Exact differential equation

Short description: Type of differential equation subject to a particular solution methodology

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 ${\mathbb{ℝ}}^2$ 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:

$I(x,y)+J(x,y)y^{\prime }(x)=0$

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

$\frac{dF}{d{x}_0}=\frac{\partial F}{\partial x_0}+\sum _{i=1}^n\frac{\partial F}{\partial x_i}\frac{d{x}_i}{d{x}_0}.$

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

$xdx+ydy=0.$

First order exact differential equations

Identifying first order exact differential equations

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:

$M_y(x,y)=N_x(x,y)⟺\{\begin{array}{l}\mathrm{\exists }\psi (x,y)\\ {\psi }_x(x,y)=M(x,y)\\ {\psi }_y(x,y)=N(x,y)\end{array}$

Proof

The proof has two parts.

First, suppose there is a function $\psi (x,y)$ such that ${\psi }_x(x,y)=M(x,y)\text{ and }{\psi }_y(x,y)=N(x,y)$

It then follows that $M_y(x,y)={\psi }_{xy}(x,y)\text{ and }{N}_x(x,y)={\psi }_{yx}(x,y)$

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

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

Begin by integrating the first equation with respect to $x$. 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.

$$\frac{\partial \psi }{\partial x}(x,y)=M(x,y)$$ $$\psi (x,y)=\int M(x,y)dx+h(y)$$ $$\psi (x,y)=Q(x,y)+h(y)$$

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

Now to show that it is always possible to find an $h(y)$ such that ${\psi }_y=N$. $$\psi (x,y)=Q(x,y)+h(y)$$

Differentiate both sides with respect to $y$. $$\frac{\partial \psi }{\partial y}(x,y)=\frac{\partial Q}{\partial y}(x,y)+h^{\prime }(y)$$

Set the result equal to $N$ and solve for $h^{\prime }(y)$. $$h^{\prime }(y)=N(x,y)-\frac{\partial Q}{\partial y}(x,y)$$

In order to determine $h^{\prime }(y)$ from this equation, the right-hand side must depend only on $y$. This can be proven by showing that its derivative with respect to $x$ is always zero, so differentiate the right-hand side with respect to $x$. $$\frac{\partial N}{\partial x}(x,y)-\frac{\partial }{\partial x}\frac{\partial Q}{\partial y}(x,y)⟺\frac{\partial N}{\partial x}(x,y)-\frac{\partial }{\partial y}\frac{\partial Q}{\partial x}(x,y)$$

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

Therefore, $$h^{\prime }(y)=N(x,y)-\frac{\partial Q}{\partial y}(x,y)$$ $$h(y)=\int (N(x,y)-\frac{\partial Q}{\partial y}(x,y))dy$$

$$\psi (x,y)=Q(x,y)+\int (N(x,y)-\frac{\partial Q}{\partial y}(x,y))dy+C$$

And this completes the proof.

Solutions to first order exact differential equations

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 $$\{\begin{array}{l}{\psi }_x(x,y)=M(x,y)\\ {\psi }_y(x,y)=N(x,y)\end{array}$$

This is equivalent to taking the exact differential of $\psi (x,y)$. $$\frac{\partial \psi }{\partial x}+\frac{\partial \psi }{\partial y}\frac{dy}{dx}=0⟺\frac{d}{dx}\psi (x,y(x))=0$$

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 $\psi (x,y)$.

The reasoning behind this is the following. Since $$\{\begin{array}{l}{\psi }_x(x,y)=M(x,y)\\ {\psi }_y(x,y)=N(x,y)\end{array}$$

it follows, by integrating both sides, that $$\{\begin{array}{l}\psi (x,y)=\int M(x,y)dx+h(y)=Q(x,y)+h(y)\\ \psi (x,y)=\int N(x,y)dy+g(x)=P(x,y)+g(x)\end{array}$$

Therefore, $$Q(x,y)+h(y)=P(x,y)+g(x)$$

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

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

Ergo, $$Q(x,y)=g(x)+f(x,y)\text{ and }P(x,y)=h(y)+d(x,y)$$

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

Since we already showed that $$\{\begin{array}{l}\psi (x,y)=Q(x,y)+h(y)\\ \psi (x,y)=P(x,y)+g(x)\end{array}$$

it follows that $$\psi (x,y)=g(x)+f(x,y)+h(y)$$

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

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:

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

Since both functions $I(x,y)$, $J(x,y)$ are functions of two variables, implicitly differentiating the multivariate function yields

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

Expanding the total derivatives gives that

$\frac{dI}{dx}=\frac{\partial I}{\partial x}+\frac{\partial I}{\partial y}\frac{dy}{dx}$

and that

$\frac{dJ}{dx}=\frac{\partial J}{\partial x}+\frac{\partial J}{\partial y}\frac{dy}{dx}$

Combining the $\frac{dy}{dx}$ terms gives

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

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

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

Now, let there be some second-order differential equation

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

If $\frac{\partial J}{\partial x}=\frac{\partial I}{\partial y}$ for exact differential equations, then

$\int \left(\frac{\partial I}{\partial y}\right)dy=\int \left(\frac{\partial J}{\partial x}\right)dy$

and

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

where $h\left(x\right)$ is some arbitrary function only of $x$ that was differentiated away to zero upon taking the partial derivative of $I(x,y)$ 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 $I(x,y)$ that is missing some original extra function $h\left(x\right)$ that was partially differentiated to zero.

Next, if

$\frac{dI}{dx}=\frac{\partial I}{\partial x}+\frac{\partial I}{\partial y}\frac{dy}{dx}$

then the term $\frac{\partial I}{\partial x}$ 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

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

only the term $f(x,y)$ is a term purely of $x$ and $y$. Let $\frac{\partial I}{\partial x}=f(x,y)$. If $\frac{\partial I}{\partial x}=f(x,y)$, then

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

Since the total derivative of $I(x,y)$ with respect to $x$ is equivalent to the implicit ordinary derivative $\frac{dI}{dx}$ , then

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

So,

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

and

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

Thus, the second order differential equation

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

is exact only if $g(x,y,\frac{dy}{dx})=\frac{dJ}{dx}+\frac{\partial J}{\partial x}=\frac{dJ}{dx}+\frac{\partial J}{\partial x}$ and only if the below expression

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

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

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

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

$(1-x^2)y^{″}-4x{y}^{\prime }-2y=0$

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 $y^{\prime }$. With one of the conditions for exactness met, one can calculate that

$\int (-2x)dy=I(x,y)-h\left(x\right)=-2xy$

Letting $f(x,y)=-2y$, then

$\int (-2y-2x{y}^{\prime }-\frac{d}{dx}(-2xy))dx=\int (-2y-2x{y}^{\prime }+2x{y}^{\prime }+2y)dx=\int \left(0\right)dx=h\left(x\right)$

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 $I(x,y)=-2xy+C_1$. Reduction to a first-order exact equation yields

$-2xy+C_1+(1-x^2)y^{\prime }=0$

Integrating $I(x,y)$ 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 $y^{\prime }$ term.

$-x^2+i^{\prime }\left(y\right)=1-x^2$

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

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

it was previously shown that equation is defined such that

$f(x,y)=\frac{dh\left(x\right)}{dx}+\frac{d}{dx}(I(x,y)-h\left(x\right))-\frac{\partial J}{\partial x}\frac{dy}{dx}$

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

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

where

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

and where $F(x,y,\frac{dy}{dx})$ is a function only of $x,y$ and $\frac{dy}{dx}$. Combining all $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$ terms not coming from $F(x,y,\frac{dy}{dx})$ gives

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

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

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

must be a function solely of $x$.

Example

Consider the nonlinear third-order differential equation

$y{y}^{‴}+3y^{\prime }{y}^{″}+12x^2=0$

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

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

which is indeed a function only of $x$. So, the differential equation is exact. Integrating twice yields that $h\left(x\right)=x^4+C_{1}x+C_2=I(x,y)$. Rewriting the equation as a first-order exact differential equation yields

$x^4+C_{1}x+C_2+y{y}^{\prime }=0$

Integrating $I(x,y)$ with respect to $x$ gives that $\frac{x^5}{5}+C_1{x}^2+C_{2}x+i\left(y\right)=0$. Differentiating with respect to $y$ and equating that to the term in front of $y^{\prime }$ in the first-order equation gives that $i^{\prime }\left(y\right)=y$ and that $i\left(y\right)=\frac{y^2}{2}+C_3$. The full implicit solution becomes

$\frac{x^5}{5}+C_1{x}^2+C_{2}x+C_3+\frac{y^2}{2}=0$

The explicit solution, then, is

$y=\pm \sqrt{C_1{x}^2+C_{2}x+C_3-\frac{2x^5}{5}}$

See also

• Exact differential
• Inexact differential equation

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