2. 1st Order

A differential equation is a first order ODE if it contains a first derivative

\dfrac{dy}{dx} + f(x)y = g(x)

Let the integrating factor (IF) be $\mu(x)$ such that multiplying both sides by $\mu(x)$ gives

\mu(x)\dfrac{dy}{dx} + \mu(x)f(x)y = g(x)\mu(x)

we want $\mu(x)$ to an equation such that

\dfrac{d(\mu(x)y)}{dx} = \mu(x)\dfrac{dy}{dx} + \mu(x)f(x)y

therefore by product rule we can say that

\mu'(x) = \mu(x)f(x), \quad f(x) = \dfrac{\mu'(x)}{\mu(x)}

solving for $\mu(x)$

ln(\mu(x)) = \int f(x) \; dx

\mu(x) = e^{\textstyle \int f(x)\, \; dx}

Note: $f(x)$ isn't always integratable

Solve

\dfrac{dy}{dx} + \dfrac{2xy}{x^2 - 1} = 5x^2 - 1

$f(x) = \dfrac{2x}{x^2 -1}$ , $\quad g(x) = 5x^2 - 1$

$\mu(x) = e^{\textstyle \int \frac{2x}{x^2 - 1}\, \; dx} \quad = \quad e^{ln(x^2 - 1)} \quad = \quad x^2 - 1$

multiply both sides of the ODE by $\mu(x)$

(x^2 - 1)\dfrac{dy}{dx} + 2xy = (x^2 - 1)(5x^2 - 1)

the left side of the equation can now be integrated easily

(x^2 - 1)y = \int{5x^4 - 6x^2 + 1\, dx}

therefore

y = \dfrac{x^5 - 2x^3 + x + c}{x^2 - 1}