First order differential equation. algebra part?

Question

Hi I'm reading my textbook on first order diff equation and i cant see to get the algebra from this part.

y'/x^2 - 2y/x^3 = ....

d/dx [y/x^2] = ...

How does that work out? Can someone explain.

Answer 1

Reverse the equations and you have differentiation by chain rule:
d/dx [y/x^2] =
d/dx[yx^-2] =
-2yx^-3 + y'x^-2 =
y'/x^2 - 2y/x^3