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

$$y + (2 y + x) y' = 0\;$$

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

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

An equation of the form

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

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

• test whether or not is exact, and
• if so, find the function F.

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:

$$\frac{\partial^2 F}{\partial x\partial y} = \frac{\partial^2 F}{\partial y\partial x}$$

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

$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

In our example that is indeed the case:

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

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:

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

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

A few points to notice:

• y is considered as constant when you differentiate or integrate with respect to x.
• Integrating $\partial F/\partial x$ with respect to x gives F + a ``constant'', but there's no reason for this ``constant'' to be the same for different values of y. Therefore we have K(y) rather than just K.

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

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

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:

• Test that the equation really is exact: you must have $\partial M/\partial y = \partial N/\partial x$.
• Integrate M with respect to x: write $$K(y) = F(x,y) - \int M(x,y)\, dx$$.
• Differentiate this with respect to y: $$K'(y) = N(x,y) - \frac{\partial}{\partial y} \int M(x,y)\, dx$$. Make sure this is a function of y only, with no x in it (after whatever simplifications are necessary). If it does depend on x, you've made a mistake somewhere.
• Integrate $K'(y)\;$ to get K(y).
• Now we have $F(x,y) = \int M(x,y) \, dx + K(y)$, i.e. the general solution is $\int M(x,y)\, dx + K(y) = C$ where C is an arbitrary constant.