4.4.1.1 Introduction and terminology used

An ode $F(x,y,y^{\prime },y^{\prime \prime })=0$ is called exact if there exists a function $R(x,y,y^{\prime })$ with order one less that of the ode, such that

$$F(x,y,y^{\prime },y^{\prime \prime })=\frac{d}{dx}R(x,y,y^{\prime })$$

Which also implies that $R=c$ some constant, because $F=0$. In the above $R(x,y,y^{\prime })$ is called the first integral of the ode $F$ (also called the reduced ode), because

$$\begin{array}{}\text{(1A)}& R=\int Fdx+c\end{array}$$

An important property of first integral is the following. If we write the ode $F(x,y,y^{\prime },y^{\prime \prime })=0$ as  $y^{\prime \prime }=\mathrm{\Phi }(x,y,y^{\prime })$ which we can always do, then

$$\begin{array}{}\text{(1B)}& R_x+y^{\prime }{R}_y+\mathrm{\Phi }{R}_{y^{\prime }}=0\end{array}$$

Lets see how this works. Given the ode $y^{\prime \prime }+x{y}^{\prime }+y=0$ which is exact as is from the exactness test $p{y}^{\prime \prime }+q{y}^{\prime }+r=0$ which is $p^{\prime \prime }-q^{\prime }+r=0$, hence $p=1,q=x,r=1$, therefore $-1+1=0$ which is true. Therefore we can write because we can write $y^{\prime \prime }+x{y}^{\prime }+y=0={(y^{\prime }+B\left(x\right)y)}^{\prime }$ and find that $B=x$, Hence

$$y^{\prime \prime }+x{y}^{\prime }+y={(y^{\prime }+xy)}^{\prime }$$

Where $y^{\prime }+xy=0$ is the reduced ode.

$$R=y^{\prime }+xy$$

For the original ode $y^{\prime \prime }+x{y}^{\prime }+y=0$, it can be written as $y^{\prime \prime }=-(x{y}^{\prime }+y)$, therefore $\mathrm{\Phi }=-(x{y}^{\prime }+y)$. Eq (1B) now becomes

$$\begin{array}{rl}{R}_x+y^{\prime }{R}_y+\mathrm{\Phi }{R}_{y^{\prime }}& =0\\ y+y^{\prime }x-(x{y}^{\prime }+y)\left(1\right)& =0\\ y+y^{\prime }x-x{y}^{\prime }-y& =0\\ 0& =0\end{array}$$

Verified. Here is another example. Given the ode ${(x-1)}^2{y}^{\prime \prime }+4y^{\prime }x+2y-2x=0$, this is exact because we can write ${(x-1)}^2{y}^{\prime \prime }+4y^{\prime }x+2y-2x=\frac{d}{dx}((2x+2)y+(x^2-2x+1)y^{\prime }-x^2)$, hence the first integral (or the reduced ode) is $R=(2x+2)y+(x^2-2x+1)y^{\prime }-x^2$. The original ode can be written as $y^{\prime \prime }=-\frac{(4y^{\prime }x+2y-2x)}{{(x-1)}^2}$, therefore $\mathrm{\Phi }=-\frac{(4y^{\prime }x+2y-2x)}{{(x-1)}^2}$. Eq (1B) becomes

$$\begin{array}{rl}{R}_x+y^{\prime }{R}_y+\mathrm{\Phi }{R}_{y^{\prime }}& =0\\ (2y+2x{y}^{\prime }-2y^{\prime }-2x)+y^{\prime }(2x+2)-\left(\frac{4y^{\prime }x+2y-2x}{{(x-1)}^2}\right)(x^2-2x+1)& =0\\ 0& =0\end{array}$$

Verified. Equations (1A) and (1B) are important as they will be used to determined an integrating factor when the ode is not exact.