Exact Differential Equations

Many differential equations have solutions that can be written in implicit form:

F(x,y) = C

Now instead of starting with the differential equation and finding the solution, suppose we look at this ``backwards'': start with the solution F(x,y) = C and differentiate it, obtaining a differential equation. For example, the solutions

y2 + x y = C

would lead to the differential equation

y + (2 y + x) y' = 0\; \end{displaymath}

In general, we have (with y = y(x))

0 = \frac{d}{dx} F(x, y) = \frac{\partial F}{\partial x}(x,y)
+ \frac{\partial F}{\partial y}(x,y) \frac{dy}{dx} \end{displaymath}

An equation of the form

\frac{\partial F}{\partial x}(x,y)
+ \frac{\partial F}{\partial y}(x,y) \frac{dy}{dx} = 0\end{displaymath}

is said to be an exact differential equation. Thus $ y + (2 y + x) y' = 0\; $ is exact, and we could make as many examples as we want by taking an arbitrary (differentiable) F and differentiating.

Now let's look ``forwards'' again: we have a differential equation which we suspect might be exact. We need to be able to

Thus we want to be able to look at the differential equation $ y + (2 y + x) y' = 0\; $ and see that $y = \partial F/\partial x$and $2 y + x = \partial F/\partial y$ where F = x y + y2, and therefore that the general solution is x y + y2 = C.

The test for exactness comes from the equality of mixed partial derivatives:

\frac{\partial^2 F}{\partial x\partial y} = 
 \frac{\partial^2 F}{\partial y\partial x} \end{displaymath}

So if $M(x,y) + N(x,y) y' = 0\;$ is exact, we must have

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \end{displaymath}

In our example that is indeed the case:

\frac{\partial y}{\partial y} = 1 =
\frac{\partial (2y+x)}{\partial x} \end{displaymath}

It turns out that (under suitable conditions) this test is all you need. The details are as follows:

Suppose M, N, $\partial M/\partial y$ and $\partial N/\partial x$are continuous in a rectangular region R: a < x < b, c < y < d. Then the differential equation $M + N y' = 0\;$ is exact in R (i.e. there is a function F defined in R with $\partial F/\partial x = M$ and $\partial F/\partial y = N$ there) if and only if $\partial M/\partial y = 
\partial N/\partial x$ everywhere in R.

Now how do we find F when the equation is exact? By integration. In our example, we want $y = \partial F/\partial x$and $2 y + x = \partial F/\partial y$. Look at the first equation, and integrate with respect to x:

$\displaystyle\int y \, dx = \int \frac{\partial F}{\partial x}\, dx $

$\displaystyle x y = F(x,y) + K(y)$

A few points to notice:

Now we plug F(x,y) = x y - K(y) into the second equation.

$\displaystyle2 y + x = \frac{\partial}{\partial y} (x y - K(y)) 
 = x - K'(y)$
$\displaystyle K'(y) = -2 y$
$\displaystyle K(y) = -y^2 + c$

This time the ``constant'' c really is constant, and we can ignore it (since at the end the solution F(x,y) = C will have a constant in it anyway). We conclude that F(x,y) = x y + y2 = C is the general solution of our differential equation.

Thus you can use the following procedure for solving an exact equation:


Robert Israel