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

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

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

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

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 exact differential 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:

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

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

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

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 ${\textstyle {dy \over dx}}$ terms gives

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

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

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

Now, let there be some second-order differential equation

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

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

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

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

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

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 $C_{1}x$ 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

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

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

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

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

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

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

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

where

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

and where $F\left(x,y,{dy \over dx}\right)$ 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

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

Thus, the three conditions for exactness for a third-order differential equation are: the ${d^{2}y \over dx^{2}}$ 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

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

must be a function solely of $x$.

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

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

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

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

Integrating $I\left(x,y\right)$ with respect to $x$ gives that ${x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i\left(y\right)=0$. 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

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

The explicit solution, then, is

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

See also

• Exact differential
• Inexact differential equation

Further reading

• Boyce, William E.; DiPrima, Richard C. (1986). Elementary Differential Equations (4th ed.). New York: John Wiley & Sons, Inc. ISBN 0-471-07894-8