$${\frac{dy}{dx} = f(x) \ g(y)}$$
Manipulate the expression into the form:
$${g(y) \ dy = f(x) \ dx}$$
\({u = \frac{y}{x}}\) is called first order homogeneous. And substitute \({y' = u + x \ u'}\).
\({u = ax + by}\) is another common substitution. And substitute \({u' = a + b \ y'}\).
The problem is linear because it is linear in y and the left side of the equation can be expressed as the derivative of y and an integrating factor.
$${\frac{dy}{dx} + f(x) \ y = g(x)}$$
Make a homogeneous substitution of the form \({v = \frac{y}{x}}\) and the resulting equation will be separable.
$${Mx \ dx + Ny \ dy = 0}$$
$${M_{xy} = N_{yx}}$$
Find an integrating factor \({u(x)}\) or \({u(y)}\) and multiply the factor to both sides such that the problem becomes exact.
For a function of the type \({u(x,y)}\) try \({x^m y^n}\) .
Use a Bernoulli substitution in the form \({v = y^{1-n}}\) where n is the power of y on the right side
of a First Order equation that is otherwise in linear form. With the substitution the problem becomes linear.
For a problem initially in the form:
$${\frac{dy}{dx} + f(x) \ y = g(x) \ y^n}$$
After the substitution the linear equation in v will be of the form:
$${\frac{dv}{dx} + (1 - n) \ f(x) \ v = (1 - n) \ g(x)}$$
$${ay'' + by' + cy = 0}$$
Solve for the roots of the Characteristic polynomial:
$${ar^2 + br + c = 0}$$
For the roots \({\alpha}\) and \({\beta}\) :
$${y = C_1 e^{\alpha t} + C_2 e^{\beta t}}$$
$${y = C_1 e^{\alpha t} + C_2 t e^{\alpha t}}$$
Assume the roots are conjugates of the form \({\alpha = j \pm ki}\). Then the solution will be of the form
$${y = C_1 e^{j t} cos(kt) + C_2 e^{j t} sin(kt)}$$
$${y = C_1 y_1 + C_2 y_2 + u_1 y_1 + u_2 y_2}$$
$${u_1 = \displaystyle \int \frac{- y_2 g(t)}{W} dt}$$
$${u_2 = \displaystyle \int \frac{y_1 g(t)}{W} dt}$$
Wronskian:
$${W = y_1 y_2' - y_2 y_1'}$$
$${y = xy' + f(y')}$$
$${\mathcal{L}\{f'(t)\} = sF(s) - f(0)}$$
$${\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \dots - s f^{(n-1)}(0) - f^{(n)}(0)}$$