View source for Exact differential equation

{{Short description|Type of differential equation subject to a particular solution methodology}}
{{Differential equations}}

In [[Mathematics|mathematics]], an '''exact differential equation''' or '''total differential equation''' is a certain kind of [[Ordinary differential equation|ordinary differential equation]] which is widely used in physics and engineering.

==Definition==
Given a simply connected and [[Open set|open]] subset ''D'' of <math>\mathbb{R}^2</math> and two functions ''I'' and ''J'' which are [[Continuous function|continuous]] on ''D'', an implicit first-order ordinary differential equation of the form

: <math>I(x, y)\, dx + J(x, y)\, dy = 0,</math>

is called an '''exact differential equation''' if there exists a continuously differentiable function ''F'', called the '''potential function''',<ref name="Walter2013">{{cite book|author=Wolfgang Walter|title=Ordinary Differential Equations|url=https://books.google.com/books?id=2jvaBwAAQBAJ&q=%22potential+function%22+exact|date=11 March 2013|publisher=Springer Science & Business Media|isbn=978-1-4612-0601-9}}</ref><ref name="Dobrushkin2014">{{cite book|author=Vladimir A. Dobrushkin|title=Applied Differential Equations: The Primary Course|url=https://books.google.com/books?id=d-5MBgAAQBAJ&q=%22potential+function%22|date=16 December 2014|publisher=CRC Press|isbn=978-1-4987-2835-5}}</ref> so that
:<math>\frac{\partial F}{\partial x} = I</math>
and
:<math>\frac{\partial F}{\partial y} = J.</math>

An exact equation may also be presented in the following form:
:<math>I(x, y) + J(x, y) \, y'(x) = 0</math>
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|exact differential]] of a function. For a function <math>F(x_0, x_1,...,x_{n-1},x_n)</math>, the exact or [[Total derivative|total derivative]] with respect to <math>x_0</math> is given by
:<math>\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}.</math>

===Example===
The function <math>F:\mathbb{R}^{2}\to\mathbb{R}</math> given by

:<math>F(x,y) = \frac{1}{2}(x^2 + y^2)+c</math>

is a potential function for the differential equation

:<math>x\,dx + y\,dy = 0.\,</math>

==First order exact differential equations==
===Identifying first order exact differential equations===
Let the functions <math display="inline">M</math>, <math display="inline">N</math>, <math display="inline">M_y</math>, and <math display="inline">N_x</math>, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region <math display="inline">R: \alpha < x < \beta, \gamma < y < \delta</math>. Then the differential equation

<math>M(x, y) + N(x, y)\frac{dy}{dx} = 0</math>

is exact if and only if

<math>M_y(x, y) = N_x(x, y)</math>

That is, there exists a function <math>\psi(x, y)</math>, called a '' potential function'', such that

<math>\psi _x(x, y) = M(x, y) \text{ and } \psi_y(x, y) = N(x, y)</math>

So, in general:

<math>
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}
</math>

====Proof====
The proof has two parts.

First, suppose there is a function <math>\psi(x,y)</math> such that <math>\psi_x(x, y) = M(x, y) \text{ and } \psi_y(x, y) = N(x, y)</math>

It then follows that <math>M_y(x, y) = \psi _{xy}(x, y) \text{ and } N_x(x, y) = \psi _{yx}(x, y)</math>

Since <math>M_y</math> and <math>N_x</math> are continuous, then <math>\psi _{xy}</math> and <math>\psi _{yx}</math> are also continuous which guarantees their equality.

The second part of the proof involves the construction of <math>\psi(x, y)</math> and can also be used as a procedure for solving first-order exact differential equations. Suppose that <math>M_y(x, y) = N_x(x, y)</math> and let there be a function <math>\psi(x, y)</math> for which
<math>\psi _x(x, y) = M(x, y) \text{ and } \psi _y(x, y) = N(x, y)</math>

Begin by integrating the first equation with respect to <math>x</math>. 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.

<math display = block>
\frac{\partial \psi}{\partial x}(x, y) = M(x, y)
</math>
<math display = block>
\psi(x, y) = \int{M(x, y) dx} + h(y)
</math>
<math display = block>
\psi(x, y) = Q(x, y) + h(y)
</math>

where <math>Q(x, y)</math> is any differentiable function such that <math>Q_x = M</math>. The function <math>h(y)</math> plays the role of a constant of integration, but instead of just a constant, it is function of <math>y</math>, since <math>M</math> is a function of both <math>x</math> and <math>y</math> and we are only integrating with respect to <math>x</math>.

Now to show that it is always possible to find an <math>h(y)</math> such that <math>\psi _y = N</math>.
<math display = block>
\psi(x, y) = Q(x, y) + h(y)
</math>

Differentiate both sides with respect to <math>y</math>.
<math display = block>
\frac{\partial \psi}{\partial y}(x, y) = \frac{\partial Q}{\partial y}(x, y) + h'(y)
</math>

Set the result equal to <math>N</math> and solve for <math>h'(y)</math>.
<math display = block>
h'(y) = N(x, y) - \frac{\partial Q}{\partial y}(x, y)
</math>

In order to determine <math>h'(y)</math> from this equation, the right-hand side must depend only on <math>y</math>. This can be proven by showing that its derivative with respect to <math>x</math> is always zero, so differentiate the right-hand side with respect to <math>x</math>. 
<math display = block>
\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)
</math>

Since <math>Q_x = M</math>,
<math display = block>
\frac{\partial N}{\partial x}(x, y) - \frac{\partial M}{\partial y}(x, y)
</math>
Now, this is zero based on our initial supposition that <math>M_y(x, y) = N_x(x, y)</math>

Therefore,
<math display = block>
h'(y) = N(x, y) - \frac{\partial Q}{\partial y}(x, y)
</math>
<math display = block>
h(y) = \int{\left(N(x, y) - \frac{\partial Q}{\partial y}(x, y)\right) dy}
</math>

<math display = block>
\psi(x, y) = Q(x, y) + \int{\left(N(x, y) - \frac{\partial Q}{\partial y}(x, y)\right) dy} + C
</math>

And this completes the proof.

===Solutions to first order exact differential equations===
First order exact differential equations of the form
<math display = block>
M(x, y) + N(x, y)\frac{dy}{dx} = 0
</math>

can be written in terms of the potential function <math>\psi(x, y)</math>
<math display = block>
\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{dy}{dx} = 0
</math>

where
<math display = block>
\begin{cases}
\psi _x(x, y) = M(x, y)\\
\psi _y(x, y) = N(x, y)
\end{cases}
</math>

This is equivalent to taking the [[Exact differential|exact differential]] of <math>\psi(x,y)</math>.
<math display = block>
\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{dy}{dx} = 0 \iff \frac{d}{dx}\psi(x, y(x)) = 0
</math>

The solutions to an exact differential equation are then given by
<math display = block>
\psi(x, y(x)) = c
</math>

and the problem reduces to finding <math>\psi(x, y)</math>.

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

The reasoning behind this is the following. Since
<math display = block>
\begin{cases}
\psi _x(x, y) = M(x, y)\\
\psi _y(x, y) = N(x, y)
\end{cases}
</math>

it follows, by integrating both sides, that
<math display = block>
\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}
</math>

Therefore,
<math display = block>
Q(x, y) + h(y) = P(x, y) + g(x)
</math>

where <math>Q(x, y)</math> and <math>P(x, y)</math> are differentiable functions such that <math>Q_x = M</math> and <math>P_y = N</math>.

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

Ergo,
<math display = block>
Q(x, y) = g(x) + f(x, y) \text{ and } P(x, y) = h(y) + d(x, y)
</math>

for some expressions <math>f(x, y)</math> and <math>d(x, y)</math>.
Plugging in into the above equation, we find that
<math display = block>
g(x) + f(x, y) + h(y) = h(y) + d(x, y) + g(x) \Rightarrow f(x, y) = d(x, y)
</math>
and so <math>f(x, y)</math> and <math>d(x, y)</math> turn out to be the same function. Therefore,
<math display = block>
Q(x, y) = g(x) + f(x, y) \text { and } P(x, y) = h(y) + f(x, y)
</math>

Since we already showed that
<math display = block>
\begin{cases}
\psi(x, y) = Q(x, y) + h(y)\\
\psi(x, y) = P(x, y) + g(x)
\end{cases}
</math>

it follows that
<math display = block>
\psi(x, y) = g(x) + f(x, y) + h(y)
</math>

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

== Second order exact differential equations ==
The concept of exact differential equations can be extended to second order equations.<ref>{{Cite book|title=Ordinary Differential Equations: An Elementary Textbook for Students of Mathematics, Engineering and the Sciences|url=https://archive.org/details/ordinarydifferen00tene_850|url-access=limited|last1=Tenenbaum|first1=Morris|last2=Pollard|first2=Harry|publisher=Dover|year=1963|isbn=0-486-64940-7|location=New York|pages=[https://archive.org/details/ordinarydifferen00tene_850/page/n131 248]|chapter=Solution of the Linear Differential Equation with Nonconstant Coefficients. Reduction of Order Method.}}</ref> Consider starting with the first-order exact equation:

:<math>I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0</math>

Since both functions {{nowrap|<math>I\left(x,y\right)</math>,}} <math>J\left(x,y\right)</math> are functions of two variables, implicitly differentiating the multivariate function yields

:<math>{dI \over dx} +\left({ dJ\over dx}\right){dy \over dx}+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

Expanding the total derivatives gives that

:<math>{dI \over dx}={\partial I\over\partial x}+{\partial I\over\partial y}{dy \over dx}</math>

and that

:<math>{dJ \over dx}={\partial J\over\partial x}+{\partial J\over\partial y}{dy \over dx}</math>

Combining the <math display="inline">{dy \over dx}</math> terms gives 

:<math>{\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</math>

If the equation is exact, then {{nowrap|<math display="inline">{\partial J\over\partial x}={\partial I\over\partial y}</math>.}} Additionally, the total derivative of <math>J\left(x,y\right)</math> is equal to its implicit ordinary derivative <math display="inline">{dJ \over dx}</math>. This leads to the rewritten equation 

:<math>{\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</math>

Now, let there be some second-order differential equation

:<math>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</math>

If <math>{\partial J\over\partial x}={\partial I\over\partial y}</math> for exact differential equations, then 

:<math>\int \left({\partial I\over\partial y}\right)dy=\int \left({\partial J\over\partial x}\right)dy</math>

and

:<math>\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)</math>

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

Next, if

:<math>{dI\over dx}={\partial I\over\partial x}+{\partial I\over\partial y}{dy \over dx}</math>

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

:<math>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</math>

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

:<math>f\left(x,y\right)={ dI\over dx}-{\partial I\over\partial y}{dy \over dx}</math>

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

:<math>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}</math>

So,

:<math>{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)</math>

and

:<math>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</math>

Thus, the second order differential equation

:<math>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</math>

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

:<math>\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</math>

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

:<math>I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0</math>

Now, however, in the final implicit solution there will be a <math>C_1x</math> term from integration of <math>h\left(x\right)</math> with respect to <math>x</math> twice as well as a <math>C_2</math>, two arbitrary constants as expected from a second-order equation. 

=== Example ===
Given the differential equation

:<math>\left(1-x^2\right)y''-4xy'-2y=0</math>

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

:<math>\int \left(-2x\right)dy=I\left(x,y\right)-h\left(x\right)=-2xy</math>

Letting <math>f\left(x,y\right)=-2y</math>, then 

:<math>\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)</math>

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

:<math>-2xy+C_1+\left(1-x^2\right)y'=0</math>

Integrating <math>I\left(x,y\right)</math> with respect to <math>x</math> yields

:<math>-x^2y+C_1x+i\left(y\right)=0</math>

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

:<math>-x^2+i'\left(y\right)=1-x^2</math>

So, <math>i\left(y\right)=y+C_2</math> and the full implicit solution becomes

:<math>C_1x+C_2+y-x^2y=0</math>

Solving explicitly for <math>y</math> yields

:<math>y= \frac{C_1x+C_2}{1-x^2}</math>

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

:<math>{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</math>

it was previously shown that equation is defined such that

:<math>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}</math>

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

:<math>{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</math>

where

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

:<math>{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</math>

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

:<math>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)</math>

must be a function solely of <math>x</math>. 

=== Example ===
Consider the nonlinear third-order differential equation

:<math>yy'''+3y'y''+12x^2=0</math>

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

:<math>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</math>

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

:<math>x^4+C_1x+C_2+yy'=0</math>

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

:<math>{x^5\over 5}+C_1x^2+C_2x+C_3+{y^2\over 2}=0</math>

The explicit solution, then, is

:<math>y=\pm\sqrt{C_1x^2+C_2x+C_3-\frac{2x^5}{5}}</math>

==See also==
* [[Exact differential]]
* [[Inexact differential equation]]

==References==
{{reflist}}

== 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}}

{{Differential equations topics}}

[[Category:Ordinary differential equations]]

{{Sourceattribution|Exact differential equation}}