First Order ODE

Solving first order ODE was introduced in 1st semester. This section extends upon it.

Exact Differential Equation

Suppose $P(x,y)$ and $Q(x,y)$ are continuous functions in a region $R$.

A differential equation of the form $P(x,y)dx + Q(x,y)dy = 0$ is exact if $\exists f(x,y)$ such that:

$$\frac{\partial f}{\partial x} = P(x,y)\;\;\text{and}\;\; \frac{\partial f}{\partial y} = Q(x,y)$$

Then $f(x,y) = C$ is the general solution. $$

Exact Equation

$P(x,y)dx + Q(x,y)dy = 0$ is an exact equation iff: $$

$$\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$$

Integrating Factor

If the differential equation $P(x,y)dx + Q(x,y)dy = 0$ is not exact, it might be possible to make it exact by multiplying it with its integrating factor $\mu(x,y)$.

If the below function is a function of $x$ only: $$

$$h(x) = \frac{1}{Q(x,y)} \left[ P_y(x,y) - Q_x(x,y) \right]$$

Then $u(x,y) = \exp(\int h(x)\,\text{d}x)$. $$

If the below function is a function of $y$ only: $$

$$k(y) = \frac{1}{P(x,y)} \left[ Q_x(x,y) - P_y(x,y) \right]$$

Then $u(x,y) = \exp(\int k(y)\,\text{d}y)$. $$