5.1.2.3 Solved by finding first intergal (applies to exact ode only)

ode internal name "higher_order_exact"

This applies only to linear higher order which are exact. Solved by finding its first integral, which will be an ode of order one less. Let look at third order ode first

\[ p_{3}y^{\prime \prime \prime }+p_{2}y^{\prime \prime }+p_{1}y^{\prime }+p_{0}y=f\left ( x\right ) \]

Notice that most books like to write the above as

\[ p_{0}y^{\prime \prime \prime }+p_{1}y^{\prime \prime }+p_{2}y^{\prime }+p_{3}y=f\left ( x\right ) \]

i.e. the index of \(p\) starts at zero for the highest derivative and go down to zero until it hits \(y\). I do not like this and prefer the index \(n\) of \(p_{n}\) to match the corresponding derivative on \(y^{\left ( n\right ) }\) as it is more clear.

In the above ode, all the \(p_{i}\) coefficients can be functions of \(x\) also. The condition of exactness is given by

\begin{equation} p_{3}^{\prime \prime \prime }-p_{2}^{\prime \prime }+p_{1}^{\prime }-p_{0}=0\tag {1}\end{equation}

If this condition is satisfied then first integral is

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

This is now second order ode which is solved for \(y\).  For a 4th order ode

\[ p_{4}y^{\prime \prime \prime \prime }+p_{3}y^{\prime \prime \prime }+p_{2}y^{\prime \prime }+p_{1}y^{\prime }+p_{0}y=f\left ( x\right ) \]

The condition is

\begin{equation} p_{4}^{\prime \prime \prime \prime }-p_{3}^{\prime \prime \prime }+p_{2}^{\prime \prime }-p_{1}^{\prime }-p_{0}=0\tag {3}\end{equation}

If the above is satisfied, then the first integral is

\begin{align} \frac {d}{dx}\left ( p_{4}y^{\prime \prime \prime }+\left ( p_{3}-p_{4}^{\prime }\right ) y^{\prime \prime }+\left ( p_{2}-p_{3}^{\prime }+p_{4}^{\prime }\right ) y^{\prime }+\left ( p_{1}-p_{2}^{\prime }+p_{3}^{\prime \prime }-p_{4}^{\prime \prime \prime }\right ) y\right ) & =f\left ( x\right ) \nonumber \\ p_{4}y^{\prime \prime \prime }+\left ( p_{3}-p_{4}^{\prime }\right ) y^{\prime \prime }+\left ( p_{2}-p_{3}^{\prime }+p_{4}^{\prime }\right ) y^{\prime }+\left ( p_{1}-p_{2}^{\prime }+p_{3}^{\prime \prime }-p_{4}^{\prime \prime \prime }\right ) y & =\int f\left ( x\right ) dx+c_{1}\tag {4}\end{align}

And so on. Hence given general higher order ode

\[ p_{n}y^{\left ( n\right ) }+p_{n-1}y^{\left ( n-1\right ) }+\cdots +p_{2}y^{^{\prime \prime }}+p_{1}y^{\prime }+p_{0}y=f\left ( x\right ) \]

The condition for exactness is

\[ p_{n}^{\left ( n\right ) }-p_{n-1}^{\left ( n-1\right ) }+p_{n-2}^{\left ( n-2\right ) }+\cdots +\left ( -1\right ) ^{n}p_{n}^{\left ( n\right ) }+\cdots =0 \]

And the first integral is

\[ p_{n}y^{\left ( n-1\right ) }+\left ( p_{n-1}-p_{n}^{\prime }\right ) y^{\left ( n-2\right ) }+\cdots +\left ( p_{1}-p_{2}^{\prime }+\cdots +\left ( -1\right ) ^{n}p_{n}^{\left ( n-1\right ) }+\cdots +p_{n}^{\left ( n-1\right ) }\right ) y=\int f\left ( x\right ) dx+c_{1}\]

5.1.2.3.1 Example 1 \(xy^{\prime \prime \prime }+\left ( x^{2}-3\right ) y^{\prime \prime }+4xy^{\prime }+2y=0\) Comparing to standard form \(p_{3}y^{\prime \prime \prime }+p_{2}y^{\prime \prime }+p_{1}y^{\prime }+p_{0}y=f\left ( x\right ) \) shows that

\begin{align*} p_{3} & =x\\ p_{2} & =x^{2}-3\\ p_{1} & =4x\\ p_{0} & =2\\ f\left ( x\right ) & =0 \end{align*}

Checking if it is exact

\begin{align*} p_{3}^{\prime \prime \prime }-p_{2}^{\prime \prime }+p_{1}^{\prime }-p_{0} & =0-2+4-2\\ & =0 \end{align*}

The first integral is therefore

\begin{align*} \frac {d}{dx}\left ( p_{3}y^{\prime \prime }+\left ( p_{2}-p_{3}^{\prime }\right ) y^{\prime }+\left ( p_{1}-p_{2}^{\prime }+p_{3}^{\prime \prime }\right ) y\right ) & =f\left ( x\right ) \\ \frac {d}{dx}\left ( xy^{\prime \prime }+\left ( x^{2}-3-1\right ) y^{\prime }+\left ( 4x-2x+0\right ) y\right ) & =0\\ \frac {d}{dx}\left ( xy^{\prime \prime }+\left ( x^{2}-4\right ) y^{\prime }+2xy\right ) & =0 \end{align*}

Hence the first integral is

\[ xy^{\prime \prime }+\left ( x^{2}-4\right ) y^{\prime }+2xy=c_{1}\]

Let us now check if this is also exact. This has form

\[ p_{2}y^{\prime \prime }+p_{1}y^{\prime }+p_{0}=f\left ( x\right ) \]

Where now

\begin{align*} p_{2} & =x\\ p_{1} & =\left ( x^{2}-4\right ) \\ p_{0} & =2x\\ f\left ( x\right ) & =c_{1}\end{align*}

Checking if it is exact

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

Show it is exact. Therefore its first integral is

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

Hence first integral is

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

This is first oder linear ode which is now easily solved.