Exact ODE

Of the form

M (x, y) d x + N (x, y) d y = 0

If a function $F (x, y)$ exists such that total differential $d F$ satisfies
$d F = M (x, y) d x + N (x, y) d y$ , then the ODE is exact.

In other words, if $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$ are both continuous functions of $x$ and $y$ on some open set $S$ , then the ODE is exact on S iff $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Theorem:

Given an ODE in the form of $M (x, y) d x + N (x y) d y = 0$

If $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$ are constants in x and y on an open set S,
then this ODE is exact on S iff $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

$d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y$
with the goal to find such an $F$

Solving an Exact ODE

Determine that the ODE is exact: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
Try to find $F (x, y)$ such that $$ dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy = M(x, y) dx + N(x, y)dy$$
Integrate $\frac{\partial F}{\partial x} = M (x, y)$ wrt $x$ , thinking of $y$ as a constant: $$ F(x, y) = \int M(x, y) dx + B(y)$$
Differentiate this w.r.t $y$ to find function $B (y)$ from: $$N(x, y) = \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}\left(\int M(x, y) dx\right) + B'(y)$$
The solution is $F (x, y) = C$ , where $C$ is an arbitrary constant

Finding an Integrating Factor

Definition

If there is a function $μ (x, y)$ such that

$μ (x, y) M (x, y) d x + μ (x, y) N (x, y) d y = 0$
is exact, then $μ (x, y)$ is an integrating factor

Finding Integrating Factor

Let $\frac{\partial f}{\partial x} = f_{x}$ and $\frac{\partial f}{\partial y} = f_{y}$

Solve $μ_{y} M + μ M_{y} = μ_{x} N + μ N_{x}$
- PDE, so generally hard to solve, simplify by assuming $μ (x, y) = μ (x)$ or $μ (x, y) = μ (y)$
If we assume $μ (x, y) = μ (x)$ , equation reduces to
$μ M_{y} = μ_{x} N + μ N_{x} ⟹ \frac{d μ}{d x} = μ (x) \frac{M_{y} - N_{x}}{N}$
Assume, $μ (x, y) = μ (y)$ , reduces to
$μ_{y} M + μ M_{y} = μ N_{x} ⟹ \frac{d μ}{d y} = μ (y) \frac{N_{x} - M_{y}}{M}$