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,^[1]^[2] 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 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.^[3] 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 ${\textstyle {dy \over dx}}$ 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 ${\textstyle {\partial J \over \partial x}={\partial I \over \partial y}}$ . Additionally, the total derivative of $J(x,y)$ is equal to its implicit ordinary derivative ${\textstyle {dJ \over dx}}$ . 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