# Second-Order In(exact) ODEs

The second total derivative of $$F(x,\ y(x))$$ is $$F_y y'' + F_{yy}(y')^2 + 2F_{xy}y' + F_{xx}$$. Thus by analogy to first-order exact ODEs, if one notices a second-order ODE where this pattern equals some expression containing no $$y$$'s, i.e., $$\frac{1}{y}y'' - \frac{1}{y^2}(y')^2 - \frac{1}{x^2} = 0$$, the LHS can be condensed, in this case turning the ODE into $$(ln|xy|)'' = 0$$, and both sides doubly integrated, in this case $$ln|xy| = Ax + B \implies y = \frac{Be^{Ax}}{x}$$. Of course, I had to reverse engineer this example from a chosen $$F(x,\ y(x))$$; otherwise it would have been likely prohibitively hard to spot.

1) I assume a second-order ODE is exact if and only if the LHS can be condensed into $$F(x,\ y(x))$$ and integrated as shown above. Is there an analogy here to the 2D-curl test for exactness?

2) Is there a method here for finding integrating factors to make inexact equations exact, at least in certain cases?

3) Is there a way to convert such ODEs into differential forms (analogous to how one "sort of" multiplies the first-order ODE through by $$dx$$, even though that's not really what's going on but in practice it basically is)?

4) There aren't many good resources on second-order exact ODEs, but the SE questions I found and this video spoken in a language I don't know never seem to involve $$(y')^2$$. Is this because it is only practical to consider the subset of cases where $$F_{yy} = 0$$, as opposed to my contrived example where $$F_{xy} = 0$$ but $$F_{yy} \neq 0$$, or are these second-order exact ODEs a different animal? If the latter, are they related to my notion of second-order exact ODEs in any way?

Consider the nonlinear second order ODE of the form$$a_2(x, y, y')y'' + a_1(x, y, y')y' + a_0(x, y, y') = 0\tag1$$

Question I: Is there an analogy here to the 2D-curl test for exactness?

Answer : The first order ODE of the form $$~M~dx~+~N~dy~=0~$$ is an exact differential equation if $$~curl (M \vec i + N \vec j)=\vec 0~$$. I have no idea if there be any form for the higher order.

Actually the necessary and sufficient condition for exactness said,

The differential equation $$a_ny^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_1y'+a_0y=X$$ where $$~a_n,~a_{n-1},~\cdots~,~a_1,~a_0~$$ and $$~X~$$ are constants or functions of $$~x~$$ only, will be exact if and only if $$a_0-a_1'+a_2''-a_3'''+\cdots+(-1)^na_n^{(n)}=0~.$$

............................................................................

Question II: Is there a method here for finding integrating factors to make inexact equations exact, at least in certain cases?

Answer : Yes, the concept of integrating factor is also admissible for the second order ODE.

Definition : An integrating factor of equation$$(1)$$ is a non zero function $$~µ(x, y, y')~$$, such that the equation $$µ(x, y, y')a_2(x, y, y')y''+ µ(x, y, y')a_1(x, y, y')y' + µ(x, y, y')a_0(x, y, y') = 0\tag2$$is exact. i.e., $$\dfrac{∂A_2}{∂y} =\dfrac{∂A_1}{∂y'},~~~~\dfrac{∂A_2}{∂x} =\dfrac{∂A_0}{∂y'},~~~ \text{and}~~~\dfrac{∂A_1}{∂x} =\dfrac{∂A_0}{∂y}$$where $$A_2(x, y, y') = µ(x, y, y')a_2(x, y, y'),\\A_1(x, y, y') = µ(x, y, y')a_1(x, y, y'),\\\text{and}~~~A_0(x, y, y') = µ(x, y, y')a_0(x, y, y')~.~~~~~~~~$$ Theorem : Assume that equation$$(1)$$ is not an exact equation. Then, it has no integrating factor of one of the forms $$~µ(x, y, y'), ~µ(x, y),~ µ(x, y'), ~~\text{or}~~ µ(y, y')~$$ if and only if$$\left(\dfrac{∂a_0}{∂y}-\dfrac{∂a_1}{∂x}\right)a_2+\left(\dfrac{∂a_2}{∂x}-\dfrac{∂a_0}{∂y'}\right)a_1+\left(\dfrac{∂a_1}{∂y'} -\dfrac{∂a_2}{∂y}\right)a_0\ne 0 ~.$$ Like the first order ODE, there are also some rules for finding integrating factor. For the details please visit the research work "Jordan Journal of Mathematics and Statistics (JJMS) 8(2), 2015, pp 155 - 167"

............................................................................

Question III: Is there a way to convert such ODEs into differential forms (analogous to how one "sort of" multiplies the first-order ODE through by dx, even though that's not really what's going on but in practice it basically is)?

Answer : Yes. For this follow the procedure given bellow.

$$(a)~~$$In equation$$(1)$$, the highest derivative is $$~a_2y''~$$, so start with $$~\frac{d}{dx}\left(a_2y'\right)~$$ and then we get the term $$~a_2y''+a_2'y'~.$$

$$(b)~~$$Now start with $$~\frac{d}{dx}\left[(a_1-a_2')y\right]~$$, we get the term $$~(a_1-a_2')y'+(a_1'-a_2'')y~.$$

$$(c)~~$$ Add the above values. We get the left hand side of the given differential equation on the right side then using the given results and then integrating, we get the first integral.

Example: Solve $$~x^2y''+3xy'+y=\dfrac 1{(1-x)^2}~.\tag3$$

Solution: Here $$~a_2=x^2,~a_1=3x,~a_0=1~$$.

The highest term is $$~x^2y''~$$, so by $$(a)$$, $$~\frac{d}{dx}\left(x^2y'\right)=x^2y''+2xy'\tag4~$$ Now by $$(2)$$, $$~\frac{d}{dx}\left[(a_1-a_2')y\right]=\frac{d}{dx}\left[xy\right]=xy'+y\tag5~$$

Adding equation $$(4)$$ and $$(5)$$ we have, $$~\frac{d}{dx}\left[x^2y'+xy\right]=x^2y''+3xy'+y=\dfrac 1{(1-x)^2}~~~~(\text{by equation} (3))~\tag6$$

This equation $$(6)$$ (converted into differential forms) is called the first integral of the given ODE. Now integrating equation $$(6)$$ we have, $$x^2y'+xy=\dfrac 1{(1-x)}+c\implies y'+\dfrac yx=\dfrac{1}{x(1-x)}+\dfrac{c}{x^2}\tag7$$Now equation $$(7)$$ is a first order linear ODE of the form $$~y'+Py=Q~.$$ You can solve it by using the method of integrating factor.

............................................................................

Answer relating to question IV: In that video(given by you), for checking the exactness, he use the theorem that I mentioned in the answer of question I. For the after process he use the procedure I discussed in the answer of question III.

Eventually I think, the whole discussion will help you for clearing your doubt.