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.

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, 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

\(\frac{dF}{dx_0}=\frac{\partial F}{\partial x_0}+\sum_{i=1}^{n}\frac{\partial F}{\partial x_i}\frac{dx_i}{dx_0}.\)

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

\(x\,dx + y\,dy = 0.\,\)

Let the functions \(M\), \(N\), \(M_y\), and \(N_x\), where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region 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:

Condensed: the full section is in Wikipedia.

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 \(\psi(x, y)\).

Condensed: the full section is in Wikipedia.

The concept of exact differential equations can be extended to second-order equations. 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

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

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 \({dy \over dx}\) terms gives

\({\partial I\over\partial x}+{dy \over dx}\left({\partial I\over\partial y}+{\partial J\over\partial x}+{\partial J\over\partial y}{dy \over dx}\right)+{d^2y \over dx^2} (J(x,y))=0\)

If the equation is exact, then \({\partial J\over\partial x}={\partial I\over\partial y}\). Additionally, the total derivative of \(J(x,y)\) is equal to its implicit ordinary derivative \({dJ \over dx}\). This leads to the rewritten equation

\({\partial I\over\partial x}+{dy \over dx}\left({\partial J\over\partial x}+{dJ \over dx}\right)+{d^2y \over dx^2} (J(x,y))=0\)

Now, let there be some second-order differential equation

\(f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^2y \over dx^2} (J(x,y))=0\)

\(\int \left({\partial I\over\partial y}\right) \, dy=\int \left({\partial J\over\partial x}\right) \, dy\)

\(\int \left({\partial I\over\partial y}\right) \, dy=\int \left({\partial J\over\partial x}\right) \, dy=I(x,y)-h(x)\)

\({dI\over dx}={\partial I\over\partial x}+{\partial I\over\partial y}{dy \over dx}\)

\(f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^2y \over dx^2} (J(x,y))=0\)

\(f(x,y)={ dI\over dx}-{\partial I\over\partial y}{dy \over dx}\)

\(f(x,y)+{\partial I\over\partial y}{dy \over dx}={dI \over dx}={d \over dx}(I(x,y)-h(x))+{dh(x) \over dx}\)

\({dh(x) \over dx}=f(x,y)+{\partial I\over\partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))\)

\(h(x) =\int\left(f(x,y)+{\partial I\over\partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))\right) \, dx\)

\(f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^2y \over dx^2} (J(x,y))=0\)

\(\int\left(f(x,y)+{\partial I\over\partial y}{dy \over dx}-{d \over dx} (I(x,y)-h(x)) \right) \, dx=\int \left(f(x,y)-{\partial \left(I(x,y)-h(x)\right)\over\partial x}\right) \, dx\)

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

Condensed: the full section is in Wikipedia.

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

\(-x^2y+C_1x+i(y)=0\)

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

\(C_1x+C_2+y-x^2y=0\)

\(y= \frac{C_1x+C_2}{1-x^2}\)

Condensed: the full section is in Wikipedia.

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

\({d^2y \over dx^2}(J(x,y))+{dy \over dx}\left({dJ \over dx}+{\partial J\over\partial x}\right)+f(x,y)=0\)

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

\({d^3y \over dx^3}(J(x,y))+{d^2y \over dx^2}{dJ \over dx}+{d^2y \over dx^2}\left({dJ \over dx}+{\partial J\over\partial x}\right)+{dy \over dx}\left({d^2J \over dx^2}+{d \over dx}\left({\partial J\over\partial x}\right)\right)+{df(x,y) \over dx}=0\)

where

\({df(x,y) \over dx}={d^2h(x) \over dx^2}+{d^2 \over dx^2} (I(x,y)-h(x))-{d^2y \over dx^2}{\partial J\over\partial x}-{dy \over dx}{d \over dx}\left({\partial J\over\partial x}\right)=F\left(x,y,{dy \over dx}\right)\)

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 \({d^2y \over dx^2}\) terms not coming from \(F\left(x,y,{dy \over dx}\right)\) gives

\({d^3y \over dx^3}(J(x,y))+{d^2y \over dx^2}\left(2{dJ \over dx}+{\partial J\over\partial x}\right)+{dy \over dx}\left({d^2J \over dx^2}+{d \over dx}\left({\partial J\over\partial x}\right)\right)+F\left(x,y,{dy \over dx}\right)=0\)

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

\(F\left(x,y,{dy \over dx}\right)-{d^2 \over dx^2} (I(x,y)-h(x))+{d^2y \over dx^2}{\partial J\over\partial x}+{dy \over dx}{d \over dx}\left({\partial J\over\partial x}\right)\)

must be a function solely of \(x\).

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

\(F\left(x,y,{dy \over dx}\right)-{d^2 \over dx^2}\left(I(x,y)-h(x)\right)+{d^2y \over dx^2}{\partial J\over\partial x}+{dy \over dx}{d \over dx}\left({\partial J\over\partial x}\right)=12x^2-0+0+0=12x^2\)

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

\(x^4+C_1x+C_2+yy'=0\)

Integrating \(I(x,y)\) with respect to \(x\) gives that \({x^5\over 5}+C_1x^2+C_2x+i(y)=0\). 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

\({x^5\over 5}+C_1x^2+C_2x+C_3+{y^2\over 2}=0\)

The explicit solution, then, is

\(y=\pm\sqrt{C_1x^2+C_2x+C_3-\frac{2x^5}{5}}\)