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.

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
:$\frac = I$
and
:$\frac = J.$

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 $F(x_0, x_1,...,x_,x_n)$, the exact or total derivative with respect to $x_0$ is given by
:$\frac=\frac+\sum_^\frac\frac.$

Example

The function $F:\mathbb^\to\mathbb$ given by

:$F(x,y) = \frac(x^2 + y^2)+c$

is a potential function for the differential equation

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

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

$M(x, y) + N(x, y)\frac = 0$

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

$\psi _x(x, y) = M(x, y) \text \psi_y(x, y) = N(x, y)$

So, in general:

$M_y(x, y) = N_x(x, y) \iff \begin \exists \psi(x, y)\\ \psi_x(x, y) = M(x, y)\\ \psi_y(x, y) = N(x, y) \end$

Proof

The proof has two parts.

First, suppose there is a function $\psi(x,y)$ such that$\psi_x(x, y) = M(x, y) \text \psi_y(x, y) = N(x, y)$

It then follows that$M_y(x, y) = \psi _(x, y) \text N_x(x, y) = \psi _(x, y)$

Since $M_y$ and $N_x$ are continuous, then $\psi _$ and $\psi _$ 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
$\psi _x(x, y) = M(x, y) \text \psi _y(x, y) = N(x, y)$

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.

$\frac(x, y) = M(x, y)$
$\psi(x, y) = \int + h(y)$
$\psi(x, y) = Q(x, y) + h(y)$

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 we $M$ 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 $y$.
$\frac(x, y) = \frac(x, y) + h'(y)$

Set the result equal to $N$ and solve for $h'(y)$.
$h'(y) = N(x, y) - \frac(x, y)$

In order to determine $h'(y)$ 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 $x$.
$\frac(x, y) - \frac\frac(x, y) \iff \frac(x, y) - \frac\frac(x, y)$

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

Therefore,
$h'(y) = N(x, y) - \frac(x, y)$
$h(y) = \int$

$\psi(x, y) = Q(x, y) + \int + C$

And this completes the proof.

Solutions to first order exact differential equations

First order exact differential equations of the form
$M(x, y) + N(x, y)\frac = 0$

can be written in terms of the potential function $\psi(x, y)$
$\frac + \frac\frac = 0$

where
$\begin \psi _x(x, y) = M(x, y)\\ \psi _y(x, y) = N(x, y) \end$

This is equivalent to taking the exact differential of $\psi(x,y)$.
$\frac + \frac\frac = 0 \iff \frac\psi(x, y(x)) = 0$

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
$\begin \psi _x(x, y) = M(x, y)\\ \psi _y(x, y) = N(x, y) \end$

it follows, by integrating both sides, that
$\begin \psi(x, y) = \int + h(y) = Q(x, y) + h(y)\\ \psi(x, y) = \int + g(x) = P(x, y) + g(x) \end$

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

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,
$Q(x, y) = g(x) + f(x, y) \text P(x, y) = h(y) + f(x, y)$

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

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

So, we can construct $\psi(x, y)$ by doing $\int$ and $\int$ 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:

:$I\left(x,y\right)+J\left(x,y\right)=0$

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

:$+\left(\right)+\left(J\left(x,y\right)\right)=0$

Expanding the total derivatives gives that

:$=+$

and that

:$=+$

Combining the $$ terms gives

:$+\left(++\right)+\left(J\left(x,y\right)\right)=0$

If the equation is exact, then Additionally, the total derivative of $J\left(x,y\right)$ is equal to its implicit ordinary derivative $$. This leads to the rewritten equation

:$+\left(+\right)+\left(J\left(x,y\right)\right)=0$

Now, let there be some second-order differential equation

:$f\left(x,y\right)+g\left(x,y,\right)+\left(J\left(x,y\right)\right)=0$

If $=$ for exact differential equations, then

:$\int \left(\right)dy=\int \left(\right)dy$

and

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

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

:$=+$

then the term $$ 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

:$f\left(x,y\right)+g\left(x,y,\right)+\left(J\left(x,y\right)\right)=0$

only the term $f\left(x,y\right)$ is a term purely of $x$ and $y$. Let $=f\left(x,y\right)$. If $=f\left(x,y\right)$, then

:$f\left(x,y\right)=-$

Since the total derivative of $I\left(x,y\right)$ with respect to $x$ is equivalent to the implicit ordinary derivative $$ , then

:$f\left(x,y\right)+$

\left(I\left(x,y\right)-h\left(x\right)\right)+

So,

:$=f\left(x,y\right)+-\left(I\left(x,y\right)-h\left(x\right)\right)$

and

:$h\left(x\right)=\int\left(f\left(x,y\right)+-\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx$

Thus, the second order differential equation

:$f\left(x,y\right)+g\left(x,y,\right)+\left(J\left(x,y\right)\right)=0$

is exact only if $g\left(x,y,\right)=+=+$ and only if the below expression

:$\int\left(f\left(x,y\right)+-\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx=\int \left(f\left(x,y\right)-\right)dx$

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.

:$I\left(x,y\right)+J\left(x,y\right)=0$

Now, however, in the final implicit solution there will be a $C_1x$ 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

:$\int \left(-2y-2xy'-\left(-2xy \right)\right)dx=\int \left(-2y-2xy'+2xy'+2y\right)dx=\int \left(0\right)dx=h\left(x\right)$

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

:$-x^2y+C_1x+i\left(y\right)=0$

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

:$C_1x+C_2+y-x^2y=0$

Solving explicitly for $y$ yields

:$y= \frac$

Higher order exact differential equations

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

:$\left(J\left(x,y\right)\right)+\left(+\right)+f\left(x,y\right)=0$

it was previously shown that equation is defined such that

:$f\left(x,y\right)=+\left(I\left(x,y\right)-h\left(x\right)\right)-$

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

:$\left(J\left(x,y\right)\right)++\left(+\right)+\left(+\left(\right)\right)+=0$

where

:$=+\left(I\left(x,y\right)-h\left(x\right)\right)--\left(\right)=F\left(x,y,\right)$
and where $F\left(x,y,\right)$
is a function only of $x,y$ and $$. Combining all $$ and $$ terms not coming from $F\left(x,y,\right)$ gives

:$\left(J\left(x,y\right)\right)+\left(2+\right)+\left(+\left(\right)\right)+F\left(x,y,\right)=0$

Thus, the three conditions for exactness for a third-order differential equation are: the $$ term must be $2+$, the $$ term must be $+\left(\right)$ and

:$F\left(x,y,\right)-\left(I\left(x,y\right)-h\left(x\right)\right)++\left(\right)$

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 $y''\left(2+\right)$ is $2y'y''$ and $y'\left(+\left(\right)\right)=y'y''$which together sum to $3y'y''$. Fortunately, this appears in our equation. For the last condition of exactness,

:$F\left(x,y,\right)-\left(I\left(x,y\right)-h\left(x\right)\right)++\left(\right)=12x^2-0+0+0=12x^2$

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

:$x^4+C_1x+C_2+yy'=0$

Integrating $I\left(x,y\right)$ with respect to $x$ gives that $+C_1x^2+C_2x+i\left(y\right)=0$. 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 $i\left(y\right)=+C_3$. The full implicit solution becomes

:$+C_1x^2+C_2x+C_3+=0$

The explicit solution, then, is

:$y=\pm\sqrt$

See also

* Exact differential
* Inexact differential equation

References

Further reading

* Boyce, William E.; DiPrima, Richard C. (1986). ''Elementary Differential Equations'' (4th ed.). New York: John Wiley & Sons, Inc.

{{Differential equations topics

Ordinary differential equations