-\left ( -2xe^{-x^{2}}\right ) \right ) y & =c\\ e^{-x^{2}}y^{\prime }-2xe^{-x^{2}}y & =c\\ y^{\prime }-2xy & =ce^{x^{2}}\end{align*}

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.