# Exact differential equation

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In mathematics, an exact differential equation or total differential equation is a certain kind of ordinary differential equation which is widely used in physics and engineering.

## Definition

Given a simply connected and open subset D of $\mathbb {R} ^{2}$ and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form
    $I(x,y)\,dx+J(x,y)\,dy=0,$
is called an exact differential equation if there exists a continuously differentiable function F, called the potential function, so that
    ${\frac {\partial F}{\partial x}}=I$
and
    ${\frac {\partial F}{\partial y}}=J.$
An exact equation may also be presented in the following form:
    $I(x,y)+J(x,y)\,y'(x)=0$
where the same constraints on I and J apply for the differential equation to be exact.
The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function $F(x_{0},x_{1},...,x_{n-1},x_{n})$, the exact or total derivative with respect to $x_{0}$ is given by
    ${\frac {dF}{dx_{0}}}={\frac {\partial F}{\partial x_{0}}}+\sum _{i=1}^{n}{\frac {\partial F}{\partial x_{i}}}{\frac {dx_{i}}{dx_{0}}}.$

### Example

The function $F:\mathbb {R} ^{2}\to \mathbb {R}$ given by
    $F(x,y)={\frac {1}{2}}(x^{2}+y^{2})+c$
is a potential function for the differential equation
    $x\,dx+y\,dy=0.\,$

## First-order exact differential equations
### Identifying first-order exact differential equations

Let the functions M{\textstyle M}, N{\textstyle N}, M$_y${\textstyle M_{y}}, and N$_x${\textstyle N_{x}}, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region R:α<x<β, γ<y<δ{\textstyle R:\alpha <x<\beta , \gamma <y<\delta }. Then the differential equation
$M(x,y)+N(x,y){\frac {dy}{dx}}=0$
is exact if and only if
$M_{y}(x,y)=N_{x}(x,y)$
That is, there exists a function $\psi (x,y)$, called a potential function, such that
$\psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)$
So, in general:
$M_{y}(x,y)=N_{x}(x,y)\iff {\begin{cases}\exists \psi (x,y)\\\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}$

#### Proof

The proof has two parts.
First, suppose there is a function $\psi (x,y)$ such that $\psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)$
It then follows that $M_{y}(x,y)=\psi _{xy}(x,y){\text{ and }}N_{x}(x,y)=\psi _{yx}(x,y)$
Since $M_{y}$ and $N_{x}$ are continuous, then $\psi _{xy}$ and $\psi _{yx}$ are also continuous which guarantees their equality.
The second part of the proof involves the construction of $\psi (x,y)$ and can also be used as a procedure for solving first-order exact differential equations. Suppose that $M_{y}(x,y)=N_{x}(x,y)$ and let there be a function $\psi (x,y)$ for which $\psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)$
Begin by integrating the first equation with respect to $x$. In practice, it doesn't matter if you integrate the first or the second equation, so long as the integration is done with respect to the appropriate variable.
${\frac {\partial \psi }{\partial x}}(x,y)=M(x,y)$ $\psi (x,y)=\int M(x,y)\,dx+h(y)$ $\psi (x,y)=Q(x,y)+h(y)$
where $Q(x,y)$ is any differentiable function such that $Q_{x}=M$. The function $h(y)$ plays the role of a constant of integration, but instead of just a constant, it is function of $y$, since $M$ is a function of both $x$ and $y$ and we are only integrating with respect to $x$.
Now to show that it is always possible to find an $h(y)$ such that $\psi _{y}=N$. $\psi (x,y)=Q(x,y)+h(y)$
Differentiate both sides with respect to $y$. ${\frac {\partial \psi }{\partial y}}(x,y)={\frac {\partial Q}{\partial y}}(x,y)+h'(y)$
Set the result equal to $N$ and solve for $h'(y)$. $h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)$
In order to determine $h'(y)$ from this equation, the right-hand side must depend only on $y$. This can be proven by showing that its derivative with respect to $x$ is always zero, so differentiate the right-hand side with respect to $x$. ${\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial x}}{\frac {\partial Q}{\partial y}}(x,y)\iff {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial y}}{\frac {\partial Q}{\partial x}}(x,y)$
Since $Q_{x}=M$, ${\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial M}{\partial y}}(x,y)$ Now, this is zero based on our initial supposition that $M_{y}(x,y)=N_{x}(x,y)$
Therefore, $h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)$ $h(y)=\int {\left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right)dy}$
$\psi (x,y)=Q(x,y)+\int \left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right)\,dy+C$
And this completes the proof.

### Solutions to first-order exact differential equations

First-order exact differential equations of the form $M(x,y)+N(x,y){\frac {dy}{dx}}=0$
can be written in terms of the potential function $\psi (x,y)$ ${\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0$
where ${\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}$
This is equivalent to taking the total derivative of $\psi (x,y)$. ${\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0\iff {\frac {d}{dx}}\psi (x,y(x))=0$
The solutions to an exact differential equation are then given by $\psi (x,y(x))=c$
and the problem reduces to finding $\psi (x,y)$.
This can be done by integrating the two expressions $M(x,y)\,dx$ and $N(x,y)\,dy$ and then writing down each term in the resulting expressions only once and summing them up in order to get $\psi (x,y)$.
The reasoning behind this is the following. Since ${\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}$
it follows, by integrating both sides, that ${\begin{cases}\psi (x,y)=\int M(x,y)\,dx+h(y)=Q(x,y)+h(y)\\\psi (x,y)=\int N(x,y)\,dy+g(x)=P(x,y)+g(x)\end{cases}}$
Therefore, $Q(x,y)+h(y)=P(x,y)+g(x)$
where $Q(x,y)$ and $P(x,y)$ are differentiable functions such that $Q_{x}=M$ and $P_{y}=N$.
In order for this to be true and for both sides to result in the exact same expression, namely $\psi (x,y)$, then $h(y)$ must be contained within the expression for $P(x,y)$ because it cannot be contained within $g(x)$, since it is entirely a function of $y$ and not $x$ and is therefore not allowed to have anything to do with $x$. By analogy, $g(x)$ must be contained within the expression $Q(x,y)$.
Ergo, $Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+d(x,y)$
for some expressions $f(x,y)$ and $d(x,y)$. Plugging in into the above equation, we find that $g(x)+f(x,y)+h(y)=h(y)+d(x,y)+g(x)\Rightarrow f(x,y)=d(x,y)$ and so $f(x,y)$ and $d(x,y)$ turn out to be the same function. Therefore, $Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+f(x,y)$
Since we already showed that ${\begin{cases}\psi (x,y)=Q(x,y)+h(y)\\\psi (x,y)=P(x,y)+g(x)\end{cases}}$
it follows that $\psi (x,y)=g(x)+f(x,y)+h(y)$
So, we can construct $\psi (x,y)$ by doing $\int M(x,y)\,dx$ and $\int N(x,y)\,dy$ and then taking the common terms we find within the two resulting expressions (that would be $f(x,y)$ ) and then adding the terms which are uniquely found in either one of them – $g(x)$ and $h(y)$.

## Second-order exact differential equations

The concept of exact differential equations can be extended to second-order equations. Consider starting with the first-order exact equation:
    $I(x,y)+J(x,y){dy \over dx}=0$
Since both functions $I(x,y)$, $J(x,y)$ are functions of two variables, implicitly differentiating the multivariate function yields
    ${dI \over dx}+\left({dJ \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0$
Expanding the total derivatives gives that
    ${dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}$
and that
    ${dJ \over dx}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}$
Combining the dy/dx {\textstyle {dy \over dx}} terms gives
    ${\partial I \over \partial x}+{dy \over dx}\left({\partial I \over \partial y}+{\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}\right)+{d^{2}y \over dx^{2}}(J(x,y))=0$
If the equation is exact, then ∂J/∂x=∂I/∂y {\textstyle {\partial J \over \partial x}={\partial I \over \partial y}}. Additionally, the total derivative of $J(x,y)$ is equal to its implicit ordinary derivative dJ/dx {\textstyle {dJ \over dx}}. This leads to the rewritten equation
    ${\partial I \over \partial x}+{dy \over dx}\left({\partial J \over \partial x}+{dJ \over dx}\right)+{d^{2}y \over dx^{2}}(J(x,y))=0$
Now, let there be some second-order differential equation
    $f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0$
If ${\partial J \over \partial x}={\partial I \over \partial y}$ for exact differential equations, then
    $\int \left({\partial I \over \partial y}\right)\,dy=\int \left({\partial J \over \partial x}\right)\,dy$
and
    $\int \left({\partial I \over \partial y}\right)\,dy=\int \left({\partial J \over \partial x}\right)\,dy=I(x,y)-h(x)$
where $h(x)$ is some arbitrary function only of $x$ that was differentiated away to zero upon taking the partial derivative of $I(x,y)$ with respect to $y$. Although the sign on $h(x)$ could be positive, it is more intuitive to think of the integral's result as $I(x,y)$ that is missing some original extra function $h(x)$ that was partially differentiated to zero.
Next, if
    ${dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}$
then the term ${\partial I \over \partial x}$ should be a function only of $x$ and $y$, since partial differentiation with respect to $x$ will hold $y$ constant and not produce any derivatives of $y$. In the second-order equation
    $f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0$
only the term $f(x,y)$ is a term purely of $x$ and $y$. Let ${\partial I \over \partial x}=f(x,y)$. If ${\partial I \over \partial x}=f(x,y)$, then
    $f(x,y)={dI \over dx}-{\partial I \over \partial y}{dy \over dx}$
Since the total derivative of $I(x,y)$ with respect to $x$ is equivalent to the implicit ordinary derivative ${dI \over dx}$ , then
    $f(x,y)+{\partial I \over \partial y}{dy \over dx}={dI \over dx}={d \over dx}(I(x,y)-h(x))+{dh(x) \over dx}$
So,
    ${dh(x) \over dx}=f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))$
and
    $h(x)=\int \left(f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))\right)\,dx$
Thus, the second-order differential equation
    $f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0$
is exact only if $g\left(x,y,{dy \over dx}\right)={dJ \over dx}+{\partial J \over \partial x}={dJ \over dx}+{\partial J \over \partial x}$ and only if the below expression
    $\int \left(f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))\right)\,dx=\int \left(f(x,y)-{\partial \left(I(x,y)-h(x)\right) \over \partial x}\right)\,dx$
is a function solely of $x$. Once $h(x)$ is calculated with its arbitrary constant, it is added to $I(x,y)-h(x)$ to make $I(x,y)$. If the equation is exact, then we can reduce to the first-order exact form which is solvable by the usual method for first-order exact equations.
    $I(x,y)+J(x,y){dy \over dx}=0$
Now, however, in the final implicit solution there will be a $C_{1}x$ term from integration of $h(x)$ with respect to $x$ twice as well as a $C_{2}$, two arbitrary constants as expected from a second-order equation.

### Example

Given the differential equation
    $(1-x^{2})y''-4xy'-2y=0$
one can always easily check for exactness by examining the $y''$ term. In this case, both the partial and total derivative of $1-x^{2}$ with respect to $x$ are $-2x$, so their sum is $-4x$, which is exactly the term in front of $y'$. With one of the conditions for exactness met, one can calculate that
    $\int (-2x)\,dy=I(x,y)-h(x)=-2xy$
Letting $f(x,y)=-2y$, then
    $\int \left(-2y-2xy'-{d \over dx}(-2xy)\right)\,dx=\int (-2y-2xy'+2xy'+2y)\,dx=\int (0)\,dx=h(x)$
So, $h(x)$ is indeed a function only of $x$ and the second-order differential equation is exact. Therefore, $h(x)=C_{1}$ and $I(x,y)=-2xy+C_{1}$. Reduction to a first-order exact equation yields
    $-2xy+C_{1}+(1-x^{2})y'=0$
Integrating $I(x,y)$ with respect to $x$ yields
    $-x^{2}y+C_{1}x+i(y)=0$
where $i(y)$ is some arbitrary function of $y$. Differentiating with respect to $y$ gives an equation correlating the derivative and the $y'$ term.
    $-x^{2}+i'(y)=1-x^{2}$
So, $i(y)=y+C_{2}$ and the full implicit solution becomes
    $C_{1}x+C_{2}+y-x^{2}y=0$
Solving explicitly for $y$ yields
    $y={\frac {C_{1}x+C_{2}}{1-x^{2}}}$

## Higher-order exact differential equations

The concepts of exact differential equations can be extended to any order. Starting with the exact second-order equation
    ${d^{2}y \over dx^{2}}(J(x,y))+{dy \over dx}\left({dJ \over dx}+{\partial J \over \partial x}\right)+f(x,y)=0$
it was previously shown that equation is defined such that
    $f(x,yt)={dht(x) \over dx}+{d \over dx}(I(x,y)-h(x))-{\partial J \over \partial x}{dy \over dx}$
Implicit differentiation of the exact second-order equation $n$ times will yield an $(n+2)$th-order differential equation with new conditions for exactness that can be readily deduced from the form of the equation produced. For example, differentiating the above second-order differential equation once to yield a third-order exact equation gives the following form
    ${d^{3}y \over dx^{3}}(J(x,y))+{d^{2}y \over dx^{2}}{dJ \over dx}+{d^{2}y \over dx^{2}}\left({dJ \over dx}+{\partial J \over \partial x}\right)+{dy \over dx}\left({d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)\right)+{df(x,y) \over dx}=0$
where
    ${df(x,y) \over dx}={d^{2}h(x) \over dx^{2}}+{d^{2} \over dx^{2}}(I(x,y)-h(x))-{d^{2}y \over dx^{2}}{\partial J \over \partial x}-{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)=F\left(x,y,{dy \over dx}\right)$
and where $F\left(x,y,{dy \over dx}\right)$ is a function only of $x,y$ and ${dy \over dx}$. Combining all ${dy \over dx}$ and ${d^{2}y \over dx^{2}}$ terms not coming from $F\left(x,y,{dy \over dx}\right)$ gives
    ${d^{3}y \over dx^{3}}(J(x,y))+{d^{2}y \over dx^{2}}\left(2{dJ \over dx}+{\partial J \over \partial x}\right)+{dy \over dx}\left({d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)\right)+F\left(x,y,{dy \over dx}\right)=0$
Thus, the three conditions for exactness for a third-order differential equation are: the ${d^{2}y \over dx^{2}}$ term must be $2{dJ \over dx}+{\partial J \over \partial x}$, the ${dy \over dx}$ term must be ${d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)$ and
    $F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2}}(I(x,y)-h(x))+{d^{2}y \over dx^{2}}{\partial J \over \partial x}+{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)$
must be a function solely of $x$.

### Example

Consider the nonlinear third-order differential equation
    $yy'''+3y'y''+12x^{2}=0$
If $J(x,y)=y$, then $y''\left(2{dJ \over dx}+{\partial J \over \partial x}\right)$ is $2y'y''$ and $y'\left({d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)\right)=y'y''$which together sum to $3y'y''$. Fortunately, this appears in our equation. For the last condition of exactness,
    $F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2}}\left(I(x,y)-h(x)\right)+{d^{2}y \over dx^{2}}{\partial J \over \partial x}+{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)=12x^{2}-0+0+0=12x^{2}$
which is indeed a function only of $x$. So, the differential equation is exact. Integrating twice yields that $h(x)=x^{4}+C_{1}x+C_{2}=I(x,y)$. Rewriting the equation as a first-order exact differential equation yields
    $x^{4}+C_{1}x+C_{2}+yy'=0$
Integrating $I(x,y)$ with respect to $x$ gives that ${x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i(y)=0$. Differentiating with respect to $y$ and equating that to the term in front of $y'$ in the first-order equation gives that $i'(y)=y$ and that $i(y)={y^{2} \over 2}+C_{3}$. The full implicit solution becomes
    ${x^{5} \over 5}+C_{1}x^{2}+C_{2}x+C_{3}+{y^{2} \over 2}=0$
The explicit solution, then, is
    $y=\pm {\sqrt {C_{1}x^{2}+C_{2}x+C_{3}-{\frac {2x^{5}}{5}}}}$