Let a differential equation take the form p(y)×y′ = q(x), where p and q are arbitrary functions. This is called a separable equation.
Assume y(x) can be chosen such that the integral of p(y), evaluated at y(x), equals the integral of q(x). Take the derivative and recreate our original differential equation.
Integration brings in a constant c, as you would expect from a first order differential equation. Write it this way.
∫ p(y) = c + ∫ q(x)
Now suppose y(x) is a solution that does not satisfy the above equation. Let the difference be d(x).
∫ p(y) = c + d(x) + ∫ q(x)
differentiate, and subtract p(y)y′ = q(x), to find d′(x) = 0, hence d(x) is constant after all. The solutions to a separable differential equation have been characterized.
Consider y′ = g(x,y). This is a perfectly good first order differential equation.
Let y(x) be a solution. Each point x,y(x) can have its own t. Thus t is also a function of x.
Let t(x) = 1/x, and let v(x) = y/x. Now g(x,y) = g(tx,ty) = g(1,v). On the left, xv′+v = y′. Rewrite the differential equation as follows.
xv′ + v = g(1,v)
xv′ = g(1,v)-v
v′ over g(1,v)-v = 1/x
This equation is separable, and can be solved through integration. Once v(x) is known, multiply by x to find y(x).