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 ${\displaystyle \mathbb {R} ^{2))$ and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form

${\displaystyle I(x,y)\,dx+J(x,y)\,dy=0,}$

is called an exact differential equation if there exists a continuously differentiable function F, called the potential function,^[1][2] so that

${\displaystyle {\frac {\partial F}{\partial x))=I}$

and

${\displaystyle {\frac {\partial F}{\partial y))=J.}$

An exact equation may also be presented in the following form:

${\displaystyle I(x,y)+J(x,y)\,y'(x)=0}$

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 ${\displaystyle F(x_{0},x_{1},...,x_{n-1},x_{n})}$, the exact or total derivative with respect to ${\displaystyle x_{0))$ is given by

${\displaystyle {\frac {dF}{dx_{0))}={\frac {\partial F}{\partial x_{0))}+\sum _{i=1}^{n}{\frac {\partial F}{\partial x_{i))}{\frac {dx_{i)){dx_{0))}.}$

Example

The function ${\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} }$ given by

${\displaystyle F(x,y)={\frac {1}{2))(x^{2}+y^{2})+c}$

is a potential function for the differential equation

${\displaystyle x\,dx+y\,dy=0.\,}$

First order exact differential equations

Identifying first order exact differential equations

Let the functions ${\textstyle M}$, ${\textstyle N}$, ${\textstyle M_{y))$, and ${\textstyle N_{x))$, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region ${\textstyle R:\alpha <x<\beta ,\gamma <y<\delta }$. Then the differential equation

${\displaystyle M(x,y)+N(x,y){\frac {dy}{dx))=0}$

is exact if and only if

${\displaystyle M_{y}(x,y)=N_{x}(x,y)}$

That is, there exists a function ${\displaystyle \psi (x,y)}$, called a potential function, such that

${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and ))\psi _{y}(x,y)=N(x,y)}$

So, in general:

${\displaystyle M_{y}(x,y)=N_{x}(x,y)\iff {\begin{cases}\exists \psi (x,y)\\\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases))}$

Proof

The proof has two parts.

First, suppose there is a function ${\displaystyle \psi (x,y)}$ such that ${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and ))\psi _{y}(x,y)=N(x,y)}$

It then follows that ${\displaystyle M_{y}(x,y)=\psi _{xy}(x,y){\text{ and ))N_{x}(x,y)=\psi _{yx}(x,y)}$

Since ${\displaystyle M_{y))$ and ${\displaystyle N_{x))$ are continuous, then ${\displaystyle \psi _{xy))$ and ${\displaystyle \psi _{yx))$ are also continuous which guarantees their equality.

The second part of the proof involves the construction of ${\displaystyle \psi (x,y)}$ and can also be used as a procedure for solving first-order exact differential equations. Suppose that ${\displaystyle M_{y}(x,y)=N_{x}(x,y)}$ and let there be a function ${\displaystyle \psi (x,y)}$ for which ${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and ))\psi _{y}(x,y)=N(x,y)}$

Begin by integrating the first equation with respect to ${\displaystyle x}$. 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.

$${\displaystyle {\frac {\partial \psi }{\partial x))(x,y)=M(x,y)}$$

$${\displaystyle \psi (x,y)=\int {M(x,y)dx}+h(y)}$$

$${\displaystyle \psi (x,y)=Q(x,y)+h(y)}$$

where ${\displaystyle Q(x,y)}$ is any differentiable function such that ${\displaystyle Q_{x}=M}$. The function ${\displaystyle h(y)}$ plays the role of a constant of integration, but instead of just a constant, it is function of ${\displaystyle y}$, since ${\displaystyle M}$ is a function of both ${\displaystyle x}$ and ${\displaystyle y}$ and we are only integrating with respect to ${\displaystyle x}$.

Now to show that it is always possible to find an ${\displaystyle h(y)}$ such that ${\displaystyle \psi _{y}=N}$.

$${\displaystyle \psi (x,y)=Q(x,y)+h(y)}$$

Differentiate both sides with respect to ${\displaystyle y}$.

$${\displaystyle {\frac {\partial \psi }{\partial y))(x,y)={\frac {\partial Q}{\partial y))(x,y)+h'(y)}$$

Set the result equal to ${\displaystyle N}$ and solve for ${\displaystyle h'(y)}$.

$${\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y))(x,y)}$$

In order to determine ${\displaystyle h'(y)}$ from this equation, the right-hand side must depend only on ${\displaystyle y}$. This can be proven by showing that its derivative with respect to ${\displaystyle x}$ is always zero, so differentiate the right-hand side with respect to ${\displaystyle x}$.

$${\displaystyle {\frac {\partial N}{\partial x))(x,y)-{\frac {\partial }{\partial x)){\frac {\partial Q}{\partial y))(x,y)\iff {\frac {\partial N}{\partial x))(x,y)-{\frac {\partial }{\partial y)){\frac {\partial Q}{\partial x))(x,y)}$$

Since ${\displaystyle Q_{x}=M}$,

$${\displaystyle {\frac {\partial N}{\partial x))(x,y)-{\frac {\partial M}{\partial y))(x,y)}$$

${\displaystyle M_{y}(x,y)=N_{x}(x,y)}$

Therefore,

$${\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y))(x,y)}$$

$${\displaystyle h(y)=\int {\left(N(x,y)-{\frac {\partial Q}{\partial y))(x,y)\right)dy))$$

$${\displaystyle \psi (x,y)=Q(x,y)+\int {\left(N(x,y)-{\frac {\partial Q}{\partial y))(x,y)\right)dy}+C}$$

And this completes the proof.

Solutions to first order exact differential equations

First order exact differential equations of the form

$${\displaystyle M(x,y)+N(x,y){\frac {dy}{dx))=0}$$

can be written in terms of the potential function ${\displaystyle \psi (x,y)}$

$${\displaystyle {\frac {\partial \psi }{\partial x))+{\frac {\partial \psi }{\partial y)){\frac {dy}{dx))=0}$$

where

$${\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases))}$$

This is equivalent to taking the exact differential of ${\displaystyle \psi (x,y)}$.

$${\displaystyle {\frac {\partial \psi }{\partial x))+{\frac {\partial \psi }{\partial y)){\frac {dy}{dx))=0\iff {\frac {d}{dx))\psi (x,y(x))=0}$$

The solutions to an exact differential equation are then given by

$${\displaystyle \psi (x,y(x))=c}$$

and the problem reduces to finding ${\displaystyle \psi (x,y)}$.

This can be done by integrating the two expressions ${\displaystyle M(x,y)dx}$ and ${\displaystyle N(x,y)dy}$ and then writing down each term in the resulting expressions only once and summing them up in order to get ${\displaystyle \psi (x,y)}$.

The reasoning behind this is the following. Since

$${\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases))}$$

it follows, by integrating both sides, that

$${\displaystyle {\begin{cases}\psi (x,y)=\int {M(x,y)dx}+h(y)=Q(x,y)+h(y)\\\psi (x,y)=\int {N(x,y)dy}+g(x)=P(x,y)+g(x)\end{cases))}$$

Therefore,

$${\displaystyle Q(x,y)+h(y)=P(x,y)+g(x)}$$

where ${\displaystyle Q(x,y)}$ and ${\displaystyle P(x,y)}$ are differentiable functions such that ${\displaystyle Q_{x}=M}$ and ${\displaystyle P_{y}=N}$.

In order for this to be true and for both sides to result in the exact same expression, namely ${\displaystyle \psi (x,y)}$, then ${\displaystyle h(y)}$ must be contained within the expression for ${\displaystyle P(x,y)}$ because it cannot be contained within ${\displaystyle g(x)}$, since it is entirely a function of ${\displaystyle y}$ and not ${\displaystyle x}$ and is therefore not allowed to have anything to do with ${\displaystyle x}$. By analogy, ${\displaystyle g(x)}$ must be contained within the expression ${\displaystyle Q(x,y)}$.

Ergo,

$${\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and ))P(x,y)=h(y)+d(x,y)}$$

for some expressions ${\displaystyle f(x,y)}$ and ${\displaystyle d(x,y)}$. Plugging in into the above equation, we find that

$${\displaystyle g(x)+f(x,y)+h(y)=h(y)+d(x,y)+g(x)\Rightarrow f(x,y)=d(x,y)}$$

${\displaystyle f(x,y)}$

${\displaystyle d(x,y)}$

$${\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and ))P(x,y)=h(y)+f(x,y)}$$

Since we already showed that

$${\displaystyle {\begin{cases}\psi (x,y)=Q(x,y)+h(y)\\\psi (x,y)=P(x,y)+g(x)\end{cases))}$$

it follows that

$${\displaystyle \psi (x,y)=g(x)+f(x,y)+h(y)}$$

So, we can construct ${\displaystyle \psi (x,y)}$ by doing ${\displaystyle \int {M(x,y)dx))$ and ${\displaystyle \int {N(x,y)dy))$ and then taking the common terms we find within the two resulting expressions (that would be ${\displaystyle f(x,y)}$ ) and then adding the terms which are uniquely found in either one of them - ${\displaystyle g(x)}$ and ${\displaystyle h(y)}$.

Second order exact differential equations

The concept of exact differential equations can be extended to second order equations.^[3] Consider starting with the first-order exact equation:

${\displaystyle I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0}$

Since both functions ${\displaystyle I\left(x,y\right)}$, ${\displaystyle J\left(x,y\right)}$ are functions of two variables, implicitly differentiating the multivariate function yields

${\displaystyle {dI \over dx}+\left({dJ \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}$

Expanding the total derivatives gives that

${\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx))$

and that

${\displaystyle {dJ \over dx}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx))$

Combining the ${\textstyle {dy \over dx))$ terms gives

${\displaystyle {\partial I \over \partial x}+{dy \over dx}\left({\partial I \over \partial y}+{\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}\right)+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}$

If the equation is exact, then ${\textstyle {\partial J \over \partial x}={\partial I \over \partial y))$. Additionally, the total derivative of ${\displaystyle J\left(x,y\right)}$ is equal to its implicit ordinary derivative ${\textstyle {dJ \over dx))$. This leads to the rewritten equation

${\displaystyle {\partial I \over \partial x}+{dy \over dx}\left({\partial J \over \partial x}+{dJ \over dx}\right)+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}$

Now, let there be some second-order differential equation

${\displaystyle f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}$

If ${\displaystyle {\partial J \over \partial x}={\partial I \over \partial y))$ for exact differential equations, then

${\displaystyle \int \left({\partial I \over \partial y}\right)dy=\int \left({\partial J \over \partial x}\right)dy}$

and

${\displaystyle \int \left({\partial I \over \partial y}\right)dy=\int \left({\partial J \over \partial x}\right)dy=I\left(x,y\right)-h\left(x\right)}$

where ${\displaystyle h\left(x\right)}$ is some arbitrary function only of ${\displaystyle x}$ that was differentiated away to zero upon taking the partial derivative of ${\displaystyle I\left(x,y\right)}$ with respect to ${\displaystyle y}$. Although the sign on ${\displaystyle h\left(x\right)}$ could be positive, it is more intuitive to think of the integral's result as ${\displaystyle I\left(x,y\right)}$ that is missing some original extra function ${\displaystyle h\left(x\right)}$ that was partially differentiated to zero.

Next, if

${\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx))$

then the term ${\displaystyle {\partial I \over \partial x))$ should be a function only of ${\displaystyle x}$ and ${\displaystyle y}$, since partial differentiation with respect to ${\displaystyle x}$ will hold ${\displaystyle y}$ constant and not produce any derivatives of ${\displaystyle y}$. In the second order equation

${\displaystyle f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}$

only the term ${\displaystyle f\left(x,y\right)}$ is a term purely of ${\displaystyle x}$ and ${\displaystyle y}$. Let ${\displaystyle {\partial I \over \partial x}=f\left(x,y\right)}$. If ${\displaystyle {\partial I \over \partial x}=f\left(x,y\right)}$, then

${\displaystyle f\left(x,y\right)={dI \over dx}-{\partial I \over \partial y}{dy \over dx))$

Since the total derivative of ${\displaystyle I\left(x,y\right)}$ with respect to ${\displaystyle x}$ is equivalent to the implicit ordinary derivative ${\displaystyle {dI \over dx))$ , then

${\displaystyle f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}={dI \over dx}={d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)+{dh\left(x\right) \over dx))$

So,

${\displaystyle {dh\left(x\right) \over dx}=f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)}$

and

${\displaystyle h\left(x\right)=\int \left(f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx}$

Thus, the second order differential equation

${\displaystyle f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}$

is exact only if ${\displaystyle g\left(x,y,{dy \over dx}\right)={dJ \over dx}+{\partial J \over \partial x}={dJ \over dx}+{\partial J \over \partial x))$ and only if the below expression

${\displaystyle \int \left(f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx=\int \left(f\left(x,y\right)-{\partial \left(I\left(x,y\right)-h\left(x\right)\right) \over \partial x}\right)dx}$

is a function solely of ${\displaystyle x}$. Once ${\displaystyle h\left(x\right)}$ is calculated with its arbitrary constant, it is added to ${\displaystyle I\left(x,y\right)-h\left(x\right)}$ to make ${\displaystyle I\left(x,y\right)}$. 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.

${\displaystyle I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0}$

Now, however, in the final implicit solution there will be a ${\displaystyle C_{1}x}$ term from integration of ${\displaystyle h\left(x\right)}$ with respect to ${\displaystyle x}$ twice as well as a ${\displaystyle C_{2))$, two arbitrary constants as expected from a second-order equation.

Example

Given the differential equation

${\displaystyle \left(1-x^{2}\right)y''-4xy'-2y=0}$

one can always easily check for exactness by examining the ${\displaystyle y''}$ term. In this case, both the partial and total derivative of ${\displaystyle 1-x^{2))$ with respect to ${\displaystyle x}$ are ${\displaystyle -2x}$, so their sum is ${\displaystyle -4x}$, which is exactly the term in front of ${\displaystyle y'}$. With one of the conditions for exactness met, one can calculate that

${\displaystyle \int \left(-2x\right)dy=I\left(x,y\right)-h\left(x\right)=-2xy}$

Letting ${\displaystyle f\left(x,y\right)=-2y}$, then

${\displaystyle \int \left(-2y-2xy'-{d \over dx}\left(-2xy\right)\right)dx=\int \left(-2y-2xy'+2xy'+2y\right)dx=\int \left(0\right)dx=h\left(x\right)}$

So, ${\displaystyle h\left(x\right)}$ is indeed a function only of ${\displaystyle x}$ and the second order differential equation is exact. Therefore, ${\displaystyle h\left(x\right)=C_{1))$ and ${\displaystyle I\left(x,y\right)=-2xy+C_{1))$. Reduction to a first-order exact equation yields

${\displaystyle -2xy+C_{1}+\left(1-x^{2}\right)y'=0}$

Integrating ${\displaystyle I\left(x,y\right)}$ with respect to ${\displaystyle x}$ yields

${\displaystyle -x^{2}y+C_{1}x+i\left(y\right)=0}$

where ${\displaystyle i\left(y\right)}$ is some arbitrary function of ${\displaystyle y}$. Differentiating with respect to ${\displaystyle y}$ gives an equation correlating the derivative and the ${\displaystyle y'}$ term.

${\displaystyle -x^{2}+i'\left(y\right)=1-x^{2))$

So, ${\displaystyle i\left(y\right)=y+C_{2))$ and the full implicit solution becomes

${\displaystyle C_{1}x+C_{2}+y-x^{2}y=0}$

Solving explicitly for ${\displaystyle y}$ yields

${\displaystyle y={\frac {C_{1}x+C_{2)){1-x^{2))))$

Higher order exact differential equations

The concepts of exact differential equations can be extended to any order. Starting with the exact second order equation

${\displaystyle {d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)+{dy \over dx}\left({dJ \over dx}+{\partial J \over \partial x}\right)+f\left(x,y\right)=0}$

it was previously shown that equation is defined such that

${\displaystyle f\left(x,y\right)={dh\left(x\right) \over dx}+{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)-{\partial J \over \partial x}{dy \over dx))$

Implicit differentiation of the exact second-order equation ${\displaystyle n}$ times will yield an ${\displaystyle \left(n+2\right)}$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

${\displaystyle {d^{3}y \over dx^{3))\left(J\left(x,y\right)\right)+{d^{2}y \over dx^{2)){dJ \over dx}+{d^{2}y \over dx^{2))\left({dJ \over dx}+{\partial J \over \partial x}\right)+{dy \over dx}\left({d^{2}J \over dx^{2))+{d \over dx}\left({\partial J \over \partial x}\right)\right)+{df\left(x,y\right) \over dx}=0}$

where

${\displaystyle {df\left(x,y\right) \over dx}={d^{2}h\left(x\right) \over dx^{2))+{d^{2} \over dx^{2))\left(I\left(x,y\right)-h\left(x\right)\right)-{d^{2}y \over dx^{2)){\partial J \over \partial x}-{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)=F\left(x,y,{dy \over dx}\right)}$

and where ${\displaystyle F\left(x,y,{dy \over dx}\right)}$ is a function only of ${\displaystyle x,y}$ and ${\displaystyle {dy \over dx))$. Combining all ${\displaystyle {dy \over dx))$ and ${\displaystyle {d^{2}y \over dx^{2))}$ terms not coming from ${\displaystyle F\left(x,y,{dy \over dx}\right)}$ gives

${\displaystyle {d^{3}y \over dx^{3))\left(J\left(x,y\right)\right)+{d^{2}y \over dx^{2))\left(2{dJ \over dx}+{\partial J \over \partial x}\right)+{dy \over dx}\left({d^{2}J \over dx^{2))+{d \over dx}\left({\partial J \over \partial x}\right)\right)+F\left(x,y,{dy \over dx}\right)=0}$

Thus, the three conditions for exactness for a third-order differential equation are: the ${\displaystyle {d^{2}y \over dx^{2))}$ term must be ${\displaystyle 2{dJ \over dx}+{\partial J \over \partial x))$, the ${\displaystyle {dy \over dx))$ term must be ${\displaystyle {d^{2}J \over dx^{2))+{d \over dx}\left({\partial J \over \partial x}\right)}$ and

${\displaystyle F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2))\left(I\left(x,y\right)-h\left(x\right)\right)+{d^{2}y \over dx^{2)){\partial J \over \partial x}+{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)}$

must be a function solely of ${\displaystyle x}$.

Example

Consider the nonlinear third-order differential equation

${\displaystyle yy'''+3y'y''+12x^{2}=0}$

If ${\displaystyle J\left(x,y\right)=y}$, then ${\displaystyle y''\left(2{dJ \over dx}+{\partial J \over \partial x}\right)}$ is ${\displaystyle 2y'y''}$ and ${\displaystyle y'\left({d^{2}J \over dx^{2))+{d \over dx}\left({\partial J \over \partial x}\right)\right)=y'y''}$which together sum to ${\displaystyle 3y'y''}$. Fortunately, this appears in our equation. For the last condition of exactness,

${\displaystyle F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2))\left(I\left(x,y\right)-h\left(x\right)\right)+{d^{2}y \over dx^{2)){\partial J \over \partial x}+{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)=12x^{2}-0+0+0=12x^{2))$

which is indeed a function only of ${\displaystyle x}$. So, the differential equation is exact. Integrating twice yields that ${\displaystyle h\left(x\right)=x^{4}+C_{1}x+C_{2}=I\left(x,y\right)}$. Rewriting the equation as a first-order exact differential equation yields

${\displaystyle x^{4}+C_{1}x+C_{2}+yy'=0}$

Integrating ${\displaystyle I\left(x,y\right)}$ with respect to ${\displaystyle x}$ gives that ${\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i\left(y\right)=0}$. Differentiating with respect to ${\displaystyle y}$ and equating that to the term in front of ${\displaystyle y'}$ in the first-order equation gives that ${\displaystyle i'\left(y\right)=y}$ and that ${\displaystyle i\left(y\right)={y^{2} \over 2}+C_{3))$. The full implicit solution becomes

${\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+C_{3}+{y^{2} \over 2}=0}$

The explicit solution, then, is

${\displaystyle y=\pm {\sqrt {C_{1}x^{2}+C_{2}x+C_{3}-{\frac {2x^{5)){5))))}$