Exact Differential Equations

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

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

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

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

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)):

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

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

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

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

$F(x,f(x))=c.$

For an initial value problem

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

we can locally find a potential function by

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

Solving

$F(x,y)=c$

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. 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

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

Licensing

Content obtained and/or adapted from:

• Exact differential equation, Wikipedia under a CC BY-SA license