Exact differential equation

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


 * $$x\,dx + y\,dy = 0.\,$$

Identifying first order exact differential equations
Let the functions $M$, $N$ , $M_y$ , and $N_x$ , where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region $R: \alpha < x < \beta, \gamma < y < \delta$. 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.

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 $$\psi _y = N$$.

Differentiate both sides with respect to $$y$$.

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

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 $$x$$.

Since $$Q_x = M$$,

Now, this is zero based on our initial supposition that $$M_y(x, y) = N_x(x, y)$$

Therefore,

And this completes the proof.

Solutions to first order exact differential equations
First order exact differential equations of the form

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

where

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

The solutions to an exact differential equation are then given by

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

it follows, by integrating both sides, that

Therefore,

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,

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

and so $$f(x, y)$$ and $$d(x, y)$$ turn out to be the same function. Therefore,

Since we already showed that

it follows that

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:


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

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


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

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 ${dy \over dx}$ terms gives


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

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


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

Now, let there be some second-order differential equation


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

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


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

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


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

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


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

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


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

So,


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

and


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

Thus, the second order differential equation


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

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


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

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


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

Now, however, in the final implicit solution there will be a $$C_1x$$ term from integration of $$h\left(x\right)$$ 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


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

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


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

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


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

So, $$h\left(x\right)$$ is indeed a function only of $$x$$ and the second order differential equation is exact. Therefore, $$h\left(x\right)=C_1$$ and $$I\left(x,y\right)=-2xy+C_1$$. Reduction to a first-order exact equation yields


 * $$-2xy+C_1+\left(1-x^2\right)y'=0$$

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


 * $$-x^2y+C_1x+i\left(y\right)=0$$

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


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

So, $$i\left(y\right)=y+C_2$$ and the full implicit solution becomes


 * $$C_1x+C_2+y-x^2y=0$$

Solving explicitly for $$y$$ yields


 * $$y= \frac{C_1x+C_2}{1-x^2}$$

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


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

it was previously shown that equation is defined such that


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

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


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

where


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

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 $${d^2y \over dx^2}$$ terms not coming from $$F\left(x,y,{dy \over dx}\right)$$ gives


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

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


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

must be a function solely of $$x$$.

Example
Consider the nonlinear third-order differential equation


 * $$yy'+3y'y+12x^2=0$$

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


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

which is indeed a function only of $$x$$. So, the differential equation is exact. Integrating twice yields that $$h\left(x\right)=x^4+C_1x+C_2=I\left(x,y\right)$$. Rewriting the equation as a first-order exact differential equation yields


 * $$x^4+C_1x+C_2+yy'=0$$

Integrating $$I\left(x,y\right)$$ with respect to $$x$$ gives that $${x^5\over 5}+C_1x^2+C_2x+i\left(y\right)=0$$. Differentiating with respect to $$y$$ and equating that to the term in front of $$y'$$ in the first-order equation gives that $$i'\left(y\right)=y$$ and that $$i\left(y\right)={y^2\over 2}+C_3$$. The full implicit solution becomes


 * $${x^5\over 5}+C_1x^2+C_2x+C_3+{y^2\over 2}=0$$

The explicit solution, then, is


 * $$y=\pm\sqrt{C_1x^2+C_2x+C_3-\frac{2x^5}{5}}$$