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 edit

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

${\displaystyle x\,dx+y\,dy=0.\,}$ 

Identifying first order exact differential equations edit

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 edit

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.

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 ${\displaystyle \psi _{y}=N}$ .

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

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

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 ${\displaystyle x}$ .

Since ${\displaystyle Q_{x}=M}$ ,

Therefore,

And this completes the proof.

Solutions to first order exact differential equations edit

First order exact differential equations of the form

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

where

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

The solutions to an exact differential equation are then given by

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

it follows, by integrating both sides, that

Therefore,

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,

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

Since we already showed that

it follows that

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 ${\displaystyle h(y)}$ .

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 edit

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

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

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 edit

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

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

• Exact differential
• Inexact differential equation

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