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Exact differential equation

## Summary

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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 R2 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

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

${\displaystyle x\,dx+y\,dy=0.\,}$

## Existence of potential functions

In physical applications the functions I and J are usually not only continuous but even continuously differentiable. Schwarz's Theorem then provides us with a necessary criterion for the existence of a potential function. For differential equations defined on simply connected sets the criterion is even sufficient and we get the following theorem:

Given a differential equation of the form (for example, when F has zero slope in the x and y direction at F(x,y)):

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

with I and J continuously differentiable on a simply connected and open subset D of R2 then a potential function F exists if and only if

${\displaystyle {\frac {\partial I}{\partial y}}(x,y)={\frac {\partial J}{\partial x}}(x,y).}$

## Solutions to exact differential equations

Given an exact differential equation defined on some simply connected and open subset D of R2 with potential function F, a differentiable function f with (x, f(x)) in D is a solution if and only if there exists real number c so that

${\displaystyle F(x,f(x))=c.}$

For an initial value problem

${\displaystyle y(x_{0})=y_{0}}$

we can locally find a potential function by

{\displaystyle {\begin{aligned}F(x,y)&=\int _{x_{0}}^{x}I(t,y_{0})dt+\int _{y_{0}}^{y}J(x,t)dt\\&=\int _{x_{0}}^{x}I(t,y_{0})dt+\int _{y_{0}}^{y}\left[J(x_{0},t)+\int _{x_{0}}^{x}{\frac {\partial I}{\partial y}}(u,t)\,du\right]dt.\end{aligned}}}

Solving

${\displaystyle F(x,y)=c}$

for y, where c is a real number, we can then construct all solutions.

## 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:

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

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

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

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

## Higher order exact differential equations

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

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

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