Exact and Linear ODEs

An ODE is said to be exact if it can be represented as the derivative of a multivariate function, i.e. in the following format:

$$dF(x,y)=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}=0$$

$$M(x,y)dx+N(x,y)dy=0$$

Sketch of Proof §

Given differential equation,

$$M(x,y)dx+N(x,y)dy=0$$

we first assume that there exists a function $F(x,y)$ such that

$$\frac{\partial F}{\partial x}=M(x,y);\frac{\partial F}{\partial y}=N(x,y).$$

By Schwarz’s theorem, it’s known that the order of partial differentiation does not matter:

\

Therefore, assuming that the above holds, the following should be true for an exact equation:

$$M_y=Nx$$

Integrating Factor §

If a DE is not exact, we may be able to multiply the equation by an integrating factor $\text{micro}$ to render it exact; the following then holds:

$$\frac{\partial (\mu M)}{\partial y}=\frac{\partial (\mu N)}{\partial x}.$$

IF in terms of x (independent var) §

By the product rule,

$$\begin{array}{rl}\mu \cdot \frac{\partial M}{\partial y}& =\frac{\partial \mu }{\partial x}\cdot N+\mu \frac{\partial N}{\partial x}.\\ & \\ \frac{\partial \mu }{\partial x}=\frac{d\mu }{dx}& =\mu \left(\frac{M_y-N_x}{N}\right)\\ & \\ \frac{1}{\mu }d\mu & =\frac{M_y-N_x}{N}dx\\ & \\ \ln (\mu )& =\int \frac{M_y-N_x}{N}dx\\ & \\ \mu (x)& =\exp \left(\int \frac{M_y-N_x}{N}dx\right).\end{array}$$

In other words, as a shortcut we may observe that if

$$\frac{M_y-N_x}{N}$$

is expressed solely in terms of $x$, we may find a valid $\text{micro}$ .

IF in terms of y (dependent var) §

Linear Equations §

A linear equation is of the form

$$\frac{dy}{dx}+P(x)\cdot y=Q(x).$$

The integrating factor can then be found by

\micro = \exp \left ( \int P(x) \ dx \right ). $$S