Exact ODEs

When a variable in a multi variable function $f(x,y)$ has variables that rely on another e.g. $y = y(t), x = x(t)$ then the derivative of that function is known as the total derivative and is given as

$$\frac{d}{dt}f(x,y) = \frac{\delta f}{\delta x} \frac{dx}{dt} + \frac{\delta f}{\delta y} \frac{dy}{dt}$$

This can also occur for when one variable is the function of another e.g. $x = x, y = y(x)$, the total derivative of such is

$$\frac{d}{dx}f(x,y) = \frac{\delta f}{\delta x} + \frac{\delta f}{\delta x} \frac{dy}{dx}$$ (1)

Exact ODEs take the form

$$M(x,y) + N(x,y)\frac{dy}{dx}=0$$

Where M and N must be separated by a plus. As you can see, this is in the same form as a the derivative of the function from equation (1), thus we can assume that

$$\frac{\delta f}{\delta x} = M \quad \frac{\delta f}{\delta x} = N$$

In order to verify this assumption we can test an 'exactness condition'

$$\frac{\delta^2f}{\delta x \delta y} = \frac{\delta^2 f}{\delta y \delta x}$$

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

We must now find a function to satisfy these conditions, however this is best left to an example.

$$x + y^2 + 2xy\frac{dy}{dx} = 0$$

$$M = x + y^2 \quad N = 2xy$$

$$\frac{\delta M}{\delta y} = 2y$$

$$\frac{\delta N}{\delta x} = 2y$$

This satisfies the 'exactness condition', and so we can begin to find the function that solves the ODE.

$$M = \frac{\delta f}{\delta x}$$

$$\int M dx = f = \frac{1}{2}x^2 + xy^2 + g(y)$$

$$N = \frac{\delta f}{\delta y}$$

$$\int N dy = f = xy^2 + h(x)$$

For $\int M dx = \int N dy$ to be true their constants must satisfy each other, thus $g(y) = 0, h(x) = \frac{1}{2}x^2$

$$f(x,y) = \frac{1}{2}x^2 + xy^2 + c = 0$$